Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

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Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Ratio Text Book Questions and Answers

Same shape Textbook Page No. 116

In both the rectangles below, the length is 1 centimetre more than the breadth.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 1
But not only the length of these are different, they look different also. In the larger rectangle, the sides look almost the same. Draw a rectangle of length 50 centimetres and breadth 49 centimetres in a larger sheet of paper. It looks almost a square, doesn’t it?
In the first rectangle above, length is double the breadth. Now look at this rectangle.
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Again the length is double the breadth. Even though it is larger than the first, they have the same shape, right?
Answer:
As the length is double the breadth in the second rectangle. The first and the second rectangle will have the same shape.

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Changing scale

Look at this photo:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 7
The shorter side is 2 centimetres and the longer side, 3 centimetres; that is, the longer side is 1\(\frac{1}{2}\) times shorter side.
Suppose we make the shorter side 3 centimetres and longer side 4.5 centimetres.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 8
Still, the longer side is 1\(\frac{1}{2}\) times the shorter side.
Now suppose we change the shorter side to 3 centimetres and increase the longer side also by 1 centimetre, making it 4 centimeters.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 9
Does the picture look right?
Answer:
Yes, the picture looks right.
Suppose if we change the shorter side to 3 centimetres and the longer side to 4 centimetres. The longer side is 1\(\frac{1}{3}\) times the shorter side.

TV Math

The sizes of TV sets are usually given as 14 inch, 17 inch, 20 inch and so on. What does it mean?
The TV screen is a rectangle: and these are lengths of the diagonals of the screen.
Does it determine the size of the screen? Rectangles of different width and lengths can have the same diagonal:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 10
In the modern TV sets, the ratio of length to width is 16 : 9. In the earlier days, it was 4 : 3.
See their difference in two TV screens of the same diagonal length.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 11

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Flags Textbook Page No. 119

When we draw our National Flag, not only should the colours be right, the ratio of width to length should also be correct. This ratio is 2 : 3.
That is, in drawing our flag, if the length is taken as 3 centimetres, the width should be 2 centimetres.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 14
This ratio is different for flags of other countries. For example, in the flag of Australia, this ratio is 1 : 2.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 15
And in the flag of Germany, it is 3 : 5.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 16

Without fractions Textbook Page No. 120

When quantities like length are measured using a definite unit, we may not always get counting numbers; and it is this fact which led to the idea of fractions.

In comparing the sizes of two quantities, one question is whether both can be given as counting numbers, using a suitably small unit of measurement. It is this question that leads to the idea of ratio.

For example, suppose the length of two objects are found as \(\frac{2}{5}\) and \(\frac{3}{5}\), when measured using a string. If \(\frac{1}{5}\) of the string is taken as the unit, we can say the length of the first is 2 and that of the second is 3. This is the meaning of
saying the ratio of the lengths is 2 : 3.
Suppose the length of two objects are \(\frac{1}{3}\) and \(\frac{1}{5}\) of the string.
To get both lengths as counting numbers, what fraction of the string can be taken as a unit of measurement?

Circle relations

Look at the pieces of circles below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 19

The smaller piece is \(\frac{1}{4}\) of the circle and the larger piece is \(\frac{1}{2}\) of the circle.
So, the larger piece is twice the size of the smaller. That is, the ratio of the sizes of small and large pieces is 1 : 2.
Now look at these pieces:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 20
What is the ratio of their sizes?
Let’s measure each using \(\frac{1}{4}\) of circles.
The smaller figure has two such pieces. What about the larger?
Answer:
Let’s measure each using \(\frac{1}{4}\) of circles.
The larger figure has three such pieces.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 21
So what is the ratio of the sizes of these two?
Answer:
Let’s measure each using \(\frac{1}{4}\) of circles.
The smaller piece has two such pieces and the larger piece has three such pieces.
2×\(\frac{1}{4}\) : 3×\(\frac{1}{4}\)
\(\frac{2}{4}\) : \(\frac{3}{4}\)
Therefore, the ratio will be 2 : 3

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Moving Ratio Textbook Page No. 122
Have you taken apart toy cars or old clocks? There are many toothed wheels in such mechanisms. See this picture:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 22
It is a part of a machine. In the two whole wheels we see, the smaller has 13 teeth and the larger has 21. So during the time that the smaller circle makes 21 revolutions, the larger one would have made only 13 revolutions.
The speed of rotation of machines is controlled by arranging such wheels with different number of teeth.

Sand and Cement

In construction of a building, sand and cement are used in definite ratios, but the ratios are different depending on the purpose. When one bowl of cement and five bowls of sand are mixed, the ratio of cement to sand is 1 : 5

When one sack of cement is mixed with five sacks of sand, the ratio is the same. But to set a brick wall, this much cement may not be needed. In this case, the ratio may be 1 : 10 or 1 : 12.

Ratio of Parts Textbook Page No. 124

We can use ratio to compare parts of a whole also. For example, in the picture below, the lighter part is \(\frac{3}{8}\) of the circle, and the darker part is \(\frac{5}{8}\) of the circle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 23
These two parts together make up the whole circle. The ratio of the sizes of these parts is 3 : 5.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 24
So here, the ratio 3 : 5 indicate the two fractions \(\frac{3}{8}\) and \(\frac{5}{8}\).

Generally in such instances, a ratio of the numbers indicates fractions of equal denominators adding up to 1.

Meaning of ratio Textbook Page No. 125

If we know only the ratio of two quantities, we can’t say exactly how much they are; but they can be compared in several ways.
For examples, suppose the volumes of two pots are said to be in the ratio 2 : 3. We can interpret this in the following ways:

  • To fill the smaller pot, we need \(\frac{2}{3}\) of the larger pot.
  • To fill the bigger pot, we need \(\frac{3}{2}\) = 1\(\frac{1}{2}\) times the smaller pot.
  • Whether we take \(\frac{1}{2}\) of the water in the smaller pot, or \(\frac{1}{3}\) of the water in the larger pot, we get the same amount of water.
  • If we fill both pots and pour them into a large pot, \(\frac{2}{5}\) of the total amount of water
    is from the smaller pot and \(\frac{3}{5}\) from the larger.

If the length of two pieces of rope are said to be in the ratio 3 : 5, what all interpretations like this can you make?
Answer:
If the length of two pieces of rope is 3 : 5.Then the following interpretations can be made.
The smaller rope is \(\frac{3}{5}\) of the length of the larger rope.
The smaller rope will be \(\frac{3}{8}\) of the total rope length.

Three measures

Look at this triangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 25
In it, the longest side is double the shortest. The side of medium length is one and a half times the shortest.
Saying this with ratios, the lengths of the shortest and the longest sides are in the ratio 1 : 2 and those of the shortest and the medium are in the ratio 2 : 3.
What is the ratio of the lengths of the medium side and the longest side?
We can say this in another way: if we use a string of length 1.5 centimetres, the shortest side is 2, the medium side is 3 and the longest side is 4.
The can be shortened by saying the sides are in the ratio 2 : 3 : 4.

Triangle Math Textbook Page No. 127

How many triangles are there with ratio of sides 2 : 3 : 4?
The lengths of sides can be 2 cm, 3 cm, 4cm.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 26
Or they can be 1 cm, 1.5 cm, 2 cm
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 27
We can take metres instead of centimetres.
Thus we have several such triangles.
In all these, what fraction of the perimeter is the shortest side?
And the side of medium length?
The longest side?
The perimeter of a triangle is 80 centimetres and its sides are in the ratio 5 : 7 : 8. Can you compute the actual lengths of sides?
Answer:
Let the ratio be x.Then the sides will be 5x, 7x and 8x.
Given that perimeter is 80 cm.
5x + 7x + 8x = 80
20x = 80
x = 4
Since the value of x is 4, the lengths will be 5×4=20cm, 7×4=28cm and 8×4=32cm.

What if the perimeter is 1 metre?
Answer:
1 metre equals 100 centimetres.
Let the ratio be x.Then the sides will be 5x, 7x and 8x.
Given that perimeter is 100 cm.
5x + 7x + 8x = 100
20x = 100
x = 5
Since the value of x is 5, the lengths will be 5×5=25cm, 7×5=35cm and 8×5=40cm.

Length and width Textbook Page No. 116

Look at these rectangles:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 3
Is there any common relation between the lengths and widths of all these?
In all these, the length is twice the width, right? (We can also say width is half the length)
In the language of mathematics, we state this fact like this:
In all these rectangles, the width and length are in the ratio one to two.
In writing, we shorten the phrase “one to two” as 1 : 2; that is
In all these rectangles, the width and length are in the ratio 1 : 2.

In the rectangle of width 1 centimetre and length 2 centimetres, the length is twice the width. In a rectangle of width 1 metre and length 2 metres also, we have the same relation. So, in both these rectangles, width and length are in the ratio one to two (1 : 2).
We can state this in reverse: in all these rectangles, length and width are in the ratio two to one (2 : 1).
What is the width to length ratio of the rectangle below?
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 4
Answer:
The length of the given rectangle is 6cm and width is 2cm.
Therefore, the width to length ratio for the above rectangle will be 1 : 3.

And what about this rectangle?
Answer:
The length of the given rectangle is 4.5cm and width is 1.5cm.
Therefore, the width to length ratio for the above rectangle will be 1 : 3.

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 5
In both, the length is 3 times the width. So what is the ratio of the width to the length?
Answer:
In the above two rectangles, the length is thrice the width.
The ratio of the width to the length is 1 : 3.

How do we say this as a ratio?
Answer:
The width and length are in the ratio one to three.

1 metre means loo centimetres. So in such a rectangle, the ratio of the width to the length is 1: 50.
Now look at these rectangles:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 6

In both, the length is one and a half times the width.
How do we say this as a ratio?
We can say one to one and a half; but usually we avoid fractions when we state ratios.
Suppose we take the width as 2 centimetres.
What is 1\(\frac{1}{2}\) times 2?
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 12
So, we say that in rectangles of this type, the ratio of width to length is two to three and write 2 : 3.
Can’t we say here that the ratio is 4 : 6 ?
Nothing wrong in it; but usually ratios are stated using the least possible counting numbers.
So how do we state using ratios, the fact that the length of a rectangle is two and a half times the width?

If the width is 1 centimetre, then the length is 2\(\frac{1}{2}\) centimetres.
What if the width is 2 centimetres?
Length would be 5 centimetres.
So, we say that width and length are in the ratio 2 to 5.

What if the length is one and a quarter times the width?
If the width is 1 centimetre, the length is 1\(\frac{1}{4}\) centimetre.
If the width is 2 centimetres then the length is 2\(\frac{1}{2}\) centimetres.
Still we haven’t got rid of fractions.
Now if width is 4 centimetres, what would be the length?

So, in all such rectangles, the width and length are in the ratio 4 : 5.
Do you notice another thing in all these?
If we stretch or shrink both width and length by the same factor, the ratio is not changed. For example, look at this table.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 13
In all these, the length is 3 times the width; or the reverse, the length is \(\frac{1}{3}\) of the width.
In terms of ratio, we say width and length are in the ratio 1 : 3 or length and width are in the ratio 3 : 1.
In terms of ratio, we say width and length are in the ratio 1: 3 or length and width are in the ratio 3: 1.

• For all rectangles of dimensions given below, state the ratio of width to length, using the least possible counting numbers:

• width 8 centimetres, length 10 centimetres
Answer:
The simplest form for the given numbers 8:10 is 4 : 5
Therefore, the ratio of width to length will be 4 : 5

• width 8 metres length 12 metres
Answer:
The simplest form for the given numbers 8:12 is 4 : 3
Therefore, the ratio of width to length will be 2 : 3

• width 20 centimetres length 1 metre
Answer:
The given numbers cannot be written in simplest form.
Therefore, the ratio of width to length will be 20 : 1

• width 40 centimetres length 1 metre
Answer:
The given numbers cannot be written in simplest form.
Therefore, the ratio of width to length will be 40 : 1

• width 1.5 centimetres length 2 centimetres
Answer:
The simplest form for the given numbers 1.5:2 is 3 : 4
Therefore, the ratio of width to length will be 3 : 4

• In the table below two of the width, length and their ratio of some rectangles are given. Calculate the third and fill up the table.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 17
Answer:
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_1

• What does it mean to say that the width to length ratio of a rectangle is 1: 1 ? What sort of a rectangle is it?
Answer:
The width to length ratio of a rectangle is 1 : 1.
Here the width of the rectangle and length of the rectangle is same.
If the width and the length of rectangle is same, then it is a square.

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Other quantities Textbook Page No. 120

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 18

There are two ropes, the shorter \(\frac{1}{3}\) metre long and the longer, \(\frac{1}{2}\) metre long. What is the ratio of their lengths?

We can do this in different ways.
We can check how much times \(\frac{1}{3}\) is \(\frac{1}{2}\).
\(\frac{1}{2}\) ÷ \(\frac{1}{2}\) = \(\frac{3}{2}\)
So, the length of the longer rope is \(\frac{3}{2}\) times that of the shorter.
That is, 1\(\frac{1}{2}\) times.
If the length of the shorter rope is taken as 1, the length of the longer is 1\(\frac{1}{2}\); if 2, then 3.
So the length of the shorter and longer rope are in the ratio 2 : 3.

We can also think in another manner. As in the case of the width and length of rectangles, we can imagine the length to be stretched by the same factor; and the ratio won’t change.
Suppose we double the length of each piece of rope.
Then the length of the shorter is \(\frac{2}{3}\) metres and that of the longer 1 metre. This doesn’t remove fractions.

By what factor should we stretch to get rid of fractions? How about 6 ?
6 times \(\frac{1}{3}\) is 2.
6 times \(\frac{1}{2}\) is 3.
The shorter is now 2 metres and the longer is 3 metres. So the ratio is 2 : 3.
There is yet another way; we can write
\(\frac{1}{3}\) = \(\frac{2}{6}\) \(\frac{1}{2}\) = \(\frac{3}{6}\)
That is, we can think of the shorter rope as made up of 2 pieces of \(\frac{1}{6}\) metres each and the longer one as 3 of the same \(\frac{1}{6}\) metres. In this way also, we can calculate the ratio as 2 to 3.
Now look at this problem : to fill a can we need only half the water in a bottle.

To fill a larger can, we need three quarters of the bottle. What is the ratio of the volume of the smaller can to the larger can ?
Here, we can write
\(\frac{1}{2}\) = \(\frac{2}{4}\)
So, if we pour \(\frac{1}{4}\) of the bottle 2 times, we can fill the smaller can; to fill the larger can, \(\frac{1}{4}\) of the bottle should be poured 3 times. So the volumes of the smaller and larger cans are in the ratio 2 : 3.

Another problem: Raju has 200 rupees with him and Rahim has 300. What is the ratio of the money Raju and Rahim have?
If we imagine both to have the amounts in hundred rupee notes, then Raju has 2 notes and Rahim, 3 notes. So the ratio is 2 : 3.

Let’s change the problem slightly and suppose Raju has 250 rupees and Rahim, 350 rupees.
If we think of the amounts in terms of 50 rupee notes, then Raju has 5 and Rahim has 7.
The ratio is 5 : 7.
What if the amounts are 225 and 325 rupees?

Imagine each amount as packets of 25 rupees. Raju has 225 ÷ 25 = 9 packets and Rahim has 325 ÷ 25 = 13 packets. The ratio is 9 : 13.
Let’s look at one more problem: in a class, there are 25 girls and 20 boys. What is the ratio of the number of girls to the number of boys?
If we split the girls and boys separately into groups of 5, there will be 5 groups of girls and 4 groups of boys. So the ratio is 5 to 4.

In this way, calculate the required ratios and write them using the least possible counting numbers, in these problems:

• Of two pencils, the shorter is of length 6 centimetres and the longer, 9 centimetres. What is the ratio of the lengths of the shorter and the longer pencils?
Answer:
Given that the length of shorter pencil is 6 centimetres and the longer is 9 centimetres.
The simplest form of 6 and 9 is 2 and 3.
In terms of ratio, the ratio of the lengths of the shorter and the longer pencils 2 : 3.

• In a school, there are 120 boys and 140 girls. What is the ratio of the number of boys to the number of girls?
Answer:
Given that there are 120 boys and 140 girls in a school.
The simplest form of 120 and 140 is 6 and 7.
In terms of ratio, it will be 6 : 7

• 96 women and 144 men attended a meeting. Calculate the ratio of the number of women to the number of men.
Answer:
Given that 96 women and 144 men attended a meeting.
The simplest form of 96 and 144 is  and 22 and 3.
In terms of ratio, it will be 2 : 3

• When the sides of a rectangle were measured using a string, the width was \(\frac{1}{4}\) of the string and the length \(\frac{1}{3}\) of the siring. What is the ratio of the width to the length’?
Answer:
Given that the width of rectangle is \(\frac{1}{4}\) of the string and the length is \(\frac{1}{3}\) of the siring.
The ratio of the width to the length will be 3 : 4

• To fill a larger bottle, 3\(\frac{1}{2}\) glasses of water are needed and to fill a smaller bottle, 2\(\frac{1}{4}\) of glasses are needed. What is the ratio of the volumes of the large and small bottles?
Answer:
Given that to fill a larger bottle 3\(\frac{1}{2}\) glasses of water are needed and to fill a smaller bottle, 2\(\frac{1}{4}\) of glasses are needed.
3\(\frac{1}{2}\) equals \(\frac{7}{2}\) and 2\(\frac{1}{4}\) equals \(\frac{9}{4}\).
The ratio of the volumes of the large and small bottles will be 14 : 9

Ratio of mixtures Textbook Page No. 123

To make idlis , Ammu’s mother grinds two cups of rice and one cup of urad dal. When she expected some guests the next day, she took four cups of rice. How many cups of urad should she take?
To have the same consistency and taste, the amount of urad must be halfthe amount of rice.
So for four cups of rice, there should be two cups of urad.
We can say that the quantities, rice and urad must be in the ratio 2 :1
Now another problem on mixing: to paint the walls of Abu’s home, first 25 litres of green and 20 litres of white were mixed.

When this was not enough, 15 litres of green was taken. How much white should be added to this?
To get the same final colour, the ratio of green and white should not change.
In what ratio was green and white mixed first?
That is, for 5 litres of green, we should take 4 litres of white.
To maintain the same ratio, for 15 litres of green, how many litres of white should we take?
How many times 5 is 15?
So 3 times 4 litres of white should be mixed.
That is 12 litres.
To get the same shade of green, how many litres of green should be mixed with 16 litres of white?

Now try these problems:

• For 6 cups of rice, 2 cups of urad should be taken to make dosas. How many cups of urad should taken with 9 cups of rice?
Answer:
Given that 6 cups of rice, 2 cups of urad should be taken to make dosas.
In terms of ratio, the ratio of the rice to urad will be 3 : 1.
Here, the number of rice cups is thrice the number of urad cups.
Therefore, to maintain the same ratio, for 9 cups of rice, 3 cups of urad should be taken.

• To set the walls of Nizar’s house, cement and sand were used in a ratio 1:5. He bought 45 sacks of cement. How many sacks of sand should he buy?
Answer:
Given that to set the walls of Nizar’s house, cement and sand were used in a ratio 1:5.
Here, 1 resembles the number of units for cement sacks and 5 resembles the number of unit for sand sacks.
The number of sand sacks required is 5 times the number of cement sacks.
If there are 45 sacks of cement, then the number of sand sacks required will be 45×5=225.

• To paint a house, 24 litres of paint was mixed with 3 litres of turpentine. How many litres of turpentine should be mixed with 32 litres of paint?
Answer:
Given that to paint a house, 24 litres of paint was mixed with 3 litres of turpentine.
In terms of ratio, the ratio of the paint to turpentine will be 8 : 1.
Let x be the number of litres of turpentine that should be mixed with 32 litres of paint.
\(\frac{8}{1}\) = \(\frac{32}{x}\)
8x = 32
x = \(\frac{32}{8}\)
x = 4
Therefore, 4 litres of turpentine should be mixed with 32 litres of paint.

• In the first ward of a panchayat, the male to female ratio is 10 : 11. There are 3311 women. How many men are there? What is the total population in that panchayat?
Answer:
Given that In the first ward of a panchayat, the male to female ratio is 10 : 11.
Let the number of men be m and there are 3311 women.
\(\frac{10}{11}\) = \(\frac{m}{3311}\)
11 × m  = 10 × 3311
m = \(\frac{10 × 3311}{11}\)
m = 3010
Therefore, the number of mens will be 3010.
The total population in that panchayat will be the sum of number of mens and womens, which is 3010+3311=6321

• In a school, the number of female and male teachers are in the ratio 5 : 1. There are 6 male teachers. How many female teachers are there?
Answer:
Given that, In a school, the number of female and male teachers are in the ratio 5 : 1.
Here, the number of female teachers is five times the number of male teachers.
If there are 6 male teachers.Then the female teachers will be 5×6=30.
Thus, there are 30 female teachers.

• Ali and Ajayan set up a shop together. Ali invested 5000 rupees and Ajayan, 3000 rupees. They divided the profit for a month in the ratio of their investments and Ali got 2000 rupees. How much did Ajayan get? What is the total profit?
Answer:
Given that, Ali invested 5000 rupees and Ajayan, 3000 rupees.
In terms of ratio, the ratio of Ali to Ajayan will be 5 : 3
Ali got 2000 rupees.Let m be the amount Ajayan should get.
\(\frac{5}{3}\) = \(\frac{2000}{m}\)
5×m = 2000×3
m = \(\frac{2000×3}{5}\)
m = 1200
Ajayan will get 1200 rupees.
The total profit will be the sum of maount got by Ali and Ajayan, which is 2000+1200=3200 rupees.

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Division Problem Textbook Page No. 125

We have seen that to make idlis, rice and urad are to be taken in the ratio 2:1. In 9 cups of such a mixture of rice and urad, how many cups of rice are taken?
2 cups of rice and 1 cup of urad together make 3 cups of mixture.
Here we have 9 cups of mixture.
How many times 3 is 9 ?
To maintain the same ratio, both rice and urad must be taken 3 fold.
So 6 cups of rice and 3 cups of urad.

Another problem: In a co-operative society there are 600 members are male and 400 are female. An executive committee of 30 members is to be formed with the same male to female ratio as in the society. How many male and female members are to be there in the committee?
In the society, the male to female ratio is 3 : 2.
3 men and 2 women make 5 in all.
Here we need a total of 30.
How many times 5 to 30?
So there should be 3 × 6 = 18 men and 2 × 6 = 12 women in the committee.

One more problem: a rectangular piece of land is to be marked on the school ground for a vegetable garden. Hari and Mary started making a rectangle with a 24 metre long rope. Vimala Teacher said it would be nice, if the sides are in the ratio 3 : 5. What should be the length and width of the rectangle?
The length of the rope is 24 metres and this is the perimeter of the rectangle.
Answer:
Given that, Hari and Mary started making a rectangle with a 24 metre long rope and their teacher suggested to make it in 3 : 5 ratio.
Let the ratio be x. Then the sides will be 3x and 5x.
Perimeter will be the total length of the rope which is 24 metre long.
3x + 5x = 24
8x = 24
x = \(\frac{24}{8}\)
x = 3
Therefore, the ength and width of the rectangle will be 9 metre and 15 metre.

If we take the length and width as 3 meters and 5 metres, what would be the perimeter?
Answer:
If the length and width is 3 meters and 5 metres respectively.
Then the perimeter will be 3+5=8 metres.

How much of 16 is 24?
\(\frac{24}{16}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\)
So width should be 1\(\frac{1}{2}\) times 3 metres; that is
3 × 1\(\frac{1}{2}\) = 4\(\frac{1}{2}\) metres.
And length should be 1\(\frac{1}{2}\) times 5 metres, that is,
5 × 1\(\frac{1}{2}\) = 7\(\frac{1}{2}\) metres

Now try these problems:

• Suhra and Sita started a business together. Suhra invested 40000 rupees and Sita, 30000 rupees. They made a profit of 7000 rupees which they divided in the ratio of their investments. How much did each get?
Answer:
Suhra invested 40000 rupees and Sita, 30000 rupees in a business together.
In terms of ratio, the ratio of Suhra to Sita will be 4 : 3.
They made a profit of 7000 rupees.
Let the ratio be x, then the investment will be 4x and 3x.
4x + 3x = 7000
7x = 7000
x = 1000
Therefore, Suhra’s profit will be 4×1000=4000 rupees
Sita’s profit will be 3×1000=3000 rupees.

• John and Ramesh took up a job on contract. John worked 7 days and Ramesh, 6 days. They got 6500 rupees as wages which they divided in the ratio of their investments. How much did each get?
Answer:
Given that John worked 7 days and Ramesh, 6 days.
In terms of ratio, the ratio for the number of days worked by John to Ramesh will be 7 : 6
Let the ratio be x, then the investment will be 7x and 6x.
7x + 6x = 6500
13x = 6500
x = \(\frac{6500}{13}\)
x = 500
Therefore, John will get 7×500 = 3500 rupees
Ramesh will get 6×500 = 3000 rupees

• John and Ramesh took up a job on contract. John worked 7 days and Ramesh, 6 days. They got 6500 rupees as wages which they divided in the ratio of the numbers of days each worked. How much did each get?
Answer:
Given that John worked 7 days and Ramesh, 6 days.
In terms of ratio, the ratio for the number of days worked by John to Ramesh will be 7 : 6
Let the ratio be x, then the investment will be 7x and 6x.
7x + 6x = 6500
13x = 6500
x = \(\frac{6500}{13}\)
x = 500
Therefore, John will get 7×500 = 3500 rupees
Ramesh will get 6×500 = 3000 rupees

• Angles of a linear pair are in the ratio 4 : 5. What is the measure of each angle?
Answer:
Angles of a linear pair are in the ratio 4 : 5.
Let the ratio be x,then it will be 4x and 5x.
4x + 5x = 90
9x = 90
x =10
4x = 4 × 10 = 40
5x = 5 × 10 = 50
Thus, the angles measurement will be 40 degrees and 50 degrees.

• Draw a line AB of length 9 centimetres. A point P is to be marked on it, such that the lengths of AP and PB are in the ratio 1 : 2. How far from A should P be marked? Compute this and mark the point.
Answer:
Length of AB is 9 centimetres.
Point P is marked in 1 : 2 ratio.
Let the ratio be x, then it will be 1x and 2x, the total is 3x.
\(\frac{1x}{3x}\) × 9 = 3 centimetres
\(\frac{2x}{3x}\) × 9 = 6 centimetres
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_2
P should be marked 3 cm away from A.

• Draw a line 15 centimetres long. A point is to be marked on it, dividing the length in the ratio 2 : 3. Compute the distances and mark the point.
Answer:
Length of AB is 15 centimetres.
Point P is marked in 2 : 3 ratio.
Let the ratio be x, then it will be 2x and 3x, the total is 5x.
\(\frac{2x}{5x}\) × 15 = 6 centimetres
\(\frac{3x}{5x}\) × 15 = 9 centimetres
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_3
P should be marked 6 cm away from A.

• Sita and Soby divided some money m the ratio 3 : 2 and Sita got 480 rupees. What is the total amount they divided?
Answer:
Sita and Soby divided some money ‘m’ in the ratio 3 : 2 and Sita got 480 rupees.
Let the ratio be m, then the ratio of money Sita got will be 3m.
3m = 480
m = \(\frac{480}{3}\)
m = 160
Money Soby will get 2×160 = 320 rupees.

• In a right triangle, the two smaller angles are in the ratio 1 : 4. Compute these angles.
Answer:
In a right triangle, the two smaller angles are in the ratio 1 : 4.
Right triangle means 90 degrees.
Let the ratio be x, total will be 1x + 4x = 5x.
\(\frac{1x}{5x}\)×90 = 18 degrees
\(\frac{4x}{5x}\)×90 = 72 degrees
Then the angles will be 18 degrees and 72 degrees.

• Draw a rectangle of perimeter 30 centimetres and lengths of sides in the ratio 1 : 2. Draw two more rectangles of the same perimeter, with lengths of sides in the ratio 2 : 3 and 3 : 7. Compute the areas of all three rectangles.
Answer:
A rectangle of perimeter 30 centimetres is drawn with lengths of sides in the ratio 1 : 2
Let the ratio be x, the total will be 1x + 2x = 3x.
One of the side will be \(\frac{1x}{3x}\)×30 = 10 centimetres
The other side will be \(\frac{2x}{3x}\)×30 = 20 centimetres
Therefore, the rectangle with 10 centimetres and 20 centimetres is drawn as shown below.
Area of this rectangle will be 10×5=50 square centimetres.
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_7

A rectangle of perimeter 30 centimetres is drawn with lengths of sides in the ratio 2 : 3
Let the ratio be x, the total will be 2x + 3x = 5x.
One of the side will be \(\frac{2x}{5x}\)×30 = 12 centimetres
The other side will be \(\frac{3x}{5x}\)×30 = 18 centimetres
Therefore, the rectangle with 12 centimetres and 18 centimetres is drawn as shown below.
Area of this rectangle will be 9×6=54 square centimetres.
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_8

A rectangle of perimeter 30 centimetres is drawn with lengths of sides in the ratio 3 : 7
Let the ratio be x, the total will be 3x + 7x = 10x.
One of the side will be \(\frac{3x}{10x}\)×30 = 9 centimetres
The other side will be \(\frac{7x}{10x}\)×30 = 21 centimetres
Therefore, the rectangle with 9 centimetres and 21 centimetres is drawn as shown below.
Area of this rectangle will be 10.5×4.5=47.25 square centimetres.
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_9

Kerala Syllabus 7th Standard Maths Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 7th Standard Maths Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes are part of Kerala Syllabus 7th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 7th Standard Maths Textbooks Solutions Pdf of Kerala Class 7 Part 1 and Part 2.

Kerala State Syllabus 7th Standard Maths Textbooks Solutions

Kerala Syllabus 7th Standard Maths Guide

Kerala State Syllabus 7th Standard Maths Textbooks Solutions Part 1

Kerala State Syllabus 7th Standard Maths Textbooks Solutions Part 2

We hope the given Kerala Syllabus 7th Standard Maths Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes will help you. If you have any queries regarding SCERT Kerala State Board Syllabus Class 7th Maths Textbooks Answers Guide Pdf of Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

You can Download Square and Square Root Questions and Answers, Activity, Notes, Kerala Syllabus 7th Standard Maths Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Square and Square Root Text Book Questions and Answers

Triangular numbers Textbook Page No. 80

See the dots arranged in triangles:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 1
How many dots are there in each?
1, 3, 6
How many dots would be there in the next triangle?
Such numbers as 1, 3, 6, 10, … are called triangular numbers.
The first triangular number is 1.
The second is 1 + 2 = 3.
The third is 1 + 2 + 3 = 6.
What is the tenth triangular number?
Answer:
From the above logic we need to add 10 numbers sum to obtain the 10th triangular number.
To find the 10th triangular number T10 we need to add 1+ 2+3+4+5+6+7+8+9+10 = 55
Therefore, 10th triangular number T10 is 55.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Squares and triangles Textbook Page No. 81

Look at these pictures:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 4
Each square is divided into two triangles.
Let’s translate this into numbers:
4 = 1 + 3
9 = 3 + 6
16 = 6 + 10
Check whether the same pattern continues. What do we see?
All perfect squares after 1 are the sums of two consecutive triangular numbers.
What is the sum of the seventh and eighth triangular numbers?
Answer:
Yes adding consecutive triangular numbers gives us perfect squares. So, we get the next pattern 9+16 = 25 which in turn is a perfect square.
7th Triangular number can be obtained as T7 = 1+2+3+4+5+6+7 = 28
8th Triangular number can be obtained as T8 = 1+2+3+4+5+6+7+8 = 36
Sum of Seventh and Eighth Triangular Numbers = T7+T8
= 28+36
= 64
Thus, the sum of 7th, 8th Triangular Numbers is 64

Increase and decrease Textbook Page No. 82

Look at this number pattern:
1 = 1
4 = 1 + 2 + 1
9 = 1 + 2 + 3 + 2 + 1
16 = 1 + 2 + 3 + 4 + 3 + 2 + 1
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 5
Can you split some more perfect squares like this?
Answer:
Writing some more perfect squares we have 25 = 1+2+3+4+5+4+3+2+1
36 = 1+2+3+4+5+6+5+4+3+2+1
49 = 1+2+3+4+5+6+7+6+5+4+3+2+1
64 = 1+2+3+4+5+6+7+8+7+6+5+4+3+2+1

Square difference

We have see that
22 = 12 + (1 + 2)
32 = 22 + (2 + 3)
42 = 32 + (3 + 4) and so on.
We can write these in another manner also:
22 – 12 = 1 + 2
32 – 22 = 2 + 3
42 – 32 = 3 + 4
In general, the difference of the squares of two consecutive natural numbers is their sum. Now look at these:
32 – 12 = 9 – 1 = 8
42 – 22 = 16 – 4 = 12
52 – 32 = 25 – 9 = 16
What is the relation between the difference of the squares of alternative natural numbers and their sum?
Answer:
32 – 12 = (3+1) x 2 = 8
42 – 22 = (4+2) x 2 = 12
52 – 32 =  = (5+3) x 2 = 16

The difference between squares of two alternate natural numbers is always even i.e. twice the sum of two numbers that are squared.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Project Textbook Page No. 84

Last digit
Look at the last digit of squares of natural numbers from 1 to 10:
1, 4, 9, 6, 5, 6, 9, 4, 1, 0
Now, look at the last digits of squares of natural numbers from 11 to 20.
Do we have the same pattern?
Let’s look at another thing: Does any perfect square end in 2?
Which are the digits which do not occur at the end of perfect squares?
Is 2637 then a perfect square?
Answer:
Last digits of squares of natural numbers from 11 to 20 are 1, 4, 9, 6, 5, 6, 9, 4, 1, 0. Yes they have the same pattern
As we observe the digits 2, 3, 7, 8 doesn’t occur at the end of perfect squares.
As we know 2637 ends with digit 7 at the end it is not a perfect square.

To decide that a number is not a perfect square, we need only look at the last digit.
Can we decide that a number is a perfect square from its last digit alone?
Answer:
Yes, we can decide if a number is perfect square or not by seeing the last digit alone as you can see above.

Rectangle and square

Look at this picture.
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 6
Dots in a rectangle.
Can you rearrange the dots to make another rectangle?
Can you rearrange the dots to make a square? Start like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 7
How many more are needed to make a square?
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 8
How many dots were there in the original rectangle? How many in this square?
What do we see here?
42 = (3 × 5) + 1
Can we do this for all rectangular arrangements?
The numbers here are 3, 4, 5.
So, for this trick to work, what should be the relation between the number of dots in each row and column of the rectangle?
We can write this in numbers as
22 = (1 × 3) + 1
32 = (2 × 4) + 1
42 = (3 × 5) + 1
Try to continue this
Answer:
52 = (4 x 6) +1
62 = (5 x 7) +1
72 = (6 x 8) +1
82 = (7 x 9) +1
92 = (8 x 10)+1

Square root of a perfect square

784 is a perfect square. What is its square root? 784 is between the perfect squares 400 and 900; and we know that their square roots are 20 and 30. So \(\sqrt{784}\) is between 20 and 30. Since last digit of 784 is 4, its square root should have 2 or 8 as the last digit. So \(\sqrt{784}\) is either 22 or 28.
784 is near to 900 than 400. So \(\sqrt{784}\) must be 28. Now calculate 282 and check.
Given that 1369, 2116, 2209 are perfect squares, find their square roots like this.

Project Textbook Page No. 87

Digit sum

16 is a perfect square and the sum of its digits is 7.
The next perfect square 25 also has digit sum 7.
The digit sum of 36 is 9.
The sum of the digits of the next perfect square 49 is 13. If we add the digits again, the sum is 4. Find the sum of the digit sums (reduced to a single digit number) of perfect squares starting from 1.
Do you see any pattern?
Is 3324 is perfect square?
Answer:
Given number is 3324
Now Split the number and add each number 3 + 3 + 2 + 4  = 12
As the result is more than 1 number we should add it again 1 + 2 = 3
All possible numbers that are perfect square have roots of either 1, 4, 7, 9
As 3 is not in the list 3324 is not a Perfect Square.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Rows and columns Textbook Page No. 80

Look this picture:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 2
Dots in rows and columns make a rectangle.
How many dots in all?
Did you count the dots one by one?
Can you make other rectangles with 24 dots?
Is any one of these a square?
How many more dots do we need to make a square? Can you remove some dots and make a square? How many?
Can you remove some dots and make a square? How many?
Numbers which can be arranged in squares are called square numbers.
Do you see anything special about of the number of dots making a square?
Answer:
There are total 4×6=24 dots in all.
No, the dots were counted by multiplying the number of dots in rows and number of dots in column.
None of these is a square.
We need more 12 dots to make a square, 24+12=36 or 62.
Yes, we can remove some dots and make a square.
We need to remove 8 dots to make a square, 24-8=16 or 42.
The number of dots making a square is the sqaure of that number.
Kerala State Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root img_1

Squares

What are the ways in which we can write 36 as the product of two numbers?
2 × 18, 3 × 12, 4 × 9
We can also write
36 = 6 × 6
And we have seen that it can also be shortened as 36 = 62.
36 is 6 multiplied by 6 itself; that is, the second power of 6.
There is another name for this:
36 is the square of 6.
Then what is the square of 5?
What is the square of \(\frac{1}{2}\)?
Answer:
Square of 5 = 52 = 25
\(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

Perfect squares

1, 4, 9, 16,… are the squares of the natural numbers. They are called perfect squares.
What is the perfect square after 16?
Why is 20 not a perfect square?
Answer:
Prime Factorization of 20 can be written as 22 x 51
As 5 is not in pair 20 is not a perfect square.
The next perfect square after 16 is 25.

Let us look at the succession of perfect squares in another way.
To reach 4 from 1, we must add 3.
To reach 9 from 4?
We can state these as
4 – 1 = 3
9 – 4 = 5
16 – 9 = 7
All these differences are odd numbers, right?
So, the difference of two consecutive perfect squares is an odd number.
Let’s write this as,
4=1 + 3
9 = 4 + 5 = 1 + 3 + 5
16 = 9 + 7 = 1 + 3 + 5 + 7
What do we see here?
When we add consecutive odd numbers starting from 1, we get the perfect squares.
This can be seen from these pictures also.
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 3
Can you write down the squares of natural numbers upto 20, by adding odd numbers? You can proceed like this
12 = 1
22 = 1 + 3 = 4
32 = 4 + 5 = 9
42 = 9 + 7 = 16

What is the relation between the number of consecutive odd numbers from 1 and their sum?
What is the sum of 30 consecutive odd numbers starting from 1?
Answer:
Let us assume the arithmetic series 1, 3, 5, 7, 9, 11, 13…..
we know the formula for sum of arithmetic series Sn = n/2[2a+(n-1)d]
Here d = 2
first term a = 1
Substituting the inputs we have Sn= 30/2[2×1+(30-1)2]
= 30/2[2+58]
= 30/2[60]
= 900
Therefore, the sum of 30 consecutive odd numbers starting from 1 is 900.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Tricks with ten Textbook Page No. 82

The square of 10 is 100. What is the square of 100?
In the square of 1000, how many zeros are there after 1?
What about the square of 10000?
What happens to the number of zeros on squaring?
So how do we spot the perfect squares among 10, 100, 1000, 10000 and so on?
Is one lakh a perfect square?
What about ten lakhs (million)?
Now find out the squares of 20, 200 and 2000.
Is 400000000 a perfect square?
What if we put in one more zero?
Answer:
Square of a number containing x zeros will become 2 times number of zeros.
1 lakh is not a perfect square as the number of zeros is not even.
Ten Lakhs is a perfect square as it comes with 6 zeros that are even.
20 = 20 x 20 = 400
200 = 200 x 200 = 40000
2000 = 2000 x 2000 = 4000000
400000000 is a perfect square as the number of zeros 8 is even.

Now some problems. Do them all in your head.

• Find out the squares of these numbers.

• 30
Answer:
302 = 30 x 30
= 900
The square of 30 is 900.

• 400
Answer:
4002 = 400 x 400
= 160000
The square of 400 is 160000

• 7000
Answer:
70002 = 7000 x 7000
= 49000000
The Square of 7000 is 49000000

• 6 × 1025
Answer:
(6 × 1025)= (6 × 1025) x (6 × 1025)
= 36 x (1025)2
= 36 x 1050

• Find out the perfect squares among these numbers.

• 2500
Answer:
The given number 2500 can be written as 502
Hence the 2500 is a perfect square number.

• 36000
Answer:
We cannot write the given number 36000 as square of two equal numbers. Hence it is not a perfect square.

• 1500
Answer:
We cannot write the given number 1500 as square of two equal numbers. Hence it is not a perfect square.

• 9 × 107
Answer:
We cannot write the given number 9 × 107 as square of two equal numbers. Hence it is not a perfect square.

• 16 × 1024
Answer:
The given number 2500 can be written as (4x 1012)2
Hence the given number 16 × 1024 is a perfect square.

Next square

What is the square of 21?
Wait a bit before you start multiplying.
The square of 20 is 400, isn’t it? So to get the square of 21, we need only add an odd number.
Which odd number?
Let’s start from the beginning. We can write
22 = 12 + 3 = 12 + (1 + 2)
32 = 22 + 5 = 22 + (2 + 3)
42 = 32 + 7 = 32 + (3 + 4)
52 = 42 + 9 = 42 + (4 + 5)
and so on. Continuing like this, how do we write 212?
212 = 202 + (20 + 21)
That is,
212 = 400 + 41 = 441
Now we can continue as before with
222 = 441 +43 = 484
and so on.
How do we find out the square of 101?
1002 = 10000
What more should we add?
100 + 101 = 201
So, 1012 = 10000 + 201 = 10201

• Find out the squares of these numbers using the above idea.
• 51
Answer:
Using the above process we write the 512 = 502+(50+51)
= 2500+(50+51)
= 2500+101
= 2601

• 61
Answer:
It can be written as 612 = 602+(60+61)
= 3600+(60+61)
= 3600+121
= 3721

• 121
Answer:
The given number is written as 1212 = 1202+(120+121)
= 14400+(120+121)
= 14400+241
= 14641

• 1001
Answer:
It is written as 10012 = 10002+(1000+1001)
= 10002+(1000+1001)
= 1000000+2001
= 1002001

• Compute the squares of natural numbers from 90 to 100.
Answer:
902
= 90 x 90
=8100
912
= 91 x 91
= 8281
922
= 92 x 92
= 8464
932
= 93 x 93
= 8649
942
= 94 x 94
= 8836
952
= 95 x 95
= 8025
962
= 96 x96
= 8928
972
= 97 x 97
= 9409
982
= 98 x 98
= 9310
992
= 99 x 99
= 9801
1002
= 100 x 100
= 10,000

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Fraction squares Textbook Page No. 83

A fraction multiplied by itself is also a square.
What is the square of \(\frac{3}{4}\) ?
(\(\frac{3}{4}\))2 = \(\frac{3}{4}\) × \(\frac{3}{4}\) = \(\frac{3 \times 3}{4 \times 4}\) = \(\frac{9}{16}\)
That is,
(\(\frac{3}{4}\))2 = \(\frac{9}{16}\) = \(\frac{3^{2}}{4^{2}}\)
So to square a fraction, we need only square the numerator and denominator separately.

Now do these problems without pen and paper.
• Find out the squares of these numbers.
• \(\frac{2}{3}\)
Answer:
\(\frac{2}{3}\)2  = \(\frac{2}{3}\)*\(\frac{2}{3}\)
= \(\frac{2×2}{3×3}\)
= \(\frac{4}{9}\)

• \(\frac{1}{5}\) 
Answer:
\(\frac{1}{5}\)2  = \(\frac{1}{5}\)*\(\frac{1}{5}\)
= \(\frac{1×1}{5×5}\)
= \(\frac{1}{25}\)

• \(\frac{7}{3}\)
Answer:
\(\frac{7}{3}\)2  = \(\frac{7}{3}\)*\(\frac{7}{3}\)
= \(\frac{7×7}{3×3}\)
= \(\frac{49}{9}\)

• 1\(\frac{1}{2}\)
Answer:
1\(\frac{1}{2}\)2  = \(\frac{1}{2}\)*\(\frac{1}{2}\)
=1 \(\frac{1×1}{2×2}\)
=1 \(\frac{1}{4}\)
= \(\frac{5}{4}\)

• Which of the fractions below are squares?

• \(\frac{4}{15}\)
Answer:
Numerator is 4 and Denominator is 15 .
Both are not perfect Square numbers so it is not possible to write the given fraction as squares.

• \(\frac{8}{9}\)
Answer:
Numerator is 8 and Denominator is 9.
Here 8 is not a perfect Square number and Denominator is a perfect square number. so it is not possible to write the given fraction as squares.

• \(\frac{16}{25}\)
Answer:
In a given fraction, numerator is 16 and Denominator is 25 which were square numbers of 4 and 5.Hence we can Write as \(\frac{4}{5}\)2

• 2\(\frac{1}{4}\)
Answer:
When the given 2\(\frac{1}{4}\) is converted to improper fraction, we get \(\frac{9}{4}\).
Here the Numerator is 9 and Denominator is 4
We can write the numerator as perfect square i.e. (3)2
Denominator 4 can be written as (2)2
Therefore, 2\(\frac{1}{4}\) can be written as \(\frac{3}{2}\)2

• 4\(\frac{1}{9}\)
Answer:
When the given 4\(\frac{1}{9}\) is converted to improper fraction, we get \(\frac{1}{9}\).
Here the Numerator is 1 and but the Denominator is 9 as a square of number (3)2
So, we cannot write the given fraction as Squares.

• \(\frac{8}{18}\)
Answer:
Numerator is 8 and Denominator is 18.
Here numerator and Denominator are not perfect square numbers so there is no chance to write the given fraction as square of numbers.

Decimal squares

What is the square of 0.5?
We know that 52 = 25. How many decimal places would be there in the product 0.5 × 0.5?
Why?
0.5 = \(\frac{5}{10}\), right?
Can you find out the square of 0.05?
Answer:
0.052 =0.05×0.05
= 0.0025 Hence four decimal places are there.
0.0025 = \(\frac{25}{1000}\)

You have computed the squares of many natural numbers. Using that table, can you find out the square of 0.15?
Do these problems also in your head.

• Find out the squares of these numbers.
Answer:
0.152 = 0.15 x 0.15
= 0.0225

• 1.2
Answer:
1.22 = 1.2 x 1.2
= 1.44

• 0.12
Answer:
0.122 = 0.12 x 0.12
= 0.0144

• 0.013
Answer:
0.0132 = 0.013 x 0.013
= 0.000169

• Which of the following numbers are squares?

• 2.5
Answer:
We cannot write the number 2.5 as a square.

• 0.25
Answer:
0.25 = 0.5 x 0.5.
Hence the number 0.25 is written as 0.52
• 0.0016
Answer:
0.0016 = 0.04 x 0.04.
Hence the number 0.0016 is written as 0.042

• 14.4
Answer:
We cannot write the number 14.4 as square.

• 1.44
Answer:
1.44= 1.2×1.2
Hence the number 1.44 is written as 1.22

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Square product Textbook Page No. 84

What is 52 × 42?
52 × 42 = 25 × 16 = ……..
There is an easier way:
52 × 42 = 5 × 5 × 4 × 4
= (5 × 4) × (5 × 4)
= 20 × 20
= 400
= (5 x 4) x (5 x 4)
= 20 x 20
= 400

Can you find out the products below like this, without pen and paper?

• 52 × 82
Answer:
52 × 82 = 5 x 5 x 8 x 8
= (5 x 8) x (5 x 8)
= 40 x 40
= 1600

• 2.52 × 42
Answer:
2.52 × 42 = 2.5 x 2.5 x 4 x 4
= (2.5 x 4) x (2.5 x 4)
= 10 x 10
= 100

• (1.5)2 × (0.2)2
Answer:
(1.5)2 × (0.2)2 = 1.5 x 1.5 x 0.2 x 0.2
= (1.5 x 0.2) x (1.5 x 0.2)
= 0.3 x 0.3
= 0.009

What general rule did we use in all these?
The product of the squares of two numbers is equal to the square of their product.
How do we say this in algebra?
x2y2 = (xy)2, for any numbers x, y
What about for three numbers?
Answer:
let the three numbers be x, y, and z
x2y2z2 = (xyz)2, for any numbers x, y, and z.

Square factors

How do we write 30 as a product of prime numbers?
30 = 2 × 3 × 5
So how do we factorize 900?
900 = 302 = (2 × 3 × 5)2 = 22 × 32 × 52
Similarly, using the facts that 24 = 23 × 3 and 242 = 576, we get
576 = 242 = (23 × 3)2 = (23)2 × 32 = 26 × 32

Can you write each number below and its square as a product of prime powers?
• 35
Answer:
35 = 5 x 7
Thus 35 can be written as product of prime powers 51 x 71

• 45
Answer:
45 = 5 x 9
= 5 x 3 x 3
= 51 x 32
Thus, 45 can be written as 51 x 32

• 72
Answer:
72 = 24 x 3
= 8 x 3 x 3
= 2 x 2 x 2 x 3 x 3
= 23  x 32
Thus, 72 can be written as 23  x 32

• 36
Answer:
36 = 9 x 4
= 9 x 2 x 2
= 3 x 3 x 2 x 2
= 32 x 22
Thus, 36 can be written as 32 x 22

• 49
Answer:
49 = 7 x 7
= 72
Thus, 49 can be written as 72

Did you note any peculiarity of the exponents of the factors of the squares?
Answer:

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Reverse computation

We have to draw a square; and its area must be 9 square centimetres.
How do we do it?
The area of a square is the square of the side.

So if the area is to be 9 square centimetres, what should be the side?
To draw a square of area 169 square centimetres, what should be the length of a side?
For that, we must find out which number squared gives 169. Looking up our table of squares, we find 132 = 169. So we must draw a square of side 13 centimetres.

Here, given a number we found out which number it is the square of. This operation is called extracting the square root.
That is, instead of saying the square of 13 is 169, we can say in reverse that the square root of 169 is 13.
Just as we write
132 = 169
as shorthand for statement “the square of 13 is 169”, we write the statement “the square root of 169 is 13” in shorthand form as
\(\sqrt{169}\) = 13
(the extraction of square root is indicated by the symbol Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 9)
Similarly, the fact that the square of 5 is 25 can also be stated, the square root of 25 is 5. In short hand form,
52 = 25
\(\sqrt{25}\) = 5
In general
For numbers x and y, if x2 = y, then \(\sqrt{y}\) = x

Now find out the square root of these numbers:

• 100
Answer:
10 = 100
\(\sqrt{100}\) = 10
Therefore, square root of 100 is 10

• 256
Answer:
162 = 256
\(\sqrt{256}\) = 16
Thus, the square root of 256 is 16.

• \(\frac{1}{4}\)
Answer:
\(\frac{1}{4}\) = \(\frac{1}{2}\)2
Thus, the Square root of \(\frac{1}{4}\) = \(\frac{1}{2}\)

• \(\frac{16}{25}\)
Answer:
\(\frac{16}{25}\) = \(\frac{4}{5}\)2
Thus, the square root of \(\frac{16}{25}\) is \(\frac{4}{5}\).

• 1.44
Answer:
1.44 = (1.2)2
Thus, the square root of 1.44 is 1.2

• 0.01
Answer:
0.01 = (0.1)2
Thus, the square root of 0.01 is 0.1

Square root factors Textbook Page No. 87

How do we find the square root of 1225?
Since a product of squares is the square of the product, we need only write 1225 as a product of squares.
First factorize 1225 into primes:
1225 = 52 × 72
And we can write
52 × 72 = (5 × 7)2 = 352
So, 1225 = 352
From this, we get \(\sqrt{1225}\) = 35
Let’s take another example. What is the \(\sqrt{3969}\) ?
As before, we first factorize 3969 into primes.
3969 = 32 × 32 × 72
= (3 × 3 × 7)2
From this, we get \(\sqrt{3969}\) = 3 × 3 × 7 = 63

Now compute the square roots of these.
• 256
Answer:
Given number is 256
Firstly factorizing it into primes we have 256 = 2 x 128
= 2 x 2 x 64
= 2 x 2 x 2 x 32
= 2 x 2 x 2 x 2 x 16
= 2 x 2 x 2 x 2 x 2 x 8
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= (2 x 2) x (2 x 2)  x (2 x 2) x (2 x 2)
= 4 x 4 x 4 x 4
= 16 x 16
= (16)2
From this, we get \(\sqrt{256}\) = 16

• 2025
Answer:
Given number is 2025
Writing the factorization of 2025 we have
= 3 x 3 x 3 x 3 x 5 x 5
= 9 x 9 x 5 x 5
= (9 x 5)2
= (45)2
Therefore, the square root of \(\sqrt{2025}\) is 45

• 441
Answer:
Given number is 441
Writing the factorization of 441 we have
= 3 x 3 x 7 x 7
= 32 x 72
= (3 x 7)2
= (21)2

Therefore, the square root of 441 is 21.

• 921
Answer:
Writing the factorization of 921 we have 3 x 307
Thus, 921 can’t be written as perfect square and the square root of 921 isn’t a natural number.

• 1089
Answer:
Given number is 1089
Writing the factorization of it we have 1089 = 3 x 3 x 11 x 11
= 32 x 112
= (3 x 11)2
= (33)2
Therefore, square root of 1089 is 33.

• 15625
Answer:
Given number is 15625
Writing the factorization of it we have 15625 = 5 x 5 x 5 x 5 x 5 x 5
= 52 x 52 x 52
= (5 x 5 x 5)2
= (125)2
Therefore, the square root of 15625 is 125

• 1936
Answer:
Given number is 1936
Writing the factorization of 1936 we have 2 x 2 x 2 x 2 x 11 x 11
= (22 x 22 x 112)
= (2 x 2 x 11)2
= 442
Therefore, the square root of 1936 is 44.

• 3025
Answer:
Given number is 3025
Writing the factorization of 3025 we have 5 x 5 x 11 x 11
= 52 x 112
= (5 x 11)2
= 552
Therefore, the square root of 3025 is 55.

• 12544
Answer:
Given number is 12544
Writing the factorization of 12544 we have 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7
= 28 x 72
= (112)2
Therefore, Square Root of 12544 is 112.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Let’s do it!

Question 1.
The area of a square plot is 1024 square meters. What is the length of its sides?
Answer:
Given area of Square plot = 1024 Square Meters
Imagine ‘S’ is side of a Square. As we know Area of Square = Side x Side
1024 = Side x Side
1024 = Side2
Side = √1024
Side = 2 5
Hence Length of Square Plot = 32 meters

Question 2.
In a hall, 625 chairs are arranged in rows and columns, with the number of rows equal to the number of columns. The chairs in one row and one column are removed. How many chairs remain?
Answer:
Total number of chairs = 625
Number of Rows and Columns present are Equal i.e. Rows =  Columns , Let it be y.
Rows .Columns=625
y. y =625
y 2= 625
y = √625
y = √(25)2
y = 25
so Number of Rows and Number of Columns =25
When 1 Row and 1 Column is removed as below
Number of Rows =25-1=24
Number of Rows =25-1=24
When the 24 Rows and 24 Columns are arranged = 24 x 24
= 576

Question 3.
The sum of a certain number of consecutive odd numbers, starting with 1, is 5184. How many odd numbers are added?
Answer:
We know the formula of sum of arithmetic series Sn = n/2[2a+(n-1)d]
The arithmetic series 1, 3, 5, 7, 9, 11, 13 ……
Here the first term a = 1
Number of odd numbers to be added = n
Common difference d = 2
Substituting in the formula above we have
5184 = n/2[2×1+(n-1)2]
5184 = n/2[2+2n-2]
5184 = n/2[2n]
5184 = n2
n = 72
Thus, the number of odd numbers to be added is 72.

Question 4.
The sum of two consecutive natural numbers and the square of the first is 5329. What are the numbers?
Answer:
Suppose x and x+1 are two Consecutive natural numbers
As per the Given data, Square of First number x2  = 5329
When they are added it gives as  x+(x+1) +x2 = 5329
x + x+1+x2 = 5329
x + x+1+x = 5329
2x +1+x = 5329
(x+1)= 5329
(x+1) = √5329
x+1 = √(73)2
x+1 = 73
x = 73-1
x= 72
Therefore, the numbers are 72 and 73.

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

You can Download Repeated Multiplication Questions and Answers, Activity, Notes, Kerala Syllabus 7th Standard Maths Solutions Chapter 4 to help you to revise the complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

Repeated Multiplication Text Book Questions and Answers

Power of products Textbook Page No. 50

This is an old story. A poor man went to a rich man for help. Said the rich man, “I can give you a thousand rupees a day for thirty days, or I can give you one paisa the first day, two paise the third day and so on, doubling each day for thirty days. Choose which you want.”
Which is the better deal?
Let’s see.
The first offer will get the poor man 30000 rupees in 30 days.
What about the second?
1 + 2 + 4 + 8 + 16 + …
We must add the thirty numbers. Do you know how much it is? 1073741823 paise. That is more than one crore rupee!

Exponentiation
We usually do the four operations of addition, subtraction, multiplication and division in arithmetic. Exponentiation is the fifth operation. Just as multiplication by natural numbers is repeated addition, exponentiation is repeated multiplication.

The first four operations are indicated by a symbol (+, -, ×, ÷) between the numbers operated on. But no such symbol is used to indicate exponentiation.

The notation is to write on the upper right of the number multiplied, the number showing how many times to multiply, in a smaller font.
For example, 4 × 4 × 4 = 43

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

Sum of Powers Textbook Page No. 52

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication 2

What part of each square is shaded?
\(\frac{1}{2}\) in the first square.
In the second?
\(\frac{1}{2}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\)
We can look at it this way.
The unshaded part is \(\frac{1}{4}\)
So, the shaded part is
1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
What do we see here?
\(\frac{1}{2}\) + \(\frac{1}{4}\) = 1 – \(\frac{1}{4}\)
Similar reasoning gives from the third square,
\(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) = 1 – \(\frac{1}{8}\)
And from the fourth,
\(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{16}\) = 1 – \(\frac{1}{16}\)
We can continue this:
\(\frac{1}{2}\) + \(\frac{1}{2^{2}}\) + \(\frac{1}{2^{3}}\) = 1 – \(\frac{1}{2^{3}}\)
\(\frac{1}{2}\) + \(\frac{1}{2^{2}}\) + \(\frac{1}{2^{3}}\) + \(\frac{1}{2^{4}}\) = 1 – \(\frac{1}{2^{4}}\)
In general, the sum of the powers \(\frac{1}{2}\), \(\frac{1}{2^{2}}\), \(\frac{1}{2^{3}}\) and so on is equal to the last power subtracted from 1.

Another sum

\(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) = 1 – \(\frac{1}{8}\)
If we multiply the numbers on either side by 8, we get
8(\(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\)) = 8(1 – \(\frac{1}{8}\))
That is,
(8 × \(\frac{1}{2}\)) + (8 × \(\frac{1}{4}\)) + (8 × \(\frac{1}{8}\) = 8 – (8× \(\frac{1}{8}\))
4 + 2 + 1 = 8 – 1
Likewise, multiplying either side of
\(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{16}\) = 1 – \(\frac{1}{16}\)
by 16, we get
8 + 4 + 2 + 1 = 16 – 1
Rearranging the numbers, we get
1 + 2 + 4 = 8 – 1
1 + 2 + 4 + 8 = 16 – 1
That is,
2 + 4 = 8 – 2
2 + 4 + 8 = 16 – 2
Using powers, we can write this as
2 + 22 = 23 – 2
2 + 22 + 23 = 24 – 2
And we can continue.
In general, the sum of the powers 2, 22, 23 and so on, is equal to 2 subtracted from the
next power to the last.

Numbers in science Textbook Page No. 54

Science often requires large numbers. For example, the average distance between the Earth and Sun is 149000000 kilometres. This is written in scientific notation as 1.49 × 108. Similarly, the distance travelled by light in one year is about 9.46 × 1017 kilometres. This distance is called a light year. Astronomical distances are often given in terms of light years. The star nearest to the Earth is the Sun. The next nearest star is Proxima Centauri, which is about 4.22 light years away, that is, about 3.99 × 1018 kilometres. We can describe this in another way. Light rays from this star take more than four years to reach the earth. Thus what we now see is how the star was four years ago. So even if the star dies now, we would go on seeing it for four more years!

Project

Last digit
The last digit of every power of 10 is 0. What is the last digit of a power of 5?
What about 6?
Look at the last digits of powers of 4; are they all the same?
What are the last digits?
Check the last digit of the powers of other single-digit numbers.
One more question: What is the last digit of 2100?

Answer: The last digit in 2100 is 4
Explanation: Given the value is 2100
Now, we will find the value of 610
4 simple as 2¹⁰ is 1024, therefore 2¹⁰⁰ will also have 4 in one place.
Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

Increase or decrease? Textbook Page No. 56

We have seen how the powers 2, 4, 8, 16, ….of 2 grow very fast. Do the powers of other numbers also increases like this?
What are the powers of \(\frac{1}{2}\)?
\(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), ……..
But these have become smaller and smaller, right?
And the powers of \(\frac{2}{3}\)?
And the powers of \(\frac{3}{2}\)?
What are the numbers whose powers become larger and larger? And what numbers have smaller and smaller powers?
What about the powers of 1?
Answer: According to the exponent rule, any number will be raised to the power of one, it equals to the number itself.

Multiples and powers

For any natural number n and for any number (natural number or fraction) x, the notation nx or n × x is used to denote the sum of x has taken n times. And xn denotes the product of x has taken n times.

Look at the rule for adding multiples of a number by natural numbers, and multiplying powers of a number:
mx + nx = (m + n)x
xm × xn = xm + n
We can multiply a number by a fraction also, but it is not repeated addition. For this multiplication, it is still true that mx + nx = (m + n) x, even when m and n are fractions. But if n is a fraction, the symbol xn as yet has no meaning.

Multiples and powers of 2

All powers of 2 are even, but all even numbers are not powers of 2. For example, 6 is an even number, not a power of 2. But
6 = 2 + 4 = 21 + 22
We can also write
10 = 2 + 8 = 21 + 23
12 = 4 + 8 = 22 + 23
14 1 = 2 + 4 + 8 = 21 + 22 + 23
Check whether other even numbers can also be written as sums of powers of 2.
For example, can we write 100 as a sum of the powers of 2?
If we look at the powers of 2 one by one, we see that 26 = 64 is less than 100, while 27 = 128 is more than 100. We can start by writing
100 = 26 + 36
And then we note that 25 = 32 < 36 and 26 = 64 > 36, so that we can write
36 = 25 + 4 = 25 + 22
Thus
100 = 26 + 25 + 22
Now try to write 150 as a sum of powers of 2, using this technique.

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

Odd numbers and powers of 2 Textbook Page No. 59

We saw that even numbers could be written as sums of powers of 2. Any odd number, except 1, is 1 added to an even number. So, we can write it as sums of powers of 2 and 1.
For example, to split 25 like this, we first write
25 = 24 + 1
Then we write 24 as the sum of powers of 2 as before
24 = 16 + 8 = 24 + 23
Thus
25 = 24 + 23 + 1
In general, any natural number can be written as the sum of some of the numbers 1, 2, 23,…

Subtraction and division Textbook Page No. 60

We have rules for subtracting multiples of a number, just as we have rules for adding such numbers. There is also a condition that we can only subtract the smaller from the larger.
That is
mx – nx = (m – n) x, for any number x and for any natural numbers m, n with m > n
What about powers?
\(\frac{x^{m}}{x^{n}}\) = xm – n
Here, we must also state the conclusion
x ≠ 0
As in the case of an addition, the rule for the subtraction of multiplies holds, even when m and n are fractions.

Division and subtraction Textbook Page No. 61

When fractions are also taken into account, we can divide a small number by a large number also; the result would be a fraction. So we can consider the division of a larger power by a smaller power also.
\(\frac{x^{m}}{x^{n}}\) = \(\frac{1}{x^{n-m}}\), if m < n
There is no analogous rule for multiplies. There is no way we can subtract a larger number from a smaller number as yet.

Pouch problem Textbook Page No. 62

100 one-rupee coins are to be put into several pouches, such that any amount up to 100 rupees can be taken without opening any pouch. Can we do it?
Put a single coin in the first pouch. 2 coins in the second, 4 in the third and so on, using powers of 2.
1 + 2 + 4 + 8 + 16 + 32 = 64 – 1 = 63
Put the remaining 100 – 63 = 37 coins in a single pouch.
Now if the required amount is less than 63, we split it into powers of 2 (and 1 if needed) and take the corresponding pouches. For example,
35 = 32 + 2 + 1
so that we need only take the first, second and sixth pouch.
For amounts larger than 63 rupees?
For example, to take 65 rupees, first, take the last pouch of 37 rupees. We need 65 – 37 = 28 rupees more. And this can be done by splitting
28 = 16 + 8 + 4

Project

Some natural numbers can be written as sums of consecutive natural numbers.
For example,
3 = 1 + 2
7 = 3 + 4
15 = 1 + 2 + 3 + 4 + 5 = 7 + 8
But some natural numbers cannot be split like this. For example, 4 cannot be written in this form
Can you find the specially of those numbers which cannot be written as sums of consecutive natural numbers?
Answer: We can’t write every number as sums of consecutive numbers. The few example numbers are 1, 2, 4, 8

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

Perfect numbers

Factors 6 are 1, 2, 3, and 6.
The sum of the factors other than 6 itself is
1 + 2 + 3 = 6
Now let’s look at 28:
28 = 22 × 7
And so its factors can be listed as
Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication 7
And the sum of factors other than 28 itself is
1 + 2 + 22 + 7 + (2 × 7) = 7 + 7 + 14 = 28
Next, let’s look at the factors of 496
24 × 31 = 16 × 31 = 496
Since 31 is a prime number, the factors are
Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication 8
In this, the sum of the numbers in the first row is
1 + 2 + 22 + 23 + 24 = 25 – 1 = 31
(see the section, Another sum)
And the sum of the numbers, except 24 × 31 in the second row is
(1 + 2 + 22 + 23) × 31 = (24 – 1) × 31
= (24 × 31) – 31
So the sum of the factors except 496 itself is
31 + (24 × 31) – 31 = 24 × 31 = 496
Numbers like 6, 28 and 496 are called perfect numbers.

Projects

32 = 25 number of factors 6
81 = 34 number of factors 5
72 = 23 × 32 number of factors 12

Write some more numbers as products of powers of primes and also the number of factors they have.
How did you calculate the number of factors?
Is there any relation between the numbers in exponents and the number of factors?
Answer: yes. Consider a power number, the number 5 is called the base, and the number 2 is called the exponent. The exponent corresponds to the number of times the base is used as a factor.

Multiply again and again Textbook Page No. 50

Look at these figures:
Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication 1
How many cells are there in the first square?
What about the second and third squares?
If we continue like this, how many cells would be there in the next square?

Answer:
Let’s look at the problem like this:
The first square has four cells. And the next one has four such squares.
So it has 4 × 4 = 16 cells.
The third square is made up of four squares like the second one, so it has 16 × 4 = 64 cells.
What about the next?
64 × 4 = 256 cells.
We can write like this also:
Number of cells
in the first square 4
in the second square 4 × 4
in the third square 4 × 4 × 4
What about the 10th square?
4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 cells.

Instead of writing in this elaborate manner, we write it shortly as 410. And read “4 raised to 10”.
If we do the actual multiplication, we can see that it is
1048576.
Now we can easily say that the number of cells in these squares is 4, 42, and 43,… So that the number of cells in the 20th square is 420, the number of cells in the 100th square is 4100 and so on. When it becomes difficult to compute the actual numbers, we can use a computer.
The numbers 4, 42, 43, 44,.. which we saw here are called powers of 4.
42 is the second power of 4.
43 the third power of 4.
And so on. If need be, we can write 4 itself as 41 and call it the first power of 4.
In 43, the number 3 is called the exponent (or power).
We also call the second power of a number, its square and the third power, its cube.

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

Exponentiation

Just as the name of repeated addition is called multiplication, the name for repeated multiplication is exponentiation.
Let’s look at some more examples.
What are the powers of 3?
How do we compute 31, 32,33,…?
We can compute the powers one by one as
31 = 3
32 = 3 × 3 = 9
33 = 3 × 3 × 3 = 27
How do we compute 36?
Instead of computing powers one after another as above, let’s see if there is another way. For example
36 = 3 × 3 × 3 × 3 × 3 × 3
Instead of multiplying one by one, let’s take three at a time.

36 = (3 × 3 × 3) × (3 × 3 × 3)
= 27 × 27
= 729
How about 29?
29 = (2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)
= 16 × 32
= 512
Can you compute this in any other way?
Answer: No

Now compute these powers:

• 26

Answer: 64
Explanation: The given number is 26
Now, re-write the powers values in the form of,
26 = 2 x 2 x 2 x 2 x 2 x 2.
4 x 4 x 4 = 16 x 4 = 64.

• 38

Answer: 6561
Explanation: The given number is 38
Now, re-write the powers values in the form of,
38 = 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3.
9 x 9 x 9 x 9 = 81 x 81= 6561.

• 44

Answer: 256
Explanation: The given number is 44
Now, re-write the powers values in the form of,
44 = 4 x 4 x 4 x 4
16 x 16 = 256.

• 29

Answer: 512
Explanation: The given number is 29
Now, the powers values are in the form of,
29 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
4 x 4 x 4 x 4 x 2 = 16 x 16 x 2 = 256 x 2
29 = 512.

• 106

Answer: 1000000
Explanation: The given number is 106
Now, the powers values are in the form of,
106 = 10 x 10 x 10 x 10 x 10 x 10
106 = 1000000.

• 110

Answer: 1
Explanation: The given number is 110
Now, the powers values are in the form of,
110 = 1 x 1 x 1 x 1 x 1 x 1 (1 power of any value, the value is 1 only).
110 = 1

• 1004

Answer: 100000000
Explanation: The given number is 1004
Now, the powers values are in the form of,
1004 = 100 x 100 x 100 x 100
1004 = 100000000

• 020

Answer: 0
Explanation: The given number is 020
Now, the powers values are in the form of,
020 = 0 (0 power of any value, the value will be 0 only).

Power of ten Textbook Page No. 52

What are the powers of 10?
They can be written 10, 102, 103, …..
How do we compute them?
102 = 10 × 10 = 100
103 = 10 × 10 × 10 = 1000
So what is 108?
Suppose we want to compute powers of 20.
How do we compute 20?
204 = 20 × 20 × 20 × 20
= (2 × 10) × (2 × 10) × (2 × 10) × (2 × 10)
= (2 × 2 × 2 × 2) × (10 × 10 × 10 × 10)
= 16 × 10000 = 160000

What about 24 × 55?
We can write it as (2 × 2 × 2 × 2) × (5 × 5 × 5 × 5 × 5)
Reordering the numbers, this becomes,
(2 × 5) × (2 × 5) × (2 × 5) × (2 × 5) × 5
= 10 × 10 × 10 × 10 × 5
= 104 × 5 = 50000

What is 1003?
1003 = 100 × 100 × 100
We can write this as 10 × 10 × 10 × 10 × 10 × 10
1003 = 106
= 1000000

Now try these problems.

• Write hundred, thousand, ten thousand, lakh, 10 lakh and crore as powers of 10.

Answer: The hundred in powers is,10 power of 2 i.e., 102 = 10 x 10 = 100.
The thousand in powers is, 10 power of 3 i.e., 103 = 10 x 10 x 10 = 1000.
The ten thousand in powers is, 10 power of 4 i.e., 104 = 10 x 10 x 10 x 10 = 10,000.
The lakh in powers is, 10 power of 5 i.e., 105 = 10 x 10 x 10 x 10 x 10= 100000.
The ten lakh in powers is, 10 power of 6 i.e., 106 = 10 x 10 x 10 x 10 x 10 x 10= 1000000.
The crores in powers is, 10 power of 7 i.e., 107 = 10 x 10 x 10 x 10 x 10 x 10 x 10 = 10000000.

• Compute the powers.
• 304

Answer: 810000
Explanation: Given the number is 304
Now, we find the value of the given number.
304 = 30 x 30 x 30 x 30
304 = 810000

• 505

Answer: 12500000
Explanation: Given the number is 505
Now, we find the value of the given number.
505 = 50 x 50 x 50 x 50 x 50
We can rewrite it as,
(5×10) (5×10) (5×10) (5×10) (5×10)
(5 x 5 x 5 x 5 x 5) x 105
25 x 25 x 5 x 105
625 x 5 x 105
3125 x 105= 12500000
505 = 12500000

• 2003

Answer: 8000000
Explanation: Given the number is 2003
Now, we find the value of the given number.
2003 = 200 x 200 x 200
= 40000 x 200
304 = 8000000

Place value

How do we split 3675 according to place values?
(3 × 1000) + (6 × 100) + (7 × 10) + 5
Using powers of 10, we can write this as
(3 × 103) + (6 × 102) + (7 × 10) + 5

Can you split these numbers like this?
• 1221

Answer: Given the number is 1221.
Now, we can split the given number as,
1221 = (1 x 1000) + (2 x 100) + ( 2 x 10) + 1

• 60504

Answer: Given the number is 60504.
We can split the given number as,
60504 = (6 x 10000) + (5 x 100) + 4
We can also split as , (6 x 10000) + ( 0 x 1000) + ( 5 x 100) + (0 x 10 ) +4

• 4321

Answer: As given in the question, the value is 4321.
Now, we need to split, according to place values.
4321 = (4 x 1000) + ( 3 x 100) + ( 2 x 10) + 1

• 732

Answer: The given value is 732.
Now, split the given number as,
732 = (7 x 100) + ( 3 x 10) + 2

What about decimals?

Answer: Decimal is a term that describes the base-10 number system. The decimal number system consists of ten single-digit numbers such as  0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. The number after 9 is 10. An example of a decimal number is 1.48, in which 1 is the whole number, while 48 is the decimal part.

How do we split 362.574?
362.574 = (3 × 100) + (6 × 10) + 2 + (5 × \(\frac{1}{10}\)) + (7 × \(\frac{1}{100}\)) + (4 × \(\frac{1}{1000}\))

We can write this as
(3 × 102) + (6 × 10) + 2 + (5 × \(\frac{1}{10}\)) + (7 × \(\frac{1}{10^{2}}\)) + (4 × \(\frac{1}{10^{3}}\))

Try to split these numbers like this
• 437.54

Answer: The given value is 437.54
Now, split the given number as,
437.54 = (4 x 100) + ( 3 x 10) + 7 + (5 x 1/10) + (4 x 1/100)
437.54 = (4 x 100) + ( 3 x 10) + 7 + (5 x 1/10) + (4 x 1/102)
437.54 = (4 x 102) + ( 3 x 10) + 7 + (5 x 1/10) + (4 x 1/102)

• 23.005

Answer: The given value is 23.005
Now, we can split the given number as,
23.005 = (2 x 10) + 3 + (5 x 1/1000)
23.005 = (2 x 10) + 3+ (5 x1/103)

• 4567

Answer: The given value is 4567.
Now, split the given number as,
4567 = (4 x 1000) + ( 5 x 100) + (6 x 10) + 7
4567 = ( 4 x 103) + ( 5 x 102) + (6 x 10) + 7

• 201

Answer: The given value is 201.
Now, split the given number as,
201 = (2 x 100) + 1
201 = (2 x 103) + 1

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

Factorization

We can factorize any number as a product of prime numbers. For example, we can write 72 as
72 = 2 × 2 × 2 × 3 × 3
Using powers, this can be written
72 = 23 × 3 2

How do we split 1000 like this?
1000 = 2 × 2 × 2 × 5 × 5 × 5
= 23 × 53

Now try to write the numbers below as products of powers of prime numbers.

• 36
Answer: 22 × 32
Explanation: Given the number is 36.
Now, we split 36 as,
36 = 2 x 2 x 3 x 3 which is nothing but 22 × 32

• 225
Answer: 32 × 52
Explanation: Given the number is 225.
Now, we split 225 as,
225 = 3 x 3 x 5 x 5 which is nothing but 32 × 52

• 500
Answer: 22 × 53
Explanation: As given in the question the number is 500.
Now, we can split 500 as,
500 = 2 x 2 x 5 x 5 x 5 which is nothing but 22 × 53

• 784
Answer: 24 × 72
Explanation: As given in the question the number is 784.
Now, we can split 784 as,
784 = 2 x 2 x 2 x 2 x 7 x 7 which is nothing but 24 × 72

• 750
Answer: 53 × 31 x 21
Explanation: As given in the question the number is 750.
Now, we can split 750 as,
750 = 5 x 5 x 5 x 3 x 2  which is nothing but 53 × 31 x 21

• 625
Answer: 54
Explanation: As given in the question the number is 625.
Now, we can split 625 as,
625 = 5 x 5 x 5 x 5  which is nothing but 54

• 1024
Answer: 210
Explanation: As given in the question the number is 1024.
Now, we can split 1024 as,
1024 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 which is nothing but 210

Powers of a fraction Textbook Page No. 54

Look at these pictures:

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication 3

What part of the first square is coloured?
And in the second?
\(\frac{1}{4}\) of
That is
\(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)
In the third square, what part is coloured?
\(\frac{1}{4}\) × \(\frac{1}{16}\) = \(\frac{1}{64}\)
This is the product of three \(\frac{1}{4}\)‘s.

If we continue like this, what part of the next square should be coloured?
And in the fifth square?
We must multiply together five \(\frac{1}{4}\)‘s.
We can shorten this as (\(\frac{1}{4}\))5
(\(\frac{1}{4}\))5 = \(\frac{1}{4}\) × \(\frac{1}{4}\) × \(\frac{1}{4}\) × \(\frac{1}{4}\) × \(\frac{1}{4}\)
= \(\frac{1}{4 \times 4 \times 4 \times 4 \times 4}\)
= \(\frac{1}{4^{5}}\)
= \(\frac{1}{64 \times 16}\)
= \(\frac{1}{1024}\)
That is in the fifth square, only \(\frac{1}{1024}\) of the square needs to be coloured.

We can write the repeated multiplication of any fraction as a power like this. For example,
(\(\frac{3}{5}\))3 = \(\frac{3}{5}\) × \(\frac{3}{5}\) × \(\frac{3}{5}\)
= \(\frac{3 \times 3 \times 3}{5 \times 5 \times 5}\) = \(\frac{3^{3}}{5^{3}}\)
= \(\frac{27}{125}\)

Let’s examine one more example:
(2\(\frac{2}{5}\))3 = (\(\frac{12}{5}\))3
= \(\frac{12}{5}\) × \(\frac{12}{5}\) × \(\frac{12}{5}\)
= \(\frac{1728}{125}\) = 13\(\frac{103}{125}\)

Compute the powers given below:

• (\(\frac{2}{3}\))5

Answer: (\(\frac{32}{243}\))
Explanation: Given the number is (\(\frac{2}{3}\))5
Now, we find the value of the given number.
(\(\frac{2}{3}\))5 = 2 x 2 x 2 x 2 x 2 / 3 x 3 x 3 x 3 x 3
= 32/ 9 x 9 x 3
= 32/ 81 x 3
= 32/243

• (\(\frac{3}{5}\))4

Answer: (\(\frac{81}{625}\))
Explanation: Given the value is (\(\frac{3}{5}\))4
Now, we will find the value of the given value.
(\(\frac{3}{5}\))4 = 3 x 3 x 3 x 3/ 5 x 5 x 5 x 5
= 9 x 9 / 25 x 25
=81/625

• (\(\frac{1}{2}\))10

Answer: (\(\frac{1}{1024}\))
Explanation: Given the value is (\(\frac{1}{2}\))10
Now, we will find the value of the given value.
(\(\frac{1}{2}\))10 = 1/ 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 1 / 4 x 4 x 4 x 4 x 4
=1/1024

• (2\(\frac{1}{2}\))3

Answer: (15\(\frac{5}{8}\))
Explanation: Given the value is (2\(\frac{1}{2}\))3
Now, we will find the value of the given value.
(2\(\frac{1}{2}\))3= (\(\frac{5}{2}\))3
= 5 x 5 x 5 / 2 x 2 x 2
= 125 / 8
= (15\(\frac{5}{8}\))

Powers of decimal Textbook Page No. 56
What is (1 .2)2?
(1.2)2 = 1.2 × 1.2
= 1.44

Can you compute (1.5)3 like this?
Answer: Given the value is (1.5)3
Now, we will find the value.
(1.5)3 = 1.5 x 1.5 x 1.5 = 3.375

What is (0.2)4?
We know that 24 = 16
We can write 0.2 as \(\frac{2}{10}\)
(0.2)4 = (\(\frac{2}{10}\))4
= \(\frac{2^{4}}{10^{4}}\)
= \(\frac{16}{10000}\)
= 0.0016

Can’t you do this without writing it out?
Can you compute (0.3)3 in your head?
What is 33?
How many decimal places would (0.3)3 have?
123 = 1728. Can you find out (1 .2)3 and (0.12)3 using this?

Answer:
Given the values are (1 .2)3 and (0.12)3
Now, we will find the value.
First, take the (1 .2)3 value.
(1 .2)3 = (12/10)3
= (12 x 12 x 12)/(10 x 10 x 10)
= 1728/1000
(1 .2)3 = 1.728
(0.12)3 = (12/100)3
= (12 x 12 x 12)/(100 x 100 x 100)
= 1728/1000000
(0.12)3 = 0.001728
The above is the way we can compute the given values by using 123.

Compute the powers given below:

• (1.1)3

Answer: 1.331
Explanation: Given the value is (1.1)3
Now, we need to find the value.
(1.1)3 = (11/10)3
(11/10)3 = (11 x 11 x 11)/ (10 x 10 x 10)
= 1331/ 1000
(11/10)3 = 1.331

• (0.02)5

Answer: 0.0000000032
Explanation: As given in the question, the value is (0.02)5
Now, we will find the final value.
(0.02)5 = (2/100)5
(2/100)5 = (2 x 2 x 2 x 2 x 2)/ (100 x 100 x 100 x 100 x 100)
= 32/10000000000
(0.02)5 = 0.0000000032

• (0.1)6

Answer: 0.000001
Explanation: Given the value is(0.1)6
We will find the final value.
(0.1)6 = (1/10)6
(1/10)6 = (1 x 1 x 1 x 1 x 1 x 1)/(10 x 10 x 10 x 10 x 10 x 10)
= 1/1000000
(0.1)6 = 0.000001

Given that 163 = 4096, can you compute these powers?

• (1.6)3

Answer: 4.096
Explanation: As given in the question, the value is (1.6)3
Now, we will find the final value.
(1.6)3 = (16/10)3
(16/10)3 = (16 x 16 x 16)/ (10 x 10 x 10)
= 4096/1000
(1.6)3 = 4.096

• (0.16)3

Answer: 0.004096
Explanation: As given in the question, the value is (0.16)3
Now, we will find the final value.
(0.16)3 = (16/100)3
(0.16)3 = (16  x 16 x16)/ (100 x 100 x 100)
= 4096/1000000
(0.16)3= 0.004096

• (0.016)3

Answer: 0.000004096
Explanation: Given in the question, the value is (0.016)3
Now, we will find the final value.
(0.016)3 = (16/1000)3
(16/1000)3 = (16 x 16 x 16)/ (1000 x 1000 x 1000)
= 316/1000000000
(0.016)3 = 0.000004096

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

Multiplication rule

We know how to write the sum of two multiples of a number as a single multiple of the same number
(3 × 2) + (5 × 2) = (3 + 5) × 2 = 8 × 2
Why is this true?
3 × 2 = 2 + 2 + 2
5 × 2 = 2 + 2 + 2 + 2 + 2
And so
(3 × 2) + (5 × 2) = (2 + 2 + 2) + (2 + 2 + 2 + 2 + 2)
= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2
= 8 × 2
Similarly, we can find the product of powers.
For example, 23 × 25
23 = 2 × 2 × 2
23 = 2 × 2 × 2 × 2 × 2
Then,
23 × 25 = (2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 28
Suppose we multiply the third and fifth powers of some other number.
(\(\frac{2}{3}\))3 × (\(\frac{2}{3}\))5 = (\(\frac{2}{3}\) × \(\frac{2}{3}\) × \(\frac{2}{3}\)) × (\(\frac{2}{3}\) × \(\frac{2}{3}\) × \(\frac{2}{3}\) × \(\frac{2}{3}\) × \(\frac{2}{3}\))
= (\(\frac{2}{3}\))8
What if we indicate the number as
x3 × x5 = (x × x × x) × (x × x × x × x × x)
x × x × x × x × x × x × x × x = x8

Now suppose we take some other powers:
x2 × x4 = (x × x) × (x × x × x × x)
= x × x × x × x × x × x
= x6
Let’s write the exponents also as m, n.
Then in general,
Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication 4
What is the general principle here?
xm × xn = xm + n, for all numbers x and all natural numbers m, n
How do we say this in ordinary language?
What is the general principle here?
There are two facts here

  1. The product of two powers of the same number is a power of this number.
  2. The exponent of the product is the sum of the exponents of the factors.

Try the problems given below, using this idea.

  • What power of 2 is got by multiplying 25 and 23?

Answer: 8th power of 2.
Explanation: Given the powers are 25 and 23
Now, we will find the total power value.
The given values bases are equal. So, the powers will be added.
25 and 23 = 25+3
= 28 i.e., 8th power of 2.

  • What is the number 102 × 105 in ordinary language?

Answer: 7th power of 10
Explanation: Given the powers are 102 and 105
Now, we will find the total power value.
The given values bases are equal. So, the powers will be added.
102 and 105 = 102+5
= 107 i.e., 7th power of 10.

  • What power of 2 is twice 210?

Answer: 11th power of 2 (211)
Explanation: Given the powers are 210 and 21
Now, we will find the total power value.
The given values bases are equal. So, the powers will be added.
210 and 21 = 210+1
= 211 i.e.,11th power of 2.

  • What must be added to 210 to get 211?

Answer: 21
Explanation: As given in the question, the powers are 210 and the power value is 211
Now, we will find out the other power value.
Actually, the given values bases are equal. Then, the powers will be added.
But, now we find another power value. So, subtract it.
210 and 211= 211-1
21 i.e., 1st power of 2.

  • What must be added to 310 to get 311?

Answer: 31
Explanation: As given in the question, the powers are 310 and the power value is 211
Now, we will find out the other power value.
Actually, the given values bases are equal. Then, the powers will be added.
But, now we find another power value. So, subtract it.
310 and 311= 311-10
= 31 i.e., 1st power of 3.

  • The table below gives some powers of 2.

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication 5

Using this, compute these products.

• 16 × 64
Answer: 1024
Explanation: As given in the question, the values are 16 and 64.
Now, write in powers based on the table value.
The power of 16 is 24 and the power value of 64 is 26
Now, we will find out the other power value.
Actually, the given values bases are equal. Then, the powers will be added.
24and 26= 210
The value of 210 is 1024.

• 64 × 256
Answer: 16384
Explanation: As given in the question, the values are 64 and 256.
Now, write in powers based on the table value.
The power of 64 is 26 and the power value of 28
Now, we will find out the other power value.
Actually, the given values bases are equal. Then, the powers will be added.
26 and 28= 214
The value of 214 is 16384.

• 32 × 512
Answer: 16384
Explanation: As given in the question, the values are 32 and 512.
Now, write in powers based on the table value.
The power of 32 is 25 and the power value of 512 is 29
Now, we will find out the other power value.
Actually, the given values bases are equal. Then, the powers will be added.
25 and 29= 214
The value of 214 is 16384.

• 128 × 256
Answer: 32768
Explanation: As given in the question, the values are 128 and 256.
Now, write in powers based on the table value.
The power of 128 is 27 and the power value of 256 is 28
Now, we will find out the other power value.
Actually, the given values bases are equal. Then, the powers will be added.
27 and 28= 215
The value of 215 is 32768.

Division rule Textbook Page No. 59

Is there any trick to computing the quotient of two powers of the same number, as for a product?
For example, what is 45 ÷ 42?
Using the product rule,
45 = 42 × 43
So what do we get on dividing 45 by 42?
45 ÷ 42 = 42
Likewise, how do we compute 57 ÷ 53?
How do we write 57 as a multiple of 53?
57 = 53 × ______
From this we get,
57 ÷ 53 = ______

Answer: 54
Explanation: Given the values 57 and 53
Now, we will find the value of the multiple.
The bases are equal. So, the powers are added.
But, we will find multiple values. So, subtract the value of the power.
i.e., 57-3 = 54
So, the value is 57 ÷ 53 = 54

What about 823 ÷ 816?
To get 823, by what should we multiply 816?
For this, we need only find what should be added to 16 to get 23.
23 – 16 = 7
So
823 = 816 × 87
Now, can’t you find 823 ÷ 816?
This we can do for powers of fractions also.
For example, let’s divide (\(\frac{2}{3}\))16 by (\(\frac{2}{3}\))9.
As before, if we write
(\(\frac{2}{3}\))16 = (\(\frac{2}{3}\))9 × (\(\frac{2}{3}\))7

Then we can find
(\(\frac{2}{3}\))16 ÷ (\(\frac{2}{3}\))9 = (\(\frac{2}{3}\))7
Now let’s see in general what we get when we divide a power of a number by a smaller power.
Let’s write the number as x. Since the operation is a division, x should not be zero. Let’s write the larger exponent, as m and the smaller exponent as n.
Now how do we compute xm ÷ xn?
What should be added to n to make it m?
So,
xm = xn × xm – n
From this we get
\(\frac{x^{m}}{x^{n}}\) = xm – n
That is,
\(\frac{x^{m}}{x^{n}}\) = xm – n, for any non zero x and any natural numbers m, n with m > n

As in the case of multiplication, can you say this in ordinary language?
Answer: yes, in the case of multiplication. This is an ordinary language.

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

Now try these problems.

• What power of 2 do we get on dividing 25 by 23?

Answer: 22 i.e., 2nd power of 2.
Explanation: Given the values 25 and 23
Now, we will find the value.
The bases are equal. So, the powers are added.
But, we will find multiple values. So, subtract the value of the power.
i.e., 25-3 = 22
So, the value is 22

• What is 109 ÷ 104?

Answer: 105
Explanation: Given the values 109 and 104
Now, we will find the multiple values.
The bases are equal. So, the powers are added.
But, we will find multiple values. So, subtract the value of the power.
i.e., 109-4 = 105
So, the value is 109 ÷ 104 = 105

• What power of 2 is half of 210?

Answer: 29
Explanation: Given the values half of 210  i.e., the values are 210 and 21
Now, we will find the final value.
The bases are equal. So, the powers are added.
But, we will find multiple values. So, subtract the value of the power.
i.e., 210-1 = 29
So, the value is 29

• Look at the table of powers of 2 on page 58. Using it, can you compute these quotients?

• 64 ÷ 16

Answer: 22 = 4.
Explanation: Given the values 64 and 16.
Now, we will find the quotient value.
Based on the table of powers of 2. We can rewrite the given values in 2 powers.
The values are 26 and 24
While the bases are equal. So, the powers are added.
But, we will find the quotient value. So, we can subtract the power values.
i.e., 26-4 = 22
So, the quotient value is 26 ÷ 24 = 22

• 1024 ÷ 128

Answer: 23 = 8.
Explanation: Given the values 1024 and 128
Now, we will find the quotient value.
Based on the table of powers of 2. We can rewrite the given values in 2 powers.
The values are 210 and 27
While the bases are equal. So, the powers are added.
But, we will find the quotient value. So, we can subtract the power values.
i.e., 210-7 = 23
So, the quotient value is 210 ÷ 27= 23

• 16384 ÷ 2048

Answer: 23 = 8.
Explanation: Given the values 16384 and 2048.
Now, we will find the quotient value.
Based on the table of powers of 2. We can write the given values in 2 powers.
The values are 214 and 211
While the bases are equal. So, the powers are added.
But, we will find the quotient value. So, we can subtract the power values.
i.e., 214-11 = 23
So, the quotient value is 214 ÷ 211 = 23

• What is 28 × \(\frac{1}{2^{3}}\) ?

Answer: 25
Explanation: Given the values 28and 1/23.
Now, we will find the final value.
The values are 28 and 1/23
28 x \(\frac{1}{2^{3}}\) =28/23
While the bases are equal. So, the powers are added.
But, we will find the final value. So, we can subtract the power values.
i.e., 28-3 = 25
So, the quotient value is 28 x 1/23 = 25

• By what should 76 be multiplied to get 72?

Answer: 74
Explanation: Given the values 76 and 72
Now, we will find the multiplied value.
While the bases are equal. So, the powers are added.
But, we will find the multiplied value. So, we can subtract the power values.
i.e., 76-2 = 74
So, the 1/ 74 should be multiplied by 76 to get the value is 72 i.e., 76 x 1/ 74 = 72

Another division Textbook Page No. 61

Look at the last but one problem above: We have
28 × \(\frac{1}{2^{3}}\) = 28 ÷ 23 = 25
From this we get,
25 ÷ 28 = \(\frac{1}{2^{3}}\)
Likewise, from the last problem, can you find 72 ÷ 76?
We have
76 × \(\frac{1}{7^{4}}\) = 72
and from this we get
72 ÷ 76 = \(\frac{1}{7^{4}}\)
In general,
\(\frac{x^{m}}{x^{n}}\) = \(\frac{1}{x^{n-m}}\), for any non zero x and any natural numbers m, n with m < n

Now try these problems:

• Simplify

• \(\frac{2^{5} \times 2^{3}}{2^{4}}\)

Answer: 24
Explanation: Given the values (25x 23)/24.
Now, we will find the final value.
The values are 25 x 23 and 1/24
\(\frac{2^{5} \times 2^{3}}{2^{4}}\) = 28/24
While the bases are equal. So, the powers are added.
But, we will find the final value. So, we can subtract the power values.
i.e., 28-4 = 24
So, the final value is  \(\frac{2^{5} \times 2^{3}}{2^{4}}\) = 24

• \(\frac{3^{7}}{3^{2} \times 3^{4}}\)

Answer: 31
Explanation: Given the values 37/ (32 x 34)
Now, we will find the final value.
The values are 37 / 32 x 34
\(\frac{3^{7}}{3^{2} \times 3^{4}}\) = 37/36
While the bases are equal. So, the powers are added.
But, we will find the final value. So, we can subtract the power values.
i.e., 37-6 = 31
Hence, the final value is  \(\frac{3^{7}}{3^{2} \times 3^{4}}\) = 31

• \(\frac{5^{2} \times 5^{4}}{5^{5} \times 5^{4}}\)

Answer: 1/53
Explanation: Given the values \(\frac{5^{2} \times 5^{4}}{5^{5} \times 5^{4}}\)
Now, we will find the final value.
The values are  52 x 54 / 55 x 54
\(\frac{5^{2} \times 5^{4}}{5^{5} \times 5^{4}}\) = 56/59
While the bases are equal. So, the powers are added.
But, we will find the final value. So, we can subtract the power values.
i.e., 1/59-6 = 1/53
Hence, the final value is \(\frac{5^{2} \times 5^{4}}{5^{5} \times 5^{4}}\) = 1/53

• \(\frac{8^{2} \times 8^{7}}{8^{6} \times 8^{3}}\)

Answer: 1
Explanation: Given the values \(\frac{8^{2} \times 8^{7}}{8^{6} \times 8^{3}}\)
Now, we will find the final value.
The values are  82 x 87 / 86 x 83
\(\frac{8^{2} \times 8^{7}}{8^{6} \times 8^{3}}\) = 89/89
While the bases are equal. So, the powers are added.
But, we will find the final value. So, we can subtract the power values.
i.e., 89-9 = 80 = 1.
Hence, the final value is \(\frac{8^{2} \times 8^{7}}{8^{6} \times 8^{3}}\) = 1

• \(\frac{4^{3} \times 4^{5}}{4^{2} \times 4^{4}}\)

Answer: 42
Explanation: Given the values \(\frac{4^{3} \times 4^{5}}{4^{2} \times 4^{4}}\)
Now, we will find the final value.
The values are  43 x 45 / 42 x 44
\(\frac{4^{3} \times 4^{5}}{4^{2} \times 4^{4}}\) = 48/46
While the bases are equal. So, the powers are added.
But, we will find the final value. So, we can subtract the power values.
i.e., 48-6 = 42
Hence, the final value is \(\frac{4^{3} \times 4^{5}}{4^{2} \times 4^{4}}\) = 42

• \(\frac{10^{4} \times 10^{5}}{10^{6} \times 10^{7}}\)

Answer: 1/104
Explanation: Given the values \(\frac{10^{4} \times 10^{5}}{10^{6} \times 10^{7}}\)
Now, we will find the final value.
The values are  104 x 105 / 106x 107
\(\frac{10^{4} \times 10^{5}}{10^{6} \times 10^{7}}\) = 109/1013
While the bases are equal. So, the powers are added.
But, we will find the final value. So, we can subtract the power values.
i.e., 1/1013-9 = 1/104
Hence, the final value is \(\frac{10^{4} \times 10^{5}}{10^{6} \times 10^{7}}\) = 1/104

• What powers of \(\frac{1}{5}\) do we get on dividing 56 by 510?

Answer: (1/5)4
Explanation: Given the values are 56 and 510
Now, we will find the \(\frac{1}{5}\) power value.
The values are  56 / 510 
While the bases are equal. So, the powers are added.
But, we will find the value in \(\frac{1}{5}\). So, we can subtract the power values.
i.e., 1/510-6 = 1/54
Hence, the final value in \(\frac{1}{5}\) is 1/54

• What is the decimal form of the number got, on dividing 108 by 1012?

Answer: 0.0001
Explanation: Given the values are 108 and 1012
Now, we will find the value in decimal form.
The values are  108 / 1012
While the bases are equal. So, the powers are added.
But, we will find the value in decimal form. So, we can subtract the power values.
i.e., 1/1012-8 = 1/104 = 1/10000 = 0.0001
Hence, the final value in decimal form is 0.0001.

• What natural number is got by dividing (\(\frac{1}{2}\))5 by (\(\frac{1}{2}\))8?

Answer: 8
Explanation: Given the values (\(\frac{1}{2}\))5 by (\(\frac{1}{2}\))8
Now, we will find the natural number.
The values are (\(\frac{1}{2}\))5/(\(\frac{1}{2}\))8
(\(\frac{1}{2}\))5/(\(\frac{1}{2}\))8 = 1/(1/2)8-5
While the bases are equal. So, the powers are added.
But, we will find the final value. So, we can subtract the power values.
i.e., 1/(1/2)8-5 = 1/(1/2)3 = 1/(1/8) = (8/1) = 8
Hence, the final value is (\(\frac{1}{2}\))5/(\(\frac{1}{2}\))8= 8.

• By what natural number should (0.25)6 be multiplied to get (0.025)4?

Answer: 15
Explanation: As given in the question, the values (0.25)6 x (0.025)4
Now, we will find the natural number value.
The values are (0.25) x 6 and (0.025) x 4.
i.e., 0.25 x 6 = 1.5
0.025 x 4 = 0.1
So, 0.1 is multiplied by 15. We, get the value of 1.5
i.e., 15 x 0.1 = 1.5
Therefore, the natural number value is 15.

• Make a table of powers 3, and use it to find these products and quotients.

• 81 × 9

Answer: 729
Explanation: As given in the question, the values 81 x 9
Now, we will find the product value. Using the table of powers 3.
The value 81 in 3 powers is 34 and the value 9 in 3 powers is 32
The 3 power values bases are equal. But power values are different.
So, we can add the powers.
34 x 32 = 34+2 = 36
Therefore, the product value is 36 = 729

• 729 × 81

Answer: 59049
Explanation: As given in the question, the value is 729 x 81
Now, we will find the product value. Using the table of powers 3.
The value 729 in 3 powers is 36 and the value 81 in 3 powers is 34
The 3 power values bases are equal. But power values are different.
So, we can add the powers.
36 x 34 = 36+4 = 310
Hence, the product value is 310 = 59049

• 6561 ÷ 243

Answer: 27
Explanation: As given in the question, the values 6561 ÷ 243
Now, we will find the product value. Using the table of powers 3.
The value 6561 in 3 powers is 38 and the value 243 in 3 powers is 35
The 3 power values bases are equal. But power values are different.
So, we can add the powers. But in this, we will subtract the powers because of finding the quotient value.
38÷ 35 = 38-5 = 33
Hence, the quotient value is 33 = 27

• 243 × 81

Answer: 19683
Explanation: As given in the question, the value is 243 x 81
Now, we will find the product value. Using the table of powers 3.
The value 243 in 3 powers is 35 and the value 81 in 3 powers is 34
The 3 power values bases are equal. But power values are different.
So, we can add the powers.
35 x 34 = 35+4 = 39
Hence, the product value is 39 = 19683

• 2187 ÷ 9

Answer: 243
Explanation: As given in the question, the value is 2187 ÷ 9
Now, we will find the product value. Using the table of powers 3.
The value 2187 in 3 powers is 37 and the value 9 in 3 powers is 32
The 3 power values bases are equal. But power values are different.
So, we can add the powers. But in this, we will subtract the powers because of finding the quotient value.
37 ÷ 32 = 37-2 = 35
Therefore, the quotient value is 35 = 243.

• 59049 ÷ 729

Answer: 81
Explanation: As given in the question, the value is 59049 ÷ 729
Now, we will find the product value. Using the table of powers 3.
The value 59049 in 3 powers is 310 and the value 729 in 3 powers is 36
The 3 power values bases are equal. But power values are different.
So, we can add the powers. But in this, we will subtract the powers because of finding the quotient value.
310 ÷ 36 = 310-6 = 34
Hence, the quotient value is 34 = 81

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

Power of powers Textbook Page No. 62

Can we write 64 as the power of a number?
In what all ways can we do this?
26 = 64
43 = 64
82 = 64
641 = 64
Likewise, can we write 312 as the powers of other numbers?
312 = 36 × 36
= (729) × (729)
= 7292
There is another way:
312 = 38 × 34
= (34 × 34) × 34
= 81 × 81 × 81
= 813
And one more:
312 = 36 × 36
= (33 × 33) × (33 × 33)
= 27 × 27 × 27 × 27 × 27
= 274
Is there any other? Try!
In these, what does 36 × 36 mean?
It is the product of two 36, right?
We can shorten it as (36)2.
Thus
(36)2 = 36 × 36
= 36 + 6
= 36 × 2
= 312
Similarly, we can write 34 × 34 × 34 as (34)3 = 34 × 34 × 34
= 34 + 4 + 4
= 34 × 3
= 312

In the same way, we can write
(42)3 = 42 × 42 × 42
= 42 × 3
= 46
(54)6 = 54 × 6
= 524
and so on.
Now let’s look at such powers of fractions.
What does ((\(\frac{2}{3}\))2)3 mean?
(\(\frac{2}{3}\))2 + 2 + 2 =(\(\frac{2}{3}\))3 × 2 = (\(\frac{2}{3}\))6
In general, for any number x and for any natural numbers m, n
Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication 6

That is,
(xm)n = xmn, for any number x and any natural numbers, m, n.

Now can you write these as a single power?

• (42)3

Answer: 46
Explanation: As given in the question, the value is (42)3
Now, we will find the single power value.
Here, we have 2 powers for one base value.
So, we can multiple the powers.
Then the value is (42)3 = (4)2×3 = 46
Hence, the given value in single power is 46

• (33)2 × 94

Answer: 314
Explanation: As given in the question, the value is (33)2 × 94
Now, we will find the single power value.
First, write the 94 in 3 powers then we can find the value easily.
94 = (32)4
Then the given values are  (33)2 x (32)4
Now, we can multiple the powers.
So, the values are is  (3)3×2 x (3)2×4 = 36 x 38
The bases are equals. So, the powers are added.
i.e., 36+8 = 314
Hence, the given value in single power is 314

• ((\(\frac{1}{2}\))3)4

Answer: (1/2)12
Explanation: As given in the question, the value is ((\(\frac{1}{2}\))3)4
Now, we will find the single power value.
Here, we have 2 powers for one base value.
So, we can multiple the powers.
Then the value is (1/23)4 = (1/2)3×4 = (1/2)12
Hence, the given value in single power is (1/2)12

• (23)4 × 26

Answer: 218
Explanation: As given in the question, the value is (23)4 × 26
Now, we will find the single power value.
So, we can multiple the powers.
Then, the values are is  (2)3×4 x (2)6 = 212 x 26
The bases are equals. So, the powers are added.
i.e., 212+6 = 218
Hence, the given value in single power is 218

Write each of these as powers of different numbers.

• 38
Answer: (32)4, (34)2
Explanation: As given in the question, the value is 38
Now, we will write the powers of different numbers.
First, write the multiples of power.
So the value is 38 = (32)4, (34)2
(i) (32)4 = (3 x 3)4 = (9)4
(ii) (34)2 = (3 x 3 x 3 x 3)2 = (81)4
Then, the given value different power numbers are (32)4, (34)2

• 46
Answer: 642
Explanation: Given in the question, the value is 46
Now, we will write the powers of different numbers.
First, write the multiples of power.
So the value is 46 = (43)2
(43)2 = (4 x 4 x 4)2 = (64)2
Therefore, the given value different power numbers are (64)2

• 215
Answer: (32)3, (8)5
Explanation: As given in the question, the value is 215
Now, we will write the powers of different numbers.
First, write the multiples of power.
So the value is 215 = (25)3, (2)3
(i) (25)3 = (2 x 2 x 2 x 2 x 2)3 = (32)3
(ii) (2)3 = (2 x 2 x 2)3 = (8)3
Then, the given value different power numbers are (32)3, (8)3

• 512

Answer: (25)6, (15625)5, (125)4 , (625)3
Explanation: As given in the question, the value is 512
Now, we will write the powers of different numbers.
First, write the multiples of power.
So the value is 512 = (52)6, (56)2,  (53)4,  (54)3
(i)  (52)6 = (5 x 5)6 =(25)6
(ii) (56)2 = (5 x 5 x 5 x 5 x 5 x 5)2 = (15625)5
(iii) (53)4 = (5 x 5 x 5)4 = (125)4
(iv) (54)3 = (5 x 5 x 5 x 5)3 = (625)3
Then, the given value different power numbers are (25)6, (15625)5, (125)4, (625)3

Factors

What are the factors of 32?
1, 2, 4, 8, 16, 32
Except 1, all these are powers of 2. So we can write the factors of 32 as
1, 2, 22, 23, 24, 25
What about the factors of 81?
81 = 34
And so the factors are
1, 3, 32, 33, 34
Now let’s find out the factors of 72.
72 = 23 × 32
Let’s look at a definite scheme to write all factors.
First, we write 1 and all factors which are the power of 2:
1, 2, 22, 23
Multiplying these by 3 gives four more factors:
3, 2 × 3, 22 × 3, 23 × 3
Multiplying the first four factors by 32 instead, we get another four factors:
32, 2 × 32, 22 × 32, 23 × 32

Are there any more?
How about writing down the factors of 200?
200 = 8 × 25 = 23 × 52
We can list the factors in order as below.
Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication 9
What about the factors of240?
240 = 16 × 15 = 24 × 3 × 5
We can list the factors like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication 10
Now list the factors of the numbers below like this

• 64
Answer: The given number is 64.
Now, we will find the factors of the given number.
First, find the LCM of the 64. Then we can easily find the factors.
In powers value, the LCM of 64 is 26
So, the factors of 64 are 26,25, 24, 23, 22, 21, 1

• 125
Answer: The given number is 125.
Now, we will find the factors of the given number.
First, find the LCM of the 125. Then we can easily find the factors.
In powers value, the LCM of 125 is 53
So, the factors of 125 are  53, 52, 51,1

• 48
Answer: The given number is 48.
Now, we will find the factors of the given number.
First, find the LCM of the 48. Then we can easily find the factors.
In powers value, the LCM of 48 is 24 x 3.
So, the factors of 48 are  24, 23, 22, 21, 3, i.e., 3 x 24, 3 x 23, 3 x 22, 3 x 21,1

• 45
Answer: The given number is 45.
Now, we will find the factors of the given number.
First, find the LCM of the 45. Then we can easily find the factors.
In powers value, the LCM of 45 is 32x 5.
So, the factors of 45 are 32, 5, 1, 3, 5 x 32, 5 x 3

• 105
Answer: The given number is 105.
Now, we will find the factors of the given number.
First, find the LCM of the 105 Then we can easily find the factors.
In powers value, the LCM of 105 is 3 x 5 x 7
So, the factors of 105 are 1, 3, 5, 7, 3 x 5, 3 x 7, 5 x 7 , 3 x 5 x 7.

Kerala Syllabus 7th Standard Maths Solutions Chapter 4 Repeated Multiplication

Let’s do it! Textbook Page No. 65

Question 1.
If 2x = 128, what is 2x + 1?

Answer: 256
Explanation: Given the value is 2x = 128.
Now, we will find the value of the 2x + 1
So, rewriting the value 2x + 1 is,
2x + 1 = 2x x 21
Place the 2x=128 value in the above expression. We get,
2x x 21 = 128 x 2 = 256.

Question 2.
If 3x = 729, what is 3x – 1?

Answer: 243
Explanation: Given the value is 3x = 729
Now, we will find the value of the 3x – 1
So, rewriting the value 3x – 1 is,
3x -1 = 3x x 1/31
Place the 3x=729 value in the above expression. We get,
3x x 1/31 = 729 x 1/3 = 243.

Question 3.
In 3x, 3x + 1, 3x – 1, 3x + 1 which is an even number?

Answer: 3x + 1 is a even number.
Explanation: Given the values are 3x, 3x + 1, 3x – 1, 3x + 1.
Now, we will find the even number.
Let us write the values of 3 powers.
31 = 3, 32 = 9, 33 = 27, 34 = 81.
Now, check the given values with powers.
3x = 31 = 3.  It is an odd number.
3x+1 = 31+1 = 32= 9.  It is an odd number.
3x-1 = 31-1 = 30= 1 .  It is an odd number.
3x+1 = 31+1 = 3+ 1 = 4 .  It is an even number.
So, 3x+1 is an even number.

Question 4.
If 610 is computed, what would be the digit in one’s place?

Answer: The digit in one’s place is 6.
Explanation: Given the value is 610
Now, we will find the value of 610
i.e., 6 x 6 x 6 x 6 x 6 x 6 x 6 x 6 x 6 x 6 = 60466176.
Hence, the digit in one’s place is 6.

Question 5.
If 56 × \(\frac{1}{5^{x}}\) = \(\frac{1}{5^{10}}\), then what is x?

Answer: 16
Explanation: Given the value is 56 × \(\frac{1}{5^{x}}\) = \(\frac{1}{5^{10}}\)
Now, we will find the value of the x.
The value is 56 × \(\frac{1}{5^{x}}\) = \(\frac{1}{5^{10}}\),
56/ 5x = 1/510
1/ 5x-6 = 1/510
x-6 = 10
x = 10+6 = 16.

Question 6.
Simplify
(i) \(\frac{3^{5} \times 3^{6}}{3^{4} \times 3^{4}}\)

Answer: 33
Explanation: Given the values \(\frac{3^{5} \times 3^{6}}{3^{4} \times 3^{4}}\)
Now, we will find the value.
The values are \(\frac{3^{5} \times 3^{6}}{3^{4} \times 3^{4}}\)
While the bases are equal. So, the powers are added.
\(\frac{3^{5} \times 3^{6}}{3^{4} \times 3^{4}}\)= 311/38
But, we will find the final value. So, we can subtract the power values.
i.e., (3)11-8 = 3
Hence, the final value \(\frac{3^{5} \times 3^{6}}{3^{4} \times 3^{4}}\) is 33.

(ii) \(\frac{4^{7} \times 4^{8}}{4^{2} \times\left(4^{3}\right)^{5}}\)

Answer: 1/42
Explanation: Given the value is \(\frac{4^{7} \times 4^{8}}{4^{2} \times\left(4^{3}\right)^{5}}\)
Now, we will find the value.
The values are \(\frac{4^{7} \times 4^{8}}{4^{2} \times\left(4^{3}\right)^{5}}\)
While the bases are equal. So, the powers are added.
\(\frac{4^{7} \times 4^{8}}{4^{2} \times\left(4^{3}\right)^{5}}\) = 415/42 x 43 x 5
= 415/42 x 4 15
The bases are equal. So, the powers are added.
So, the value is 415/4 17
But, we will find the final value. So, we can subtract the power values.
i.e., 1/(4)17-15 = 1/(4)
Hence, the final value is \(\frac{4^{7} \times 4^{8}}{4^{2} \times\left(4^{3}\right)^{5}}\) = 1/42

(iii) \(\frac{\left(6^{4}\right)^{2} \times\left(6^{5}\right)^{3}}{\left(6^{2}\right)^{2} \times\left(6^{4}\right)^{5}}\)

Answer: 1/6
Explanation: As given in the question, the value is \(\frac{\left(6^{4}\right)^{2} \times\left(6^{5}\right)^{3}}{\left(6^{2}\right)^{2} \times\left(6^{4}\right)^{5}}\)
Now, we will find the final value.
The value is \(\frac{\left(6^{4}\right)^{2} \times\left(6^{5}\right)^{3}}{\left(6^{2}\right)^{2}
Here, 1 base has 2 powers. So, we can multiple the powers.
i.e., [latex]\frac{\left(6^{4}\right)^{2} \times\left(6^{5}\right)^{3}}{\left(6^{2}\right)^{2} \times\left(6^{4}\right)^{5}}\) =  68 x 615/ 64 x 620
While the bases are equal. So, the powers are added.
68 x 615/ 64 x 620 = 623/ 624
But, we will find the final value. So, we can subtract the power values.
i.e., 1/(6)24-23 = 1/(6)1 = 1/6
Therefore, the final value is 1/6.

Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles

You can Download Drawing Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 7th Standard Maths Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles

Drawing Triangles Text Book Questions and Answers

Circle and triangle Textbook Page No. 104
Euclid, who lived in Greece during the third century BC, is considered the Master of geometry. His book, “The Elements” is the first authoritative text on geometry.

The first result on this book is the construction of an equilateral triangle of specified length for sides. This is Euclid’s method:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 1
This figure consisting of two intersecting circles is often used as a motif in churches of medieval Europe.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 2

Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles

New figures

In the figure used for drawing an equilateral triangle, two triangles can be drawn, one up and the other down.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 8
Suppose we erase the middle line:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 9
What are the peculiarities of this quadrilateral?
Answer:
This quadrilateral represents a Rhombus.
In a Rhombus, opposite sides are parallel and opposite angles are equal.
What if we draw three circles, instead of two?
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 10
Joining the points of intersections and the centres of two circles, we get a figure like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 11
What is its name?
What can you say about the lengths of the sides?
Answer:
The figure has 6 sides, which represents a Hexagon.
It has 6 equal sides and angles.
It is made up of 6 equilateral triangles. Therefore, all the sides and angles are equal.

Inside a circle Textbook Page No. 106

You know how to divide a circle into six equal parts, using a comer of one of the set squares in the geometry box.
Joining the ends of these lines, we get a figure like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 12
Joining the ends of these lines, we get a figure like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 13
What if we join three alternate points?
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 14
If we join the points left also, we get a star like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 15

Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles

Sides and angles

Draw a triangle of sides 6 centimetres. 3 centimetres and the angle between them 60°.
Measure the other two angles.
Now draw a triangle of sides 8 centimetres and 4 centimetres (with angle between them 60° as before). Do the other angles change?
Answer:
Let us first draw a rough figure for the given measurements.
Steps for construction:
Step 1: Assume a point A and draw an angle of 60º.
Step 2: Draw base line from A at a distance of 6cm and mark the end as B.
Step 3: Now, locate C at 3cm distance from A on the above line and then join B and C.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_1
Other angles:
To find out the other angles, place the protractor on point B and measure the angle. Similarly, place the protractor on point C and measure the angle.
The approximate measured angles will be ∠B=26º and ∠C=94º
Answer:
Construction of second traingle:
Given that two sides of triangle are 8 cm and 4 cm, the angle between them is 60º.
Steps for construction:
Step 1: Assume a point A and draw an angle of 60º.
Step 2: Draw base line from A at a distance of 8cm and mark the end as B.
Step 3: Now, locate C at 4cm distance from A on the above line and then join B and C.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_2
Other angles:
To find out the other angles, place the protractor on point B and measure the angle. Similarly, place the protractor on point C and measure the angle.
The approximate measured angles will be ∠B=30º and ∠C=90º.

What is the relation between the sides here?
Let’s draw such triangles using GeoGebra. For this, first create a slider a with Min = O and Max = 10. Using the Segment with Given Length tool, draw line AB of length 2a. Use the Angle with given size to draw a line AB’ slanted at 60° through A. Next choose the Circle with Centre and Radius tool and click on A. In the dialog box, give a as the radius. Mark the point where the circle cuts the slanted line as C.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 22
Now we can hide the lines, angles and circle in the figure. Using the Polygon tool, draw triangle ABC. Select the Distance or Length tool and click on the sides of the triangle to get their lengths; select Angle tool and click within
the triangle to get the angles.
Use slider to change the value of a.
How do the sides change?
Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_37
The length of AB is twice of AC.
What about the angles?
Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_38
The angle will remain same even if the lengths are increased.

Angles and sides Textbook Page No. 108

Draw a line of 6 centimetres and draw another line slanted at 30° at one end. With the other end as centre, draw some circles of different radii:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 23
What is the minimum radius for which the circle meets the upper line?
Answer:
The minimum radius for which the circle will meet upper line will be 3.3cm.

For what all radii do the circle cut the upper line at two points?
Answer:
The circle with radii 3.3cm and 3.4cm will cut the upper line at two points.

From the given figure, we can observe that the circle with
We want to draw ΔABC with AB = 6 cm and ∠B = 30°. What should be the minimum length of AC?
Answer:
Inorder to form a triangle, the minimum lenth of AC should be 3.4cm.

For what all lengths of AC do we get two such triangles?
Answer:
For 3.3cm and 3.4cm of lengths of AC, we will get two triangles.

Let’s do this using GeoGebra. Draw line AB of length 6 units and line AB’ such that ∠BAB’ = 30°. Make a slider a and use the Circle with Centre and Radius tool to draw a circle centred at B, of radius a. Change the value of a using the slider. For what all values of a does the circles meet AB’?
Answer:
As shown in figure below, the minimum value of a which intersect AB’ will be 3.3cm. The circle will intersect  line AB’ for all the values ranging inbetween 3.3 and 6cm.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_25

Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles

Parallel triangles 

At the two ends of a line, draw angles of 70° and 80° to make a triangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 28
What is its third angle?
Answer:
Sum of angles in a triangle = 180º
Given that two angles are 70º and 80º
Let us assume the third angle as xº
70º + 80º + xº = 180º
150º + xº = 180º
xº = 180º-150º
xº = 30º
Therefore, the third angle will be 50º
Draw lines parallel to the sides of the triangle and make another triangle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 29
Measure the angles of this triangle. Draw other triangle like this and check. Do the angles change?
Answer:
For the above figure, given that two angles of a triangle are 70º and 80º.
After calculating, the third angle measurement is 30º.

An another triangle is drawn by joining the parallel lines of the first triangle.
Answer:
If the sides of two triangles are parallel, then the angles of the new triangle will be the same as the first whereas the length of the sides differ.
Hence, the angles will not change and remain same. Measurements of second triangle will also be 70º, 80ºand 30º.

Let’s do this using GeoGebra. Make slider a with Min = 0 and Max = 2. Draw a triangle using the Polygon tool. Mark a point D within it. Select the tool Dilate Object from Point by Factor and click within the triangle and on
D. In the dialog box give the value of Factor as a and give OK.

Use the slider to change a. Using the Angle tool, the angles of both triangles can be displayed.
Change the position of D to one of the vertices and see what you get.
Answer:
Even if the slider a is changed, the angles reamins the same.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_42

Unchanging relations Textbook Page No. 110

Can we draw a triangle with AB = 6 and AC = 2 BC? Let’s draw such triangle with GeoGebra. Draw line AB of the length 6. Make a slider a with suitable Min and Max values. Draw the circle of radius a centred at B and the circle of radius 2a centred at A. Mark the points C, D where these circles cut each other. Draw the lines AC and BC. Now we can hide the circles. Change the value of a using the slider. (We can also right click on the slider and select Animate). Right click on C and in the dropdown menu, check the Trace on option. What is the path of the point C? Draw the lines AD, BD. Draw the path of D also. This path can be seen more clearly by making
the change smaller. (Right click on the slider and choose Properties → Slider → Increment)
Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_26      Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_27

Draw triangles with the relation AC = 3BC or 2AC = 3AC, instead of AC = 2BC. In all these, what are the paths of C and D?
If AC = 3BC
Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_28  Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_30
If 2AC = 3BC
Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_31 Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_32
If AC = 2BC
Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_33 Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_34
What if AC = BC?
Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_35 Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_36
It will form an isosceles traingle.

Right path

Look at this picture:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 34
We can go from A to C, straight along the line AC; or we can first go to B along AB and then to C along BC. Which path is shorter?
Answer:
The line segment having less measurement than the other two will be the shorter path.

Do you see any relation connecting the sides of a triangle from this?
Answer:
Yes, two lines sharing a common endpoint form an angle. Thus it forms three angles ∠A ∠B and ∠C.

Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles

Sticky math

Take two thin sticks of the same length. Break one of them into two pieces.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 35
Can we make a triangle with these three sticks? Now break off a small bit from the long stick.
Answer:
Case 1:
Let us suppose take two sticks of same length 6cm.Given that one of them is broken into two pieces such as the two sticks will be 3cm each.
Steps of construction:
Step 1: Draw a line segment AB of length 6cm.
Step 2: Draw a circle of radius 3cm each from point A and B respectively.
Step 3: Mark point C at the point of intersection above the base line to form the required traingle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_3

Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 36
Now can you make a triangle?
Answer:
Case 2:
In an another scenario, let us assume that 1cm of long stick is broken.
Now the new measurements will be 5cm, 3cm and 3cm.
Steps of construction:
Step 1: Draw a line segment AB of length 5cm.
Step 2: Draw a circle of radius 3cm each from point A and B respectively.
Step 3: Mark point C at the point of intersection above the base line to form the required traingle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_4

Unchanging perimeter Textbook Page No. 112

Can you draw a triangle of perimeter 15 centimetres? Let’s see how we do it in GeoGebra. We first make two sliders to control the length of sides. Make sliders a and b with Min = 0 and Max = 7.5. Use the Segment with Given Length tool to draw the line AB of length a. Then what should be the sum of the lengths of the other two sides?
The perimeter should be 15. So,
AC + BC = 15 – AB = 15 – a

If the length of one of these sides is b, then what must be the length of the other? We use this to draw the other two sides. With A as centre, draw a circle of radius b, and with B as centre draw a circle of radius 15 – a – b. Mark the points C, D where the circles cut each other. Draw triangle ABC, using the Polygon tool. Choose the Distance or Length tool and click within the triangle to get its perimeter. Change the value of a and b using the sliders. Don’t we get different triangles of same perimeter?

Let’s see how we can get a nice picture with this set up. Draw the lines AD and BD also. Give the Trace On option for the lines AC, BC, AD, BD and the points C, D. Fix some convenient value for a and animate the slider for b. Look at the picture got. What are the path of C and D?
Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_39

Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles

Unchanging angles

Can you draw triangle ABC with AB = 5 and ∠C = 60°? Let’s do it with GeoGebra.
Draw line AB of length 5. Make an Angle slider a. Choose the Angles with Given Size tool and click first on B and then on A. In the pop-up window, give a as the size of the angle. Now we get a point B’ with ∠BAB’ = a. With the same tool, click first on A and then on B; in the pop-up window, give the size of the angle as 120 – a and select the clockwise option also. We now get another point A’. Join AB’ and BA’ using the Ray Through Two Points tool. Mark the point C where these lines intersect. Draw triangle ABC, using the Polygon tool. Now we can hide all lines and points other than those of the triangle. We can see the angles of the triangle by choosing the Angle tool and clicking within the triangle, change the value of a and see what happens. Choose Trace on for lines AC, BC
and the point C and animate the slider. What is the path of C?
Do this with other angles instead of 60° at C.
We can also use a slider to change the angle at C.
Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_40
Answer:
The path of C is as shown in figure below.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_41

Triangle in a rectangle Textbook Page No. 104

Remember how we drew rectangles, using a set square? Draw rectangle ABCD with AB = 5 cm and BC = 4 cm. Let’s join a pair of opposite sides.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 3

Now we have two right triangles.
Can you name them? What are the lengths of the perpendicular sides of each?
The two right triangles are BAD and BCD.
Length of perpendicular sides are 5 cm and 4 cm.
Now let’s draw a right triangle with perpendicular sides of length 6 centimetres and 8 centimetres.
First draw a pair of perpendicular lines and name the point where they meet as A.
Mark B on one of the lines, 6 centimetres from A and mark C on the other line, 8 centimetres from A.
Now we need only join B and C to get the triangle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 4
Measure the length of BC and write it next to BC. Similarly draw a right triangle with lengths of perpendicular sides 5 centimetres and 7 centimetres.
Answer:
Given: Two sides of an perpendicular triangle are 5cm and 7cm.
Construction steps:
Step 1: Draw a pair of perpendicular lines and mark the intersecting point as A.
Step 2: Next, mark B and C at a distance of 7cm and 5cm respectively from point A. Join BC to form the required traingle.

Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_43

Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles

Another triangle

In both the triangles we drew now, the length of two sides were specified; and angle between them was right. If the angle between is not right, how do we draw a triangle?
Let’s draw triangle ABC, with AB = 6 cm, AC = 8 cm and ∠A = 70°.
First draw an angle of 70°.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 5
Next mark B on one line, 6 centimetres from A and C on the other line, 8 centimetres from A.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 6
If we join B and C, we have the required triangle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 7

Now draw triangles of these specifications:

• MN = 6 centimetres, ∠M = 70°, ML = 5 centimetres.
Answer:
Given that two sides of triangle are 6 cm and 5 cm, the angle between them is 70º.
Steps for construction:
Step 1: Assume a point M and draw an angle of 70º.
Step 2: Draw base line from M at a distance of 6cm and mark the end as N.
Step 3: Now, locate L at 5cm distance from M on the above line and then join N and L.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_5

• PQ = 7 centimetres, QR = 7 centimetres, ∠Q = 50°.
Answer:
Given that two sides of triangle are 7 cm each, the angle between them is 50º.
Steps for construction:
Step 1: Assume a point Q and draw an angle of 50º.
Step 2: Draw base line from Q at a distance of 7cm and mark the end point as P.
Step 3: Now, locate R at 7cm distance from Q on the above line and then join P and R.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_6

• XY = 6.5 centimetres, ∠Y = 110°, YZ = 7.5 centimetres.
Answer:
Given that two sides of triangle are 7.5 cm and 6.5 cm, the angle between them is 110º.
Steps for construction:
Step 1: Assume a point Y and draw an angle of 110º.
Step 2: Draw base line from Y at a distance of 7.5cm and mark the end point as Z.
Step 3: Now, locate X at 6.5cm distance from Y on the above line and then join X and Z.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_7

• CD = 5 centimetres, DE = 5 centimetres, ∠D = 60°.
Answer:
Given that two sides of triangle are 5 cm each, the angle between them is 60º.
Steps for construction:
Step 1: Assume a point D and draw an angle of 60º.
Step 2: Draw base line from  at a distance of 5cm and mark the end point as C.
Step 3: Now, locate E at 5cm distance from D on the above line and then join C and E.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_8

Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles

Another angle Textbook Page No. 106

In all the triangles we have drawn so far, the lengths of two sides and the included angle were specified.
Can we draw a triangle, if some other angle is specified?
For example, we want to draw triangle XYZ with XY = 5 centimetres, YZ = 3 centimetres and ∠X= 30°.
First, we draw a rough sketch and write the specifications:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 16
To draw the actual triangle, we start by drawing the line XY of length 5 centimetres:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 17
Next, we draw an angle of 30° at X.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 18
Now we must locate Z. It should be 3 centimetres from Y; and also should be on the upper line.

All points at a distance of 3 centimetres from Y are on the circle centred at Y of radius 3 centimetres. Let’s draw this circle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 19
How many points are on both the circle and the upper line?
Taking one of them as Z, we get a triangle as specified:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 20
What if we take the other point as Z?
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 21

Suppose in this problem, we change the length of YZ to 4 centimetres?
Do we again get two triangles?
Answer:
Yes, we get two traingles if the length of Y is 4cm.

If YZ is to be 2.5 centimetres, how many triangles do we get?
Answer:
We get two traingles if the length of Y is 2.5cm.

What if we take it as 2 centimetres?
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 24
Can we draw a triangle?
Answer:
No, it is not possible to draw a triangle if YZ is taken as 2cm as it does not touches the inclination line or extended line from X.

Suppose we take the length of YZ as 6 centimetres?
Answer:
Yes, it is possible to draw a triangle if YZ is taken as 6cm.

Now try to draw triangles of these specifications.

• AB = 5 centimetres, BC = 6 centimetres, ∠A = 40°
Answer:
Given that two sides of triangle are 5 cm and 6 cm, the angle between them is 40º.
Steps for construction:
Step 1: Draw a line segment of length 5cm and mark the endpoints as A and B.
Step 2: Draw 40º from point A.
Step 3: Now to locate C, draw a circle of radius 6cm and mark the intersecting point as C. Join B and C. Two traingles can be formed with given measurements.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_9

• PQ = 8 centimetres, PR = 7 centimetres, ∠Q = 50°
Answer:
Given that two sides of triangle are 8 cm and 7 cm, the angle between them is 50º.
Steps for construction:
Step 1: Draw a line segment of length 8cm and mark the endpoints as Q and P.
Step 2: Draw 50º from point Q.
Step 3: Now,to locate R, draw a circle of radius 7cm and mark the intersecting point as R.Join P and R.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_10

• XY = 4 centimetres, YZ = 6 centimetres, ∠X = 70°
Answer:
Given that two sides of triangle are 4 cm and 6 cm, the angle between them is 70º.
Steps for construction:
Step 1:Draw a line segment of length 4cm and mark the endpoints as X and Y.
Step 2: Draw 70º from point X.
Step 3: Now,to locate Z, draw a circle of radius 6cm and mark the intersecting point as Z.Join Z and Y.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_11

Two angles 

Can we draw a triangle if one side and two angles are specified?
Let’s see how we can draw triangle STU with ST = 5 centimetres, ∠S = 60°, ∠T = 70°.
We first draw a rough sketch:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 25
Let’s start by drawing line ST of length 5 centimetres.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 26
Now we want to find the position of U.
Draw a line through S of inclination 60° and a line through T of inclination 70°.
The point where they meet is U.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 27

Draw triangles of the following specifications.

• YZ= 7 centimetres, ∠Y = 45°, ∠Z = 65°
Answer:
Given: Two angles of a triangle are 45º and 65º.
Step 1: Let us draw a line segment of length 7 cm and mark the end points as Y and Z.
Step 2: From Y draw an inclination of 45º.
Step 3: From Z draw an inclination of 65º and extend the line to intersect with the previous one. Mark the intersection point as X.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_13

• MN = 6.5 centimetres, ∠M= 60°, ∠N = 55°
Answer:
Given: Two angles of a triangle are 60º and 55º.
Step 1: Let us draw a line segment of length 6.5 cm and mark the end points as M and N.
Step 2: From M draw an inclination of 60º.
Step 3: From N draw an inclination of 55º and extend the line to intersect with the previous. Mark the intersection point as O.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_14

• Draw ΔABC with AB = 7 centimetres, ∠A = 60°, and ∠B = 60°. How much is ∠C? Measure the length of BC and CA.
Answer:
Given that: Two angles of a triangle are 60º and 60º.
Step 1: Let us draw a line segment of length 7 cm and mark the end points as A and B.
Step 2: From A draw an inclination of 60º.
Step 3: From B draw an inclination of 60º and extend the line to intersect with the previous one. Mark the intersection point as C.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_15
Calculation:
Two angles given are 60º and 60º. The sum of angles in a triangle is 180º.Therefore the third angle will be 180º-120º=60º.
If all the angles of a triangle are equal, then it is known as equilateral triangle.
Therefore, the other two sides will also be 7cm.

• Draw APQR with PQ = 4.5 centimetres, ∠P = 70° and ∠Q = 70°. How much is ∠R? Measure the length
of PR and RQ.
Answer:
Given that: Two angles of a triangle are 70º and 70º.
Step 1: Let us draw a line segment of length 4.5 cm and mark the end points as P and Q.
Step 2: From P draw an inclination of 70º.
Step 3: From Q draw an inclination of 70º and extend the line to intersect with the previous one. Mark the intersection point as R.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_17
Calculation:
Two angles given are 70º and 70º. The sum of angles in a triangle is 180º.Therefore the third angle will be 180º-140º=40º.

Suppose in the last problem above, ∠R is to be 70°, instead of ∠Q?
In the problem so far, one side and the two angles on this side were specified.
Here we are given ∠P and ∠R.
We need ∠P and ∠Q.
How do we calculate ∠Q?
∠Q = 180° – (70° + 70°) = 40°
Now can’t we draw the triangle?
Answer:
Given: ∠R=70º,∠Q=40º

Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles

Three sides Textbook Page No. 110

We can draw triangles of three specified sides also.
Let’s try ΔABC with AB = 5 cm, BC = 3 cm and AC = 4 cm.
We draw rough sketch for reference.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 30
First we draw AB of length 5 centimetres.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 31
Now we must locate C.
It is 4 centimetres away from A and 3 centimetres away from B.
All points which are 4 centimetres away from A are on the circle centred at A, of radius 4 centimetres.
Likewise, if we draw the circle centred at B of radius 3 centimetres, we get all points 3 centimetres away from B.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 32
Both the points where these circles cut are at a distance of 4 centimetres from A and 3 centimetres from B. Either of them can be used to draw the required triangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Triangle in a rectangle 33

Now can’t you draw triangles of the following specifications?
• PQ = 5 centimetres, QR = 5 centimetres, PR = 4 centimetres.
Answer:
Steps of construction:
Step 1: Draw a line segment PQ of length 5cm.
Step 2: Draw a circle of radius 4cm and 5cm from point P and Q respectively.
Step 3: Mark point R at the point of intersection above the base line to form the required traingle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_18

• XY = 7.5 centimetres, YZ = 6.5 centimetres, XZ = 5.5 centimetres.
Answer:
Steps of construction:
Step 1: Draw a line segment XY of length 7.5cm.
Step 2: Draw a circle of radius 5.5cm and 6.5cm from point X and Y respectively.
Step 3: Mark point Z at the point of intersection above the base line to form the required traingle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_19

• DE = 7 centimetres, EP = 7 centimetres, DP = 7 centimetres.
Answer:
Steps of construction:
Step 1: Draw a line segment DE of length 7cm.
Step 2: Draw a circle of radius 7cm each from point D and E respectively.
Step 3: Mark point P at the point of intersection above the base line to form the required traingle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_20

• Draw ΔABC with AB = 6 centimetres, AC = 5 centimetres, ∠A = 85°.
Answer:
Given that two sides of triangle are 6 cm and 5 cm, the angle between them is 85º.
Steps for construction:
Step 1: Draw a line segment of length 6cm and mark the endpoints as A and B.
Step 2: Draw 85º from point A.
Step 3: Now,to locate C, draw a circle of radius 5cm from A and mark the intersecting point as C.Join A and C.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_21

• Draw ΔPQR with PQ = S centimetres, ∠Q = 60° PR = 7 centimetres. Measure the third side and write the length in the figure.
Answer:
Given that two sides of triangle are 5 cm and 7 cm, the angle between them is 60º.
Steps for construction:
Step 1: Draw a line segment of length 5cm and mark the endpoints as P and Q.
Step 2: Draw 60º from point Q.
Step 3: Now,to locate R, draw a circle of radius 7cm from P and mark the intersecting point as R.Join P and R.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_22

• Draw ΔMNTwith MN = 8 centimetres, ∠M = 60° ∠N = 50°.
Answer:
Given: Two angles of a triangle are 60º and 50º.
Step 1: Let us draw a line segment of length 8 cm and mark the end points as M and N.
Step 2: From M draw an inclination of 60º.
Step 3: From N draw an inclination of 50º and extend the line to intersect with the previous one. Mark the intersection point as T.
Hence, the required triangle is constructed.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_23

• Draw ΔXYZ with XY = 6 centimetres, YZ = 7 centimetres, XZ = 7 centimetres.
Answer:
Steps of construction:
Step 1: Draw a line segment XY of length 6cm.
Step 2: Draw a circle of radius 7cm each from point X and Y respectively.
Step 3: Mark point Z at the point of intersection above the base line to form the required traingle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 8 Drawing Triangles img_24

Can you draw a triangle of sides 5 centimetres, 4 centimetres and 10 centimetres?
To draw a traingle for given measurements, we need to check if it satisfies the traingle rule.
Given that: AB=5cm, BC=4cm and AC=10cm
Case 1: AB+BC=9cm and AC is 10cm. Therefore,AB+BC<AC
Case 2: BC+AC=14cm and AB is 5cm. Therefore, BC+CA>AB
Case 3: AC+AB=15cm and BC is 4cm. Therefore, AC+AB>BC
As the rule is not satisfied for case 1. A traingle cannot be formed with given combination.

How about sides 5 centimetres, 4 centimetres and 9 centimetres?
To draw a traingle for given measurements, we need to check if it satisfies the traingle rule.
Given that: AB=5cm, BC=4cm and AC=9cm
Case 1: AB+BC=9cm and AC is 9cm. Therefore,AB+BC=AC
Case 2: BC+AC=13cm and AB is 5cm. Therefore, BC+CA>AB
Case 3: AC+AB=14cm and BC is 4cm. Therefore, AC+AB>BC
As the rule is not satisfied for case 1. A traingle cannot be formed with given combination.

Sides 5 centimetres, 4 centimetres and 8.5 centimetres?
To draw a traingle for given measurements, we need to check if it satisfies the traingle rule.
Given that: AB=5cm, BC=4cm and AC=8.5cm
Case 1: AB+BC=9cm and AC is 8.5cm. Therefore,AB+BC>AC
Case 2: BC+AC=12.5cm and AB is 5cm. Therefore, BC+CA>AB
Case 3: AC+AB=13.5cm and BC is 4cm. Therefore, AC+AB>BC
As the rule is satisfied for all the three combinations. A traingle can be formed with given measurments.

If two sides are to be 5 centimetres and 4 centimetres, the third side should be less than how much centimetres?
Answer:
To form a triangle the sum of two sides should be greater than the third side for all the three combinations.
So, according to the above rule the third side should be less than sum of the two given sides which is 9cm.
What is the relation between the lengths which can be used to draw a triangle?
Answer:
The relation between the lengths of a traingle is that the sum of two sides should always be greater than the third side.
Why is it that we cannot draw triangles of some specified lengths of sides?
Answer:
According to the traingle property or rule, the sum of two sides of a traingle should always be greater than the third side. If this condition is not satisfied then constructing a traingle is not possible as the arcs will not intersect.
Example:2cm, 5cm and 7cm. We cannot construct a traingle with these measurements as it does’nt satisfy the rule.

Now check which of the lengths below can be the sides of a triangle.

• 8 centimetres 6 centimetres 13 centimetres.
Answer:
To check if the given length can be sides of a triangle, use the below rule.
Rule: The sum of two sides should be greater than the third side for all the combinations.
Given that: AB=8cm, BC=6cm and AC=13cm
Case 1: AB+BC=14cm and AC is 13cm. Therefore,AB+BC>AC
Case 2: BC+AC=19cm and AB is 8cm. Therefore, BC+CA>AB
Case 3: AC+AB=21cm and BC is 6cm. Therefore, AC+AB>BC
As the rule is satisfied for all the three combinations. A traingle can be formed with given measurements.

• 2 centimetres 5 centimetres 8 centimetres.
Answer:
To check if the given length can be sides of a triangle, use the below rule.
Rule: The sum of two sides should be greater than the third side for all the combinations.
Given that: AB=2cm, BC=5cm and AC=8cm
Case 1: AB+BC=7cm and AC is 13cm. Therefore,AB+BC<AC
Case 2: BC+AC=13cm and AB is 2cm. Therefore, BC+CA>AB
Case 3: AC+AB=10cm and BC is 5cm. Therefore, AC+AB>BC
As the rule is not satisfied for case 1. A traingle cannot be formed with given measurements.

• 5 centimetres 4 centimetres 9 centimetres.
Answer:
To check if the given length can be sides of a triangle, use the below rule.
Rule: The sum of two sides should be greater than the third side for all the combinations.
Given that: AB=5cm, BC=4cm and AC=9cm
Case 1: AB+BC=9cm and AC is 9cm. Therefore,AB+BC=AC
Case 2: BC+AC=13cm and AB is 5cm. Therefore, BC+CA>AB
Case 3: AC+AB=14cm and BC is 4cm. Therefore, AC+AB>BC
As the rule is not satisfied for case 1. A traingle cannot be formed with given measurements.

• 4 centimetres 6 centimetres 7 centimetres.
Answer:
To check if the given length can be sides of a triangle, use the below rule.
Rule: The sum of two sides should be greater than the third side for all the combinations.
Given that: AB=4cm, BC=6cm and AC=7cm
Case 1: AB+BC=10cm and AC is 7cm. Therefore,AB+BC>AC
Case 2: BC+AC=13cm and AB is 4cm. Therefore, BC+CA>AB
Case 3: AC+AB=11cm and BC is 6cm. Therefore, AC+AB>BC
As the rule is satisfied for all the three combinations. A traingle can be formed with given measurements.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

You can Download Parallel Lines Questions and Answers, Activity, Notes, Kerala Syllabus 7th Standard Maths Solutions Chapter 2 to help you to revise the complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Parallel Lines Text Book Questions and Answers

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 1

Parallel Lines Textbook Page No. 14

We get a line by joining any two points; on the other hand, do any two lines meet at a point?
What about the lines got by extending a pair of opposite sides of a rectangle?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 2

Do they meet however much we extend them? Why?
Answer:
No, the lines got by extending a pair of opposite sides of a rectangle will not meet as they are parallel to eachother.

Explanation:
Parallel lines are lines in a plane that are always the same distance apart. Parallel lines never intersect. Perpendicular lines are lines that intersect at a right (90 degrees) angle.

Look at this quadrilateral:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 3
If we extend the top and bottom sides, do they meet?
What about the left and right sides?
Answer:
No, by extending top and bottom sides of the quadrilateral, the lines will not meet. As they are parallel to each other. Here, the left and right sides are not parallel lines.

Explanation:
Bottom and top lines are parallel lines which does not meet. Left and right lines can meet as they are not parallel lines. So if they are extended then they will meet at some point.

What if we draw a quadrilateral like this?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 4
Does any pair of opposite sides meet if we extend them? Why?
Answer:
Yes, left and right pair of lines can meet in somepoint if extended yet not the top and bottom pair of lines cannot meet as they are parallel lines to each other.

Explanation:
Left and right lines can meet in some extent yet the top and bottom lines will not meet as they are parallel lines. Lines which are at the same distance gap do not meet anywhere, as they are called parallel.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Same distance

Don’t you know how to draw a rectangle?
How do we draw a rectangle of length 5 centimeters and a breadth of 3 centimeters?
There are several ways, right?
First, draw a horizontal line 5 centimeters long and draw a vertical line at one end, 2 centimeters high:
Next at the other end of the vertical line, draw a
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 5
perpendicular 5 centimeters long. Joining the other end of this perpendicular to the end of the first line makes our rectangle:
Now extend the top and bottom sides of this rectangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 6
get a pair of parallel lines:
So if we take a line and a point 2 centimeters from it, how
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 7

do we draw a parallel to the line through the point?
First, draw the perpendicular to the line through the point:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 14
Then draw the perpendicular to this perpendicular through
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 15
the point:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 16
Instead of actually drawing the first perpendicular, we can use a ruler:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 17
Now by shifting the set square upward and putting its square corner at the point, we can draw the parallel:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 18
What if the point is below the line?
What did we see here?
Through any point not on a line, we can draw a parallel line.
How many parallels can be drawn through a point not on a line?
Answer:
For a given line and a point outside the line, only 1 parallel can be drawn through the point was stated by.

Explanation:
Only one parallel line can be drawn through a point not on a line.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Same direction
Opposite sides of a rectangle are parallel:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 19
This can be said in a different manner: Two perpendiculars to the same line are parallel.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 25

Now look at this figure:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 26
Are the slanted lines parallel’?
What would happen if these lines are extended upwards?
What if the lines are like this?
Answer:
Yes, the lines will never meet as they are parallel to each other and two lines not in the same plane that do not intersect.

Explanation:
Slanted lines are two lines not in the same plane that do not intersect. Lines are slanting if they don’t go straight across or straight up. They look like a slope and go both up and down, and across too.
Parallel lines are lines in a plane that are always the same distance apart. Parallel lines never intersect. Perpendicular lines are lines that intersect at a right (90 degrees) angle.

 

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 27

Would the lines meet if extended Upwards?
Suppose we extend them downwards?
If the lines are not to meet either way, by how much should the right line be slanted?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 28Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 29
Now draw a figure like this in your notebook:
Is there a quick and easy way to draw a line parallel to AB through P?
Answer:
No, the lines will not meet because they are parallel to each other.

Explanation:
The lines are parallel to each other which means they won’t intersect. We can draw a line parallel to AB through P by 45 degrees downwards pointing from A to P.

In the quadrilateral shown below, both pairs of opposite sides are parallel:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 30
Can you draw this with the same lengths and angle?
Answer:
Yes, we can draw this with the same lengths and angle, its called a parallogram.

Explanation:
A quadrilateral is a plane figure that has four sides or edges, and also has four corners or vertices.
A quadrilateral that has its opposite sides equal and parallel, its called as parallogram.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Parallels and angles Textbook Page No. 19

In the figure below, the top and bottom lines are parallel:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 31
How much is the angle marked on the top?
Answer:
As those lines are parallel to each other, so the marked angle will be 50°.

Explanation:
Parallel lines are always the same distance apart. And they never meet we use arrowheads to show that lines are parallel. They angles formed will be same. Angle marked on top is going to 50°.

Parallel lines should have the same slant with any other.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 38
There are other angles here. Can you figure out those?
Answer:
Here, the other angle will be a parallel line. So, the other angle will be the same.

Explanation:
Other angles are formed by the parallel lines, so they will be the same as they are corresponding angles.

First look at the three angles below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 39
What are the relations between the four angles made by two lines cutting across each other?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 40
Can’t you find out the angles at the top also like this?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 41
Answer:
We can find the angles of top diagram as they are same like the below angles shown as they are corresponding angles, Alternate angle and interior angles.

Explanation:
When two or more lines are cut by a transversal, the angles which occupy the same relative position are called corresponding angles. When the lines are parallel, the corresponding angles are congruent.
The alternate angle theorem states that when two parallel lines are cut by a transversal, then the resulting alternate interior angles or alternate exterior angles are congruent.
If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary, i.e. they add up to 180°.

In the next figure also, the lines at the top and bottom are parallel:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 42
Write down the measures of the other seven angles in this figure.
What we have seen here can be written thus:
Answer:
∠BAG = 140°
∠GAH = 40°
∠BAG = 140°
same as below angles above angles are formed as they lines are parallel lines to each other.

Explanation:
Kerala-State-Syllabus-7th-Standard-Maths-Solutions-Chapter-2-Parallel-Lines
∠BAG + ∠BAF = 180° [ As they are interior angles]
=> ∠BAG + 40°  = 180°
=> ∠BAG = 180° – 40°
=>∠BAG = 140°
∠GAH = ∠BAF [ As they are vertical angles]
=> ∠GAH = 40°
∠BAG = ∠FAH [As they are vertical angles]
=> ∠BAG = 140°

Parallel lines make equal angles with any other line.

In the figures below, there is a pair of parallel lines and a third line cutting across them. In each figure. the measure of one angle is given and another angle is mailed. Find out its measure and write it in the figure itself.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 43
Answer:
The angles are same as they are interior alternative angles formed by the parallel lines.

Explanation:
Both the angles are same as they are alternative interior angles.
Figure 1:
Both are 45°
Figure 2:
Both are 60°

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 49
Answer:
Figure 1:
Same angles as they corresponding angles 120°
Figure 2:
They are Supplementary angles 100°
Figure 3:
They are alternative interior angles of 45°

Explanation:
Figure 1:
They are corresponding angles and they are the same angles.
Figure 2:
They both are not the same yet their total sum is equal to 180°as they are Supplementary angles, whose angles sum up to 180 degrees.
=> 80° + ?? = 180°
=> ?? = 180° – 80°
=> ?? = 180°
Figure 3:
Both the angles are same as they are alternative interior angles.

Matching angles
When a line cuts across a pair of parallel lines, eight angles are formed:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 50
In the figure, the bottom line makes four angles, and the top line makes another four angles with the line cutting across them.

we can pair one angle at the bottom with one at the top in several different ways. Some such pairs are equal; others are supplementary (meaning their sum is 180°).

Let’s look at the pairs of equal angles. For convenience, they are divided into two types. Look at the pair of angles marked in the figure below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 51
Of these, the angle at the bottom is on top of the horizontal line, and on the right of the slanted line; the angle at the top also is on top of its horizontal line, and on the right of the slanted line.
There are three more pairs of angles which are at similar positions at the bottom and top:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 52
Angles in each such pairing, done according to similar positions, are called corresponding angles.

Equal angles from the bottom and top can be paired in another manner. See the angles below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 58
The bottom angle is on top of the horizontal line, and on the right of the slanted line
What about the top angle?
At the bottom of the horizontal line, and on the left of the slanted line.
We can pair the equal angles in three other ways, with the positions quite the opposite:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 59
Angles in each such pairing, done with reverse positions, are called alternate angles.

In the figure below, the pair of parallel lines and the cutting line are all named. The measure of one angle is also given. Complete the tables below by writing the names and measures of all pairs of corresponding and alternate angles:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 60
Answer:
Kerala-State-Syllabus-7th-Standard-Maths-Solutions-Chapter-2-Parallel-Lines-Matching angles

Explanation:
Corresponding angles = ∠XCB AND ∠QRC =  60°
∠BCR AND ∠QRY = 180° – 60°= 120°
∠ACY AND ∠PRY = 120° .
Alternative angles = ∠BCR AND ∠XRP = 180° – 60°= 120° .[As they are sum = 180° ]

In short,
The angles formed by a line cutting across two parallel lines can be paired in several ways, choosing one angle of the four made with one line and one of the four made with the other. All eight pairs will have equal angles. Based on the positions with respect to the lines, angles in four such pairs are called corresponding angles and four angles on the other hand are called alternate.

Supplementary angles Textbook Page No. 26

Let’s have another look at a picture of a two parallel lines cut by a third line:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 65
How much is the marked upper angle?
There is such a pair of supplementary angles on the left of the slanted line also.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 66
The angles in each ofthese two pairs are called co-interior.
There are also two pairs of co-exterior angles.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 67

In the figure below, the lines AB and PQ are parallel and the line XY cuts them at C and R. Find all the pairs of co-interior and co-exterior angles and write down their names and measures.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 68
Answer:
Kerala-State-Syllabus-7th-Standard-Maths-Solutions-Chapter-2-Parallel-Lines-Supplementary angles Textbook Page No. 26

Explanation:
Co-interior angles or Consecutive interior angles are thetwo angles that are on the same side of the transversal. Co-interior angles are the interior angles and it sums up to 180 degrees.
∠XRP, ∠BCY, ∠XRQ, ∠QRX
Co-exterior angles are two angles on the same side of the transversal in a figure where two parallel lines are intersected by transversal .
∠XCB, ∠QRY, ∠XCA, ∠YRP

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Parallel lines and triangles
See this figure:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 69
A line starting from B is to be drawn, parallel to AX.

How do we do this?
The angles at A and B are co-interior, right?
Answer:
Yes, a line from B drawn parallel to AX, angles formed are said to be co-interior angles.

Explanation:
A line parallel to AX from B, angles are said to be co-interior.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 77
Draw this in your notebook.
Next draw a slanted line through B in this figure. Let the angle with AB be 70°. This line is not parallel to AX. Let the point at which this line meets AX be named C:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 78
Now ABC is a triangle. And we know the angles at A and B.
How much is the angle at C?
The lines AC and BY are parallel. Concentrate on these and the line BC:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 79
∠ACB and ∠CBYare alternate angles:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 80
Likewise, can you compute the angle at C in the triangle below?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 81
How about extending AC and drawing a line from B parallel to it, like the first figure?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 82
We have to calculate ∠ACB.
First, we note that it is equal to ∠CBY( Why?).
To find ∠CBY, we need only know ∠ABY; and this angle together with ∠A make a co-interior pair.
So,
∠ABY = 180° – 40°= 140°
From this we get
∠CBY = 140° – 60° = 80°
Thus we find
∠ACB = ∠CBY = 80°

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 97

Now see this triangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 98
The measures of the angles are given by the letters a, b, c.
What is the relation between them?
Let’s draw parallel lines as before.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 99
From this figure, we see that ∠CBY = ∠ACB = c°
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 100
From the figure above,
∠A + ∠ABY = 180°
That is,
a + b + c = 180
What do we get from this?
The sum of the angles of any triangle is 180°

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Let’s do it! Textbook Page No. 31

Question 1.
Find out the pairs of parallel lines in the figure below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 85
Answer:
FE || LK,
AB || GH,
CD || MN.

Explanation:
Here, the given parallel lines are pairs of lines that do not intersect or meet at any point. So the parallel lines are FE || LK, AB || GH, and CD || MN.

Question 2.
In the figure below, AB and CD are parallel. Compute all the angles in the figure.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 86
Answer:
∠CAB = ∠ECD = 70°,
∠CDE = ∠ABD = 60°,
∠A = 70°,
∠B = 60°,
∠E = 50°.

Explanation:
Given that AB || CD and ∠CAB = ∠ECD = 70° corresponding angles and ∠CDE = ∠ABD = 60°. And the sum of all angles in the triangle is ∠A+∠B+∠E = 180°, so
70° + 60° + ∠E = 180°,
130° + ∠E = 180°,
∠E = 180° – 130°
= 50°.

Question 3.
In the figure below, a parallelogram is divided into four triangles by the diagonals. Calculate the angles of all these triangles.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 87
Answer:
The angles of the triangle are
∠AOB = 65°,
∠DBC = 65°.

Explanation:
Kerala-Syllabus-7th-Standard-Maths-Solutions-Chapter-2-Parallel-Lines-87-1
Given that ∠BAO = 25°, ∠AOB = 90°.  As
∠BAO + ∠AOB + ∠ABO = 180°,
25°+90°+ ∠ABO = 180°,
115° + ∠ABO = 180°,
∠ABO = 65°.
And ∠AOB = ∠DOC = 90°, as opposite angles are same.
∠AOD = 180° – 90°
= 90°.
∠AOD = ∠COB = 90°,
∠ABO = ∠CDO = 65° as these are alternate angles.
And ∠DBC will be
180° = 65°+50°+∠DBC
115° + ∠DBC = 180°
∠DBC = 180°-115°
= 65°.

Question 4.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 90
In the figure above, AB and DE are parallel. Compute the angles of both triangles.
Answer:
∠C = 40°.

Explanation:
Given that AB||DE,
∠A = 80°,
∠D = 80° as they are corresponding angles.
∠E = 60°,
∠B = 60° as they are corresponding angles.
∠A + ∠B + ∠C = 180°, so
80°+60°+ ∠C = 180°,
140°+ ∠C = 180°,
∠C = 180°-140°
= 40°.

Question 5.
In the figure below, PQ and RS are parallel. Calculate all other angles in the figure.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 91
Answer:
∠PTQ = 80°.

Explanation:
Given that PQ||RS and ∠P =  60°, ∠Q =  40°,
∠S = ∠P = 60° (Alternate angles),
∠R = ∠Q = 40° (Alternate angles).
Here, triangle PQT,
∠P + ∠Q + ∠T = 180°,
60° + 40° + ∠T = 180°,
100°+ ∠T = 180°,
∠T = 180°-100°
= 80°.
∠PTQ = 80°,
∠STR = 80° (opposite angles).

Question 6.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 92
In the figure, AB and CD are parallel. Compute the third angle.
Answer:
The third angle is ∠K = 80°.

Explanation:
Given AB||CD and EF||AB||CD,
∠AOK = ∠OKF = 35°,
∠CPK = ∠PKF = 45° (alternate angles),
∠K = ∠OKF + ∠PKF
= 35° + 45°
= 80°.

Question 7.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 93
In the figure, PR and ST are parallel. Is there any relation among the angles of the two triangles?
Answer:
∠P = ∠S corresponding angles,
∠T = ∠R corresponding angles.

Explanation:
Given that PR||ST,
here the 2 triangles are triangle PQR and triangle SQT,
∠P = ∠S are corresponding angles,
∠T = ∠R corresponding angles.

Question 8.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 94
In the figure above, AB and CD are parallel. What is the relation between the angles of the small and large triangles?
Answer:
∠C = ∠B as these are alternative angles,
∠A = ∠D as these are alternative angles,
∠CPD = ∠APB opposite angles.

Explanation:
Given that AB||CD,
Let’s consider the triangle CPB and triangle APB,
∠C = ∠B (As these are alternative angles)
∠A = ∠D (As these are alternative angles)
∠CPD = ∠APB (opposite angles)

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Question 9.
Draw a line AB and a line CD parallel to it. Draw a line EF cutting across these lines at the points Mand N. Measure and write down one of the angles so made. Calculate the other angles. Write down the pairs of corresponding angles, alternate angles, co- interior angles and co-exterior angles.
Answer:
Kerala-Syllabus-7th-Standard-Maths-Solutions-Chapter-2-Parallel-Lines-87-1-1
Explanation:
Here, the pair of corresponding angles are
∠EMB = ∠MND,
∠NMB = ∠FND,
∠EMA = ∠MNC,
∠AMN = ∠CNF.
The pair of alternate angles are
∠MNB = ∠CNF,
∠BMN = ∠ENC,
∠EMA = ∠DNF,
∠DNM = ∠NMA.
The pair of co-interior angles are
∠BMN+∠DNM = 180°,
∠AMN+∠CNM = 180°.
The pair of co-exterior angles are
∠EMB+∠DNF = 180°,
∠EMA+∠CNF = 180°.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Distance Textbook Page No. 15

Would these lines meet, if extended?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 8

Answer:
Yes, these lines are not parallel lines, so they can meet each other by extending.

Explanation:
The lines given are not parallel lines. They can meet if extended somewhere.

How about these?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 9

Answer:
No, given lines are to be parallel lines so, the lines do not intersect each other.

Explanation:
The lines look like they are parallel, which means they cannot meet each other.

Look at the distance between the lines in both the figures:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 10
What can we say about the distance between two parallel lines?
Answer:
In two figures, parallel lines are the second figure, these lines do not intersect each other.

Explanation:
In first figure, lines are not parallel, the distance between the lines will reduce and they will intersect when they are extended.
In the second figure, lines are parallel lines. The distance between the lines will be the same.

In GeoGebra, draw a quadrilateral. Use the Line through two points tool to extend the sides.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 11
Do they meet’?
Use the move tool to drag the corners of the quadrilateral. When do the sides fail to meet?
Answer:
No, the lines are going to meet each other if they are extended as they are not parallel lines.

Explanation:
The lines draw are not parallel lines, which means they can meet if extended.

Perpendicular and parallel Textbook Page No. 16

See the figure:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 12
Look at the perpendicular lines to the horizontal.
Are they parallel?

Answer:
Yes, the lines are parallel lines.

Explanation:
The lines are perpendicular which means they are parallel lines.

Now see this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 13
A perpendicular is drawn to the horizontal line and then a perpendicular to the perpendicular is drawn.
Are the horizontal lines parallel?
Answer:
Yes, the lines drawn horizontal are parallel lines.

Explanation:
The lines are drawn are perpendicular to each other, which means they are parallel lines.
They cannot meet each other.

There are tools in GeoGebra to draw parallel and perpendiculars to a line. First draw a line and mark a point on it. Select the Perpendicular line tool, and click on the line and point, to get a perpendicular to the line through the point. Such a perpendicular can be drawn through a point not on the line also. Try drawing a perpendicular to this perpendicular.

To draw a line parallel to another, the Parallel line tool is used. Mark a point not on the line. Select this tool and click on the line and the point, to get the parallel through the point. Drag the point using the Move tool. What happens if the point is on the line?

Draw line AB in GeoGebra and mark two other points C, D on it.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 20
Now select the Slider tool and click on the GeoGebra screen. In the dialog box, select the Angle option by clicking on the small circle beside it. Against the Name, type a. Now click Apply.
Select the Angle with given size tool and click on B and D. In the dialog box, type a as Angles and click OK. Now we get a point B’ Join D and B’ by a line.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 21
Now make another slider named b. Select the Angle tool and click on B and C in that order. In the dialog box, type b as the angle and click OK. Join the new point B”with C.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 22
Use the Move tool to change the values of a and b. What happens to the lines? When do they fail to meet?
Try these with a single slider for the angles at C and D.
Answer:
Angles formed from C and D lines are going to be acute and they are going to intersect if extended.

Explanation:
The lines drawn seem to be intersecting and the angles formed from C and D are going to acute.

Not a square, but… Textbook Page No. 18
You know how to draw rectangles, using a set square. What if we use another comer, instead of the square one?
_________________
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 23
Does any pair of opposite sides meet, when extended?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 24
Answer:
No, the lines won’t intersect as they are parallel lines.

Explanation:
The lines drawn are parallel, which means they cannot intersect even though they are extended.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

When parallels cut
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 32
Draw a pair of parallel lines.
And another pair cutting across them.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 33
Look at the figure made between them:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 34
What is its name?
Answer:
The name of the figure is a parallelogram.

Explanation:
The figure drawn from the two vertical and two horizontal lines is said to be a parallelogram.
A parallelogram is a quadrilateral with two pairs of parallel sides.

Rectangle and Parallelogram Textbook Page No. 20

Cut out a cardboard rectangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 35
Now cut a triangle through the bottom corner as shown below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 36
Place the triangle on the other side like this.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 37
Is this a parallelogram?
Why?
Answer:
Yes, its a parallelogram as both the lines drawn are parallel.

Explanation:
Lines drawn are parallel to each other, so it is said to be a parallelogram.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Unchanging shapes
Place two straight and slender twigs parallel to each other. Place another one across and paste them together:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 44
Now break at the middle to get two pieces:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 45
Place one piece over the other. All these angles match, don’t they?
Answer:
Yes, the angles will match as they are drawn from parallel lines.

Explanation:
The lines are parallel to each other. They are broken in between which does not affect the angles.

Angles of a Parallelogram Textbook Page No. 22

All angles in a rectangle are right angles.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 46
What about the angles in a parallelogram?
Answer:
There are four interior angles in a parallelogram and the sum of the interior angles of a parallelogram is always 360°.

Explanation:
The opposite interior angles are equal. The angles on the same side of the transversal are supplementary, which means they add up to 180 degrees. Hence, the sum of the interior angles of a parallelogram is 360 degrees.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 47
Can you find the angles in the first parallelogram?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 48
Now find the angles in the second one.
Answer:
The angles opposite to given angles are the same as they are supplementary and vertically opposite angles.

Explanation:
Second figure:
The angles opposite to given angles are the same as they are supplementary and vertical opposite angles.
Kerala-State-Syllabus-7th-Standard-Maths-Solutions-Chapter-2-Parallel-Lines-Angles of a Parallelogram Textbook Page No. 22

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Corresponding angles Textbook Page No. 23
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 53
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 54

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Alphabet angles

Draw the letter N as below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 55
What is the relation between the two angles marked?
Answer:
The two angles formed by N letter are said to be alternative angles.

Explanation:
When two parallel lines are cut by a transversal, then the resulting alternate interior angles or alternate exterior angles are congruent.

Now look at the letter M.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 56
See any relation between the angles?
Draw a line down the middle and see:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 57
Answer:
The angles in the letter M are said to be corresponding angles.

Explanation:
When two or more lines are cut by a transversal, the angles which occupy the same relative position are said to be corresponding angles .

Corresponding and alternate Textbook Page No. 25

Look at the picture.

Pairs of corresponding angles are of the same color.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 61
What about this picture?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 62
Answer:
Figure 1 is correct example for corresponding angles formed by parallel lines yet not the figure 2.

Explanation:
When two or more lines are cut by a transversal, the angles which occupy the same relative position are called corresponding angles. When the lines are parallel, the corresponding angles are congruent.

In GeoGebra draw a line AB and a parallel line through C.
Mark points D and F on these and join them.
Mark two points G and H as in the figure.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 63
Next select the Angle tool and click on G F, H in that order and then on B, D, F.
Now we can see the measure of these angles.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 64
Use the Move tool to change the position of F.
Find the other angles at F and D like this.
We can also colour these angles. Right click on any angle and select Object properties. In the menu, click on Color and choose a color. Choose the same color for equal angles at the top and bottom.
Answer:
Kerala-State-Syllabus-7th-Standard-Maths-Solutions-Chapter-2-Parallel-Lines-Corresponding and alternate Textbook Page No. 25

Explanation:
The angles formed by parallel lines and corresponding angles are said to be equal and same. I have used blue color and red color to represent them in the figure.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Sums of angles Textbook Page No. 27

Look at the parallelogram:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 70
Can you write the measures of the other three angles?
What is the sum of all the angles?
Answer:
Kerala-State-Syllabus-7th-Standard-Maths-Solutions-Chapter-2-Parallel-Lines-Sums of angles Textbook Page No. 27

Explanation:
Opposite angles in a parallelogram are said to be congruent.
=> ∠A = ∠C = 60°
=> ∠B = ∠D = 120°
Consecutive angles in a parallelogram are supplementary.
=> ∠A + ∠B = 180°
=> 60° + ∠B = 180°
=> ∠B = 180° – 60°
=> ∠B = 120°

Now look at this parallelogram:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 71
No angle is given.
But can’t you say what the sum of the two angles on the left is?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 72
What about those on right?
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 73
So, what is the sum of all four angles?
Answer:
The sum of all four angles in parallelogram are 360°.

Explanation:
Consecutive angles are the angles that formed when a transversal intersects two parallel lines.
Consecutive angles in parallelogram are supplementary.
=> The sum of two angles on top left and bottom right are said to be 180°.
=> Left = 180°.
=> Right = 180°.
=> 180° + 180° = 360°.

Triangle and parallel lines Textbook Page No. 28

Draw a triangle like this in a piece of card board.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 74
Now place a long and thin stick along the side BC and stick a pin through it at C.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 75
Rotate the stick upwards till it is parallel to AB.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 76
Now what angle does the stick make with BC?
And with AC?
So, how much is the angle at C in the triangle?
Answer:
∠ACB = 80°
∠C in the triangle = 180°

Explanation:
Sum of angles in a triangle = 180°
=> ∠A + ∠ACB + ∠B = 180°
=> 60° + ∠ACB + 40° = 180°
=> ∠ACB + 100°= 180°
=> ∠ACB = 180° – 100°
=> ∠ACB = 80°
Name the line at C.
Kerala-State-Syllabus-7th-Standard-Maths-Solutions-Chapter-2-Parallel-Lines-Triangle and parallel lines Textbook Page No. 28
∠DCA = ∠A = 60° [Alternative angles]
∠ECB = ∠B = 40° [Alternative angles]
=> ∠ECB + ∠DCA + ∠ACB = ∠C
=>40° +  60° + 80° = ∠C
=>180° = ∠C

Parallelogram and triangle Textbook Page No. 29

Look at this parallelogram:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 83
What is the relation between the angles marked red?
And the angles marked green?
And those marked in different colours?
Answer:
Both green and red color angles are opposite angles and are equal.

Explanation:
Opposite angles in a parallelogram are said to be congruent.
=> Angles of green color are the same.
=> Angles of red color are the same.

Now join the opposite corners to make two triangles:
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 84

What is the relation between the angles marked blue?
And those marked yellow?
What is the sum of three angles of different colours?
What is the sum of the three angles of each triangle?
Answer:
The angles marked in blue and yellow color are alternative angles and are said to be equal.

Explanation:
The angles marked in blue and yellow color are alternative angles and are said to be equal.
Sum of three angles of different colours in parallologram are said to be 360°
Sum of three angles of each triangle said to be 180°

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Theorem and proof

How do we conclude that the sum of the angles in any triangle is 180°? Is it enough if we draw several triangles, measure the angles and check the sum? How can we say for sure that for a triangle not among these, the sum is 180°?

In any triangle, we can draw a line through one vertex, parallel to the opposite side. And then using the relations between angles made by parallel lines, we can see that the sum of the angles of a triangle is 180°.
By doing this, we achieve much.

  • Even if we change the triangle, the arguments used do not change. So, the conclusion of these arguments also is true for the changed triangle.
  • Properties of parallel lines can be easily recognized. But the fact that sum of the angles of a triangle is 180°, is not immediately obvious. This is an example of establishing complex ideas, starting
    from simple truths.
  • When arguments are linked one after another using ideas of parallel lines, we not only get the theorem that sum of the angles of a triangle is 180°, but also see why it is so.

Unchanging relation

In GeoGebra, use the polygon tool to draw triangle ABC. Using the Angle tool, we can get the measures of its angles.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 88
Now draw a line DE and mark a point F on it. Select the Angle with given size tool and click on E and F in that order. In the dialog box, type α as the angle and click OK. We get a new Point E’. With the same tool, click on E’ and F and type in β as the angles. We get a new point E”. Click on E” and F and type γ as the angle to get another point E”’. Join FE’ and FE”. In this picture, we have ∠EFE’ = ∠A; ∠E’FE” = ∠B; ∠E” FE”’ = ∠C. Give the same colours to equal angles.

Use the Move tool to change the triangle’s angles. The angles in the figure on the right also change. What remains unchanged?
Answer:
When the triangle’s angles are changed, the angles in the figure on the right also changes.
The points D, E”’ and E remains unchanged.

Explanation:
The sum of angles in a triangle equals 180°. Therefore, the three angles ∠EFE’ = ∠A; ∠E’FE” = ∠B; ∠E” FE”’ = ∠C will also be equal to 180° or a straight angle. Thus, D, E”’ and E which forms the straight line remains unchanged even all the three angles are changed.

Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines

Drawing parallelograms Textbook Page No. 32

Let’s draw a parallelogram using GeoGebra.
First draw two lines AB and BC. Use the Parallel line tool to draw the line through C parallel to AB and the line through A parallel to BC. Mark the point D where the lines meet. Use the Polygon tool to complete the parallelogram ABCD.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 89
Lines sticking out may be deleted.
Now right click on AB and select Trace on. Do this for BC also. Select the Move tool, click within the parallelogram and drag upwards. What do you get?
Answer:
When you drag upwards from AB And BC, we are going to get a another parallel line forming parallelogram.

Explanation:
When we extended lines from AB And BC, we are going to get parallel lines. These lines can be formed in another parallelogram having DC as one of the side.

Drawing Pictures

Try to draw these pictures using GeoGebra.
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 95
Kerala Syllabus 7th Standard Maths Solutions Chapter 2 Parallel Lines 96
Use the Regular Polygon tool to draw the large triangle.
Answer:

Explanation:

Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

You can Download Unchanging Relations Questions and Answers, Activity, Notes, Kerala Syllabus 7th Standard Maths Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

Unchanging Relations Text Book Questions and Answers

Measures and relations     Textbook Page No. 36

We can draw squares of different sizes. The perimeter and area change with the length of sides. But in all the squares, the perimeter is always four times the length of side; and the area is the product of the length of side by
itself.

There are many such situations where even though the measures change, certain relations between them do not change. For example, if we take several objects made of iron, their volume and weight will be different. But if we divide the weight by the volume, we get the same number, 7.8. This is called the density of iron. Similarly if we do this for objects made of copper, weight divided by volume gives 8.9. This is the density of copper.

In such contexts, the unchanging relation between measures is denoted using letters.
For example, if we write the weight of an object made of iron as w and the volume as y, then we can write
w = 7.8v
If we take copper instead of iron, the relation will be
w = 8.9v
In general, if the weight of an object is w, its volume is y and if it is made of a material of density d, then their relationship can be written as
w = dv

Measures and numbers

Numbers were invented by Man to indicate and compare various measurements. For example, instead of saying, “a large group of people”, if we say, “a group of hundred people”, one gets a clear picture. Or instead of saying “walked some distance”, we can be more precise and say, “walked two and a half kilometers”.

Length, weight, and time are measured directly using instruments; area, volume, and density are not directly measured but are got by computations. For that, we need operations with numbers. For example, to find the volume of a rectangular block we must measure its length, width, and height and calculate the product.

In course of time, men started thinking about operations with pure numbers without linking them to measurement. For example, after discovering that to compute the perimeter of a rectangle, instead of measuring all four sides and adding, we need only measure two different sides and then double their sum they went on to discover the number principle that instead of multiplying two numbers separated by two and adding, we need only multiply their sum by two.
Much later, we started writing this using letters as
2x + 2y = 2(x + y)

Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

Number theory Textbook Page No. 38

We have seen that general properties of operations on numbers can be stated in shorthand, using letters.
For example, the fact that
zero is added to any number gives the same number.
Can be shortened to
x + 0 = x, for any number x

Similarly, the fact that
To find the sum of two numbers, they can be added in any order can be written
x + y = y + x, for all number x and y.
Simple ideas like these which are easily recognized need not be shortened in this manner.
But look at this:

Adding to a number, one more than itself, and adding one to double the number gives the same result.
It is much more convenient to shorten the elaborate statement as
x + (x + 1) = 2x + 1, for any number x.
Here we must note an important thing. It is easy to remember such shorthand formulas, but to use them in time, we must know their meaning clearly.

Folding umbrellas are easy to carry; but if we don’t know how to open them, we are sure to get wet.

Two operations, one result

Adding one to twice a number is an arithmetical operation and the number got by doing it is the result. For example, doing this on 3 gives 7, and on 10 gives 21.
Adding one to a number and then adding the sum to the original number is another operation. For example, this operation on 4 gives 4 + (4 + 1) = 9. These two operations done on the same number give the same result. It is this fact that we write in algebra as
x + (x + 1) = 2x + 1

In this, the short form x + (x + 1) on the left side means, adding a number to one more than the number. The form 2x + 1 on the right means, doubling a number and adding one to it. The equal sign says that these two operations lead to the same result.

Similarly, we can write the operation of doubling two numbers separately and adding 2x + 2y in algebra. The algebraic form of the operations of doubling the sum of two numbers is 2 (x + y). The general principle that these two operations on the same pairs of numbers give the same result is written
2x + 2y = 2(x + y) for all numbers x, y.

Many general properties of numbers have a similar form, starting that apparently different operations giving the same results.

Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

Arithmetic and algebra Textbook Page No. 40

Studies on numbers are generally known as arithmetic, and stating them using letters is algebra.
In arithmetic, the operations of adding 3 and 7 are written as 3 + 7. The sum, or the result of the operation, is 10. And we write,
3 + 7 = 10
combining the operation and the result.

In algebra, the operation of adding two numbers can be written x + y. How do we write the sum? We cannot find it without knowing the actual numbers added. So we can only write x + y for the sum also.

But the fact that
Any number added to itself is double the number that can be written in algebra as x + x = 2x, for all numbers x.

Note that this is not a general principle, but is the very definition of multiplication by 2.

Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

Difference of difference

Finding the sum of three numbers is very natural. So three is actually no need to keep in mind the algebraic expression
(x + y) + z = x + (y + z)
Its only use is that it sometimes makes computation easier. For example, in calculating 29 + 37 + 63, if we can quickly see 37 + 63 = 100, then the total can be easily formed as 129. (To add in the given order may require pen and paper).
But we must be careful with subtraction.
The meaning of
(10 – 3) – 2
is, subtract 3 from 10 and then subtract 2 from the resulting 7; that is, the final result is 5.
What about this?
10 – (3 – 2)
First, subtract 2 from 3 to get 1; then subtract this 1 from 10, to get 9.
In other words, these operations give different results. But the result of both (10 – 3) – 2 and 10 – (3 + 2) is 5. We must keep in mind the general result
(x – y) – z = x – (y + z)
and it’s meaning:
instead of subtracting two numbers one after another, we need only subtract their sum.

Theory and practice Textbook Page No. 42

To do 25 + 20 – 15, we can first add and then subtract to get 45 – 15 = 30; or we can first do the subtraction and then addition to get 25 + 5 = 30.
But in 25 + 10 – 15, it is easy to see that we cannot do the subtraction first.

In operations like this with actual numbers, it is easy to see that certain operations cannot be done. But when we write them in algebra, we must also state the conditions which make them true.
That is why in writing
(x + y) – z = x + (y – z) we also write the condition y > z.

Less and more

Look at these differences:
10 – 9 = 1
10 – 8 = 2
10 – 7 = 3
10 – 6 = 4
When we subtract less, we get more, right?
How much more?
When one is subtracted, we get one more; when two are subtracted, we get two more.
In general
When less is subtracted, the result is more; whatever less is subtracted, that much more is the result.
Let’s write this in algebra. First note that if x and y are two numbers, then y subtracted from x is x – y.
Now taking another number z, the number y – z is z less than y; so x – (y – z) is z more than x – y.
Thus,
x – (y – z) = (x – y) + z

Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

Sum and difference Textbook Page No. 44

What happens when we add the sum and difference of two numbers?
The difference is the smaller number subtracted from the larger; the sum is the small number added to the larger.

For example, taking the numbers 7 and 3, the sum is 7 + 3 and the difference is 7 – 3. Without reducing these to 10 and 4, if we write their sum, we get
(7 + 3) + (7 – 3)
Here the larger number 7 is added twice; the smaller 3 is added once and then subtracted.
So that net result is 7 + 7 = 14.
In other words, by changing the order of operations, we find
(7 + 3) + (7 – 3) = (7 + 7) + (3 – 3) = 14
It is this fact that we write as the general algebraic rule,
(x + y) + (x – y) = (x + x) + (y – y) = 2x

Different ways

Look at this problem:
The price of a book and pen together is 16 rupees. The price of the book is 10 rupees more than that of the pen. What is the price of each?
If we forget about books and pens and look at the prices as mere numbers, the problem is like this:

The sum of two numbers is 16 and their difference is 10. What are the numbers?
Double the larger number is 16 + 10 = 26; so the larger number is 13. And then the smaller number is 16 – 13 = 3. Thus the price of the book is 13 rupees and the price of the pen is 3 rupees.

There is another way. Instead of a book and pen, suppose two books are bought. Since the price of the book is 10 rupees more, we would have to give 10 more rupees; that is, 16 + 10 = 26 rupees.
This is the price of two books. So, the price of one book is 13 rupees.

Calendar math Textbook Page No. 46

Take any month’s calendar and mark four numbers making a square:
Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations 2
In the picture, the sum of the chosen numbers is 8 + 9 + 15 + 16 = 48. Divide this by 4 and subtract 4. We get the first number 8, right?
Why does this happen?

Answer:
If we take the first number as x, the numbers marked are
Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations 3
Their sum is
x + (x + 1) + (x + 7)(x + 8) = 4x + 16
We can write this as
4x + 16 = (4 × x) + (4 × 4)
= 4(x + 4)
This is adding 4 to the first number and then multiplying by 4.
So to get back the first number we need to divide by 4 and then subtract 4.

Another trick
Take any month’s calender and mark nine numbers in a square, instead of four.
Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations 5
Their sum in the picture above is 144. And it is 9 times the middle number 16.
Mark other numbers like this and check whether this happens everytime.
To see why this is so, take the middle number as x. We can fill in the other numbers like this.
Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations 6
In this, if we match up pairs like x – 8, x + 8, we can see without any computation that the sum is 9x. That is 9 times the middle number.

Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

Measures and numbers Textbook Page No. 36

One side of a square is 3 centimeters. What is its perimeter?
Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations 1
What about a square of side 5 centimeters?
The perimeter of any square is four times the length of a side, right? And don’t you remember how we wrote it in short form using letters?

If we write s for the length of a side of a square and p for the perimeter, then we can write
p = 4 × s
We also know that when we write such a relation between numbers using letters, we don’t write the multiplication symbol × (and why). So, we write the relation between the length of side s and perimeter p of a square as
p = 4s

What about a rectangle, instead of a square?
If we know the lengths of two unequal sides, how do we find out the perimeter?
If we denote the length of each side as l and b and the perimeter as p, how do we write the relation between p, l, and b?
How do we write the relation between the lengths of sides of a rectangle and its area using letters?

Answer: Given a square figure with a side of 3cm.
Next, the side of 5cm square is, P = 4 x 5 = 20 sq.cm
The perimeter of a square is 4 x S.
The length of two unequal sides is called a rectangle. By using the formula, we can find the perimeter.
The formula for the perimeter of a rectangle is P = 2(l+b). This is a relationship between p, l, and b.
The area of a rectangle is A = l x b.

Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

Number relations

Look at these sums:
1 + 2 = 3
2 + 3 = 5
3 + 4 = 7
We add two consecutive natural numbers.
Now look at these:
(2 × 1) + 1 = 3
(2 × 2) + 1 = 5
(2 × 3) + 1 = 7
We double natural numbers and add 1.
How come we end up with the same numbers?
Let’s take a natural number and do the first operation. For example, if we start with 7, the next natural number is 8; and the sum
7 + 8 = 15
Suppose we write the 8 here as 7 + 1? We see that
7 + 7 + 1 = (2 × 7) + 1 = 15
This we can do for any natural number instead of 7.

If we take any natural number and add the next natural number, or if we double the first number and add 1, we get the same number as the result.

Can we do this only with natural numbers?

Answer: Let us consider that, the natural number is x.
Add the next natural number as , x+1 i.e., x+(x+1).
If we double the first number is 2x.
Add one to the first number i.e, 2x+1
We get the same result which this equation.
The equation is x+(x+1) = 2x+1.
Consider the x value as 1. Then the equation is,
1+(1+1) = 2(1) + 1
1+(2) = 2+1
3 =3.
Observed that, we get the same values.

For example let’s take the fraction, half. There is no meaning in saying the next fraction; but we can say, the number got by adding one to it. That is, half and one make one and a half. The half we started with, together with the one and a half we got now makes two.
On the other hand, doubling half makes one; and one added to it makes 2. That is,
\(\frac{1}{2}\) + 1\(\frac{1}{2}\) = \(\frac{1}{2}\) + (\(\frac{1}{2}\) + 1) = (2 × \(\frac{1}{2}\)) + 1

This computation is right, whatever fraction we start with. So we can extend the fact given above.

If we add to any number, one more than itself, or if we double the number and add one to it, we get the same number as the result.

This general property of numbers can be denoted in a shorter form using letters. For this, let’s write x for the number we start with. One added to it is x + 1; and the sum of these is x + (x + 1). Next, x doubled is written as 2x; and one added to this is 2x + 1. So the general property we have discovered can be written thus;
x + (x + 1) = 2x + 1, for every number x.

This mathematical shorthand for writing number-related facts using letters is called algebra.
Let’s look at a simple example. Suppose, to one number we add another and then subtract the added number. What do we have now? The first number back, right?

If we take the first number as x and the added number (which is later subtracted) as y, then we can write in algebra what happens as
(x +y) – y = x, for all numbers x, y.

Note that this is a general principle, which applies to all numbers. Certain properties hold only for specific numbers; for example both 2 + 2 and 2 x 2 give 4. But x + x = x x x is not a general principle (if we replace 3 for 2, it becomes incorrect).

Now do each of the following operations on several numbers and describe the results in a different form. Write each such relation in ordinary language. Then write it as an algebraic expression, using letters.

  • Add to a number, two more than the number.
  • Add one to a number and subtract two.
  • From a number, subtract another, and then add twice the subtracted number.
  • Add to a number the double of itself.
  • Add two consecutive natural numbers and find the number, one less than this.
  • Add two consecutive odd numbers and subtract the even number in their middle.
  • To a number, add another and subtract the first.
  • To a number, add another, and then add this sum to the first number.
  • Subtract two times a number from five times the number.
  • Add two times a number with three times the same number.

Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

However, we add

What is 38 + 25 + 75?
We can add in the given order:
38 + 25 = 63
63 + 75 = 138
We can also add like this:
25 + 75 = 100
38 + 100 = 138
We don’t need pen and paper to do it the second way. Now try this:
29 + \(\frac{1}{3}\) + \(\frac{2}{3}\)
To do this easily, which two numbers would you add first?
What do we see from these two sums?
Answer:

To find the sum of three numbers, either we can find the sum of the first two and add this to the third, or we can find the sum of the last two and add this to the first. This we can state in another form

Instead of adding to one number, two numbers one after another, we need only add their sum.

We can show the order of operations using brackets. For example, we can write the first sum like this:
(38 + 25) + 75 = 38 + (25 + 75)
And the second sum like this
(29 + \(\frac{1}{3}\)) + \(\frac{2}{3}\) = 29 + (\(\frac{1}{3}\) + \(\frac{2}{3}\))
So we can write the general principle of adding three numbers using algebra:
(x + y) + z = x + (y + z), for all number x, y, z.
Now suppose we went to calculate 36 + 25 + 64.
Isn’t it easier to add 36 and 64 first?
First write 25 + 64 as 64 + 25 and then write the full sum as (36 + 64) + 25.
That is, the addition of numbers can be done in any order.

Now try to find these sums in your head:

• 49 + 125 + 75
Answer: 249

Explanation: Given the numbers 49, 125, and 75.
Now, we will find the sum value.
The principle for these numbers is,
(x+y)+z = x+(y+z)
So, the values is (49+125)+75 = 49+(125+75)
174+75 = 49+(200)
249 = 249

• 347 + 63 + 37
Answer: 447

Explanation: Given the numbers 347, 63, and 37.
Now, we will find the sum value.
The principle for these numbers is,
(x+y)+z = x+(y+z)
So, the values is (347+63)+37 = 347+(63+37)
410+37 = 347+(100)
447= 447

• 88 + 72 + 12
Answer: 172

Explanation: Given the numbers 88, 72, and 12.
Now, we will find the sum value.
The principle for these numbers is,
(x+y)+z = x+(y+z)
So, the values is (88+72)+12 = 88+(72+12)
160+12 = 88+(84)
172 = 172

Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

• \(\frac{1}{4}\) + 1 \(\frac{3}{4}\) + 2
Answer: 4

Explanation: Given the numbers 1/4, (1)(3/4), and 2.
Now, we will find the sum value.
1/4+ 7/4+2.
The principle for these numbers is,
(x+y)+z = x+(y+z)
So, the value is (1/4+7/4)+2 = 1/4+(7/4+2)
8/4+2 = 1/4+(15/4)
2+2 = 16/4
4 = 4

• 15.5 + 0.25 + 0.75
Answer: 16.5

Explanation: Given the numbers 15.5, 0.25, and 0.75.
Now, we will find the sum value.
Using the principle for these numbers is,
(x+y)+z = x+(y+z)
So, the values is (15.5+0.25)+ 0.75 = 15.5+(0.25+0.75)
15.30+0.75 = 15.5+(1)
16.5 = 16.5

• 8.2 + 3.6 + 6.4
Answer: 18.2

Explanation: Given the numbers 8.2, 3.6, and 6.4.
Now, we will find the sum value.
The principle for these numbers is,
(x+y)+z = x+(y+z)
So, the values are (8.2+3.6)+6.4= 8.2+(3.6+6.4)
11.8+6.4 = 8.2+(10.0)
18.2 = 18.2

Addition and subtraction Textbook Page No. 40

We have seen the general principle of adding three numbers.
What if we subtract repeatedly, instead of adding?
Answer: Instead of adding, if we subtract three numbers we get the difference value.

Look at this problem:
Unni had 500 rupees with him and gave 100 rupees to Appu. Sometime later, Abu borrowed 50 rupees from him. Now, how much money does Unni have?
After the loan to Appu, he had
500 – 150 = 350 rupees
And after lending money to Abu, he had
350 – 50 = 300 rupees

We can also think in a different way. The total money he lent is
150 + 50 = 200 rupees
So what he finally has is
500 – 200 = 300 rupees
In other words, whether we do (500 – 150) – 50 or 500 – (150 + 50), we get the same number as the result. Similarly, can you do this in your head?
218 – 20 – 80
How can we state what we have seen here as a general principle?
Answer: The general principle for subtracting 3 numbers is (x-y)-z = x-(y+z), then we get an equal value (L.H.S = R.H.S).

Instead of subtracting from one number, two numbers one after another, we need only subtract the sum of these two numbers.
And in algebra?
(x – y) – z = x – (y + z), for all numbers x, y, z.
Instead of adding or subtracting two numbers in succession, suppose we add one number and subtract another.
Look at this problem.
There were 38 children when the class started. 5 came in late. Sometime later, 3 went to attend the Math Club meeting. How many are in the class now?
Let’s do this in the order of events. When 5 more joined, there were
38 + 5 = 43

And when 3 children left, there were 43 – 3 = 40
If instead, we look at the events together, we can compute like this: 5 children came in and 3 others left. So the final increase in number is only
5 – 3 = 2
In the beginning, there were 38 children.
So the total number now is
38 + 2 = 40

Thus, instead of adding one number and then subtracting another, we can subtract the second number from the first and add. For example,
(108 + 25) – 15 = 108 + (25 – 15) = 118
We should be a bit careful here. To compute like this, the number subtracted should be less than the number added. For example, look at this problem:
25 + 10 – 15
To do this, we cannot first subtract 15 from 10.
In algebra
(x + y) – z = x + (y – z) , for all numbers x, y, z with y > z.

Using these ideas, try to do these problems without pen and paper:

• (135 – 73) – 27
Answer: 35

Explanation: Given the numbers 135, 73, and 27.
Now, we will find the difference value.
The principle for this expression is,
(x-y)-z = x-(y+z)
So, the values are (135-73)-27= 13-(73+27)
62-27 = 135-(100)
35 = 35
The value is 35.

• (37 – 1\(\frac{1}{2}\)) – \(\frac{1}{2}\)
Answer: 35

Explanation: Given the numbers 37, (1)(1/2), and 1/2.
Now, we will find the difference value.
The value of (1)(1/2) in fraction form is 3/2
The principle for this given expression is,
(x-y)-z = x-(y+z)
So, the values are (37-3/2)-1/2= 37-(3/2+1/2)
(71-1)/2 = 37-(4/2)
70/2 = (74-4)/2
35 = 35

• (298 – 4.5) – 3.5
Answer: 290

Explanation: Given the numbers 298, 4.5, and 3.5.
Now, we will find the difference value.
The principle for this given expression is,
(x-y)-z = x-(y+z)
So, the values are (298-4.5)-3.5
293.5 – 3.5
290.

• (128 + 79) – 29
Answer: 178

Explanation: Given the numbers 128, 79, and 29.
Now, we will find the value.
The principle for a given expression is, (x+y)-z  where y>z.
So, the values are (128+79)-29
207 – 29
178.

• (298 + 4.5) – 3.5
Answer: 299

Explanation: Given the numbers 298, 4.5, and 3.5.
Now, we will find the value.
The principle for this expression is, (x+y)-z  where y>z.
So, the values are (298+4.5)-3.5
302.5 – 3.5
299

• (149 + 3\(\frac{1}{2}\)) – 2 \(\frac{1}{2}\)
Answer: 150

Explanation: Given the numbers 149, 3(1/2), and (2)1/2.
Now, we will find the value.
The principle for this expression is, (x+y)-z  where y>z.
So, the values are (149+ 7/2)-(2)1/2
149 + 7/2 -5/2
149+2/2
149+1 = 150.

Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

Subtracting and adding

Look at this problem.
Gopu had 110 rupees in his savings box. He took out 15 rupees to buy a pen. He got a pen for 10 rupees. He
returned 5 rupees to the box. How much is in the box now?

Let’s first compute in the order of what Gopu did. When he took out 15 rupees, the box had
110 – 15 = 95 rupees
Since he put back 5 rupees, the box now has 95 + 5 = 100 rupees
After all, this have happened, we can also think like this: he took out 15 rupees and put back 5 rupees; so the actual decrease in the box is only
15 – 5 = 10 rupees
So the box now has
110 – 10 = 100 rupees
Writing the first computation as (110 – 15) + 5 and the second as 110 – (15 – 5), what we see is that (110 – 15) + 5 = 110 – (15 – 5)
That is, instead of subtracting a number and then adding another, we need only subtract the difference of the second from the first. For example,
(29 – 17) + 7 = 29 – (17 – 7) = 19
Can we do this in all problems of subtracting and then adding? For example, can we rewrite
(29 – 7) + 17
in this manner?
So, the change of operation is written in algebra as
(x – y) + z = x – (y – z) , for all numbers x, y, z with y > z

Now use this idea to calculate these without pen and paper.

• (135 – 73) + 23
Answer: 85

Explanation: Given the numbers 135, 73, and 23.
Now, we have to find the value.
The principle for these equation is, (x-y)+z = x-(y-z).
So, after substituting the values  (135-73)+23
62 +23
85.

• (38 – 8\(\frac{1}{2}\)) + \(\frac{1}{2}\)
Answer: 30

Explanation: Given the numbers 38, 8(1/2), and 1/2.
Now, we have to find the value.
The principle for these equation is, (x-y)+z = x-(y-z).
So, after substituting the values  (38-17/2)+1/2
76 -17/2 +1/2
59/2+1/2
60/2
30.

• (19 – 6.5) + 5.5
Answer: 18

Explanation: Given the numbers 19, 6.5, and 5.5.
Now, we have to find the value.
The principle for these equation is, (x-y)+z = x-(y-z).
So, after substituting the values  (19-6.5)+5.5
12.5 +5.5
18

• (135 – 6.5) + 5.5
Answer: 134

Explanation: Given the numbers 135, 6.5, and 5.5.
Now, we have to find the value.
The principle for these equation is, (x-y)+z = x-(y-z).
So, after substituting the values  (135-6.5)+5.5
128.5 +5.5
134.

• 135 – (35 – 18)
Answer: 118

Explanation: Given the numbers 135, 35, and 18.
Now, we have to find the value.
The principle for these equation is, x-(y-z) = (x-y)+z.
So, after substituting the values  135-(35-18)
135-(17)
118

• 4.2 – (3.2 – 2.3)
Answer: 3.3

Explanation: Given the numbers 4.2, 3.2, and 2.3.
Now, we have to find the value.
The principle for these equation is, x-(y-z) = (x-y)+z).
So, after substituting the values 4.2 -(3.2-2.3)
4.2 – 0.9
3.3

Sums and differences Textbook Page No. 44

Athulya often teases her classmates with her new discoveries. Today she had a new trick. “Think of any two numbers and give me their sum and difference; I can tell you the numbers.”
“Sum is 10 and difference is 2”, Rahim started.
“Easy! numbers are 6 and 4”, Athulya said.
“Sum 16, difference 5”, mischievous Jessy challenged. Athulya thought for a moment and said,
“Nice try! Numbers are 10 \(\frac{1}{2}\) and 5\(\frac{1}{2}\) ”
How did Athulya find the numbers?
How do we find two numbers, using their sum and difference?

Let’s take the numbers x and y. Then the sum is x + y. If we take the larger number as x, the difference is x – y. Using these, we have to find x and y.
To find x from x + y, we need only subtract y:
(x + y) – y = x
But we don’t know y.
What if we add x again?
(x + y) – y + x = x + x = 2x
Subtracting y and then adding x is the same as adding x and then subtracting y, right?
(x + y) + (x – y) = 2x
What does this mean?
Answer: Adding sum and difference gives twice the larger number.

For example, Rahim’s sum is 10 and the difference is 2. Their sum is 12. This is twice his larger number. So the larger number is 6, and his smaller number is 10 – 6 = 4.

Now let’s look at what Jessy said: sum 16, difference 5; their sum is 21. So the larger number is half of this and so it is 10 \(\frac{1}{2}\) ; the smaller is 16 – 10 \(\frac{1}{2}\) = 5 \(\frac{1}{2}\).
Do you get Athulya’s trick?
We can see another thing here. Subtract the difference from the sum:
(x + y) – (x – y) =(x+y)-x+y
= x + y – x + y = x – x + y + y = 2y
What does this mean?
Answer: Subtracting the difference from the sum gives twice the smaller number.

For example, in Rahim’s case, the sum is 10, and the difference is 2. So, double the smaller number is 10 – 2 = 8, and so the smaller number is half of this, which is 4.
The sum and difference of some pairs of numbers are given below. Can you find the numbers?

  • Sum 12 difference 8

Answer: 2
Explanation: Given that, the sum is 12 and the difference is 8.
The smaller double number is 12 – 8 = 4.
The smaller number is half of this, which is 2.

  • Sum 140, difference 80

Answer: 30
Explanation: Given that, the sum is 140 and the difference is 80.
The smaller double number is 140 – 80 = 60.
The smaller number is half of this, which is 30.

  • Sum 23, difference 11

Answer: 6
Explanation: Given that, the sum is 23 and the difference is 11.
The smaller double number is 23 – 11 = 12.
The smaller number is half of this, which is 6.

  • Sum 20, difference 5

Answer: 7.5
Explanation: Given that, the sum is 20 and the difference is 5.
The smaller double number is 20 – 5 = 15.
The smaller number is half of this, which is 7.5.

Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations

Addition and multiplication

We have seen that twice a number is added to thrice the number gives five times the number. (The last problem in the section Number relations)
What is the algebraic form of this statement?
2x + 3x = 5x, for every number x.

This we can state in another manner.
Instead of multiplying a number by 2 and 3 separately and adding, we need only multiply this number by 5.
For example,
(2 × 16) + (3 × 16) = 5 × 16 = 80
In this, suppose we take other numbers instead of 2 and 3?
Look at this problem:

In a math conference, two rooms are used for discussion. In one room there are 40 people and in the other, 35. At tea-time each is to be given 2 biscuits. How many biscuits are needed?
In the first room, we need
40 × 2 = 80
And in the second,
35 × 2 = 70
So altogether, we need
80 + 70 = 150
We can also think like this: The total number of people is
40 + 35 = 75
So the number of biscuits needed is
75 × 2 = 150
What do we see here? Instead of multiplying 2 by 40 and 35 separately and then adding, we need only multiply their sum 75 by 2.
This can be done in multiplication by fractions also. For example, if we add half of 4 and half of 6, we get 2 + 3 = 5; half of the sum 10 is also 5.

What general principle do we see in all these?

Multiplying two numbers by a number separately and adding give the same result as multiplying their sum by the number.

How do we write this in algebra?
xz + yz = (x + y) z, for all numbers x, y, z
What about subtraction?
Multiplying two numbers by a number separately and subtracting give the same result as multiplying their difference by the number.
In algebra,
xz – yz = (x – y) z, for all numbers x, y, z

Now try these problems.

• (63 × 12) + (37 × 12)
Answer: 120

Explanation: Given the expression is (63 x 12) + (37 x 12)
The expression is in the form of xz + yz.
The algebraic form of xz + yz = (x+y)z.
Based on this, the given expression is,
(63 x 12) + (37 x 12) = (63+37) x 12
100 x 12 = 120.
So, the final value is 120.

• (15 × \(\frac{3}{4}\)) + (5 × \(\frac{3}{4}\))
Answer: 15

Explanation: Given the expression is (15 x 3/4) + (5 x 3/4)
The expression is in the form of xz + yz.
The algebraic form of xz + yz = (x+y)z.
Based on this, the given expression is,
(15 x 3/4) + (5 x 3/4) = (15+5) x 3/4
20 x 3/4 = 5 x3 = 15
So, the final value is 15.

• (\(\frac{1}{3}\) × 20) + (\(\frac{2}{3}\) × 20)
Answer: 20

Explanation: Given the expression is (1/3 x 20) + (2/3 x 20)
The expression is in the form of xz + yz.
The algebraic form of xz + yz = (x+y)z.
Based on this, the given expression is,
(1/3 x 20) + (2/3 x 20) = (1/3+2/3) x 20
(3/3) x 20 = 1 x 20= 20
Hence, the final value is 20.

• (65 × 11) – (55 × 11)
Answer: 110

Explanation: Given the expression is (65 x 11) – (55 x 11)
The expression is in the form of xz – yz.
The algebraic form of xz – yz = (x-y)z.
Based on this, the given expression is,
(65 x 11) – (55 x 11) = (65-55) x 11
10 x 11 = 110
Therefore, the value is 110.

• (2\(\frac{1}{2}\) × 23) – (1\(\frac{1}{2}\) × 23)
Answer: 23

Explanation: Given the expression is (2(1/2) x 23) – (1(1/2) x 23)
The expression is in the form of xz – yz.
The algebraic form of xz – yz = (x-y)z.
Based on this, the given expression is,
(2(1/2) x 23) – (1(1/2) x 23) = (5/2-3/2) x 23
2/2 x 23 = 1 x 23 = 23.
Therefore, the value is 23

• (13.5 × 40) – (3.5 × 40)
Answer: 400

Explanation: Given the expression is (13.5 x 40) – (3.5 x 40)
The expression is in the form of xz – yz.
The algebraic form of xz – yz = (x-y)z.
Based on this, the given expression is,
(13.5 x 40) – (3.5 x 40) = (13.5-3.5) x 40
10 x 40 = 400
Therefore, the value is 400.

Let’s do it! Textbook Page No. 47

Question 1.
From the square below, take any 9 numbers making a square. Find the relation between their sum and the number in the middle of the square. Justify this relation using algebra.
Kerala Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations 4
Now try this with 25 numbers in a square.

Answer: (i) Given the square, which consists of 1 to 36 numbers.
First, take 9 numbers in given square box numbers.
The taken 9 numbers are below,


Now, we will find the sum of those 9 numbers.
Then the sum is, 9+10+11+15+16+17+21+22+23 = 144.
Now, the middle number is 16.
So, 9 times the middle number is 16 x 9 = 144 (Because the table consists of 9 numbers).
The relation between the middle number and the sum is, that the sum of 9 numbers in the given square is 9 times the middle number.

(ii) Now, try this with 25 numbers in a square box.
The 25 numbers are,
Kerala State Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles_Table(i)
Now, we will find the sum of 25 numbers.
Then the sum is, 1+2+3+4+5+7+8+9+10+11+13+14+15+16+17+19+20+21+22+23+25+26+27+28+29= 375.
Now, the middle number is 15.
So, 25 times the middle number is 25 x 15 = 375 (Because the table consists of 25 numbers).
Therefore the relation between the middle number and the sum, that is the sum of 25 numbers in the given square box is 25 times the middle number.

Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts

You can Download Pie Charts Questions and Answers, Activity, Notes, Kerala Syllabus 7th Standard Maths Solutions Chapter 14 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts

Pie Charts Text Book Questions and Answers

Pie charts Textbook Page No. 188

Pie charts are usually drawn to show a single numerical fact split into categories and to makes comparisons between these.
Such a picture is drawn with the size of each part depending on the number it represents.

Food for thought

Pie is the name of a baked dish very much popular in England and USA.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 4
It is divided and shared by cutting into slices as shown in the picture. The name pie chart originates from this.

Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts

Bar and Circle Textbook Page No. 190

The household expenses of Renu’s family is shown as a bar graph and as a pie-chart below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 5

Look at the bar graph. We can easily see the amounts spent under each head and compare one another, but we cannot directly see what fraction of the total each is. That is easier to see in the pie chart. But the actual amount of each is not easy to see in it. Thus, each type of picture has certain advantages and defects.

Let’s make a table

All students of Class 7 in a school are in one of the various clubs. A pie chart showing the number of students in each club is shown below.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 8
There are 50 children in Vidyarangam. Make a table showing the number of children in each club.
Answer:
Given data:
Number of children in vidyarangam=50
Solution:
From the pie chart, we can observe that science club is half of the vidyarangam club. Therefore, number of children in science club will be \(\frac{1}{2}\)×50=25.
Number of children in social and english club seems to be equal to the science club. Therefore, number of children in social and english club will be 25.
Number of children in maths club are three times of the science club. Therefore, number of children in maths club will be 3×25=75.
From our assumptions, we can make the below table.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_14

Piechart with computers Textbook Page No. 192

Pie charts can be easily drawn using a computer.

Open Libre Office Calc program and type in the details as below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 9
Click on any cell and choose Insert → chart → pie. This makes a pie-chart. Change the numbers and see what happens to the picture.
Answer:
Given data:
From the above table, we get the number of students from each club and the total number of students belonging to all the clubs.
Total number of students: 30+20+25+15+10=100
To draw a pie chart, we need to calculate the angle of each sector.
Formula:
Angle of sector = 360º×\(\frac{Given value}{Total sum of all the values}\)
Solution:
Angle of sector for maths club is calculated as 360º×\(\frac{30}{100}\) = 108º
Angle of sector for science club is calculated as 360º× \(\frac{20}{100}\) = 72º
Angle of sector for social science club is calculated as 360º× \(\frac{25}{100}\) = 90º
Angle of sector for vidhyarangam club is calculated as 360º×\(\frac{15}{100}\) = 54º
Angle of sector for english club is calculated as 360º×\(\frac{10}{100}\) = 36º
Steps of Construction:
Step 1: Draw a circle with any convenient radius and mark the center as ‘O’.
Step 2: Next, mark a point A anywhere on the circumference of the circle and join OA.
Step 3: To represent maths club sector, construct ∟AOB=108º.
Step 4: To represent science club sector, construct ∟BOC=72º.
Step 5: To represent social science club sector, construct ∟COD=90º.
Step 6: To represent vidyarangam club sector, construct ∟DOE=54º.
Step 7: To represent english club sector, construct ∟EOA=36º.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_3

Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts

Electricity

The pie chart below shows the details of electricity supplied by the Kerala State Electricity Board during 2011-2012.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 13
What all facts do we get from this?
Answer:
The above pie chart depicts supplied electricity in different sectors.
From the given pie chart, it is clearly observed that the electricity supplied to Household sector is more as it covers the major part when compared to others.
Industrial sector is on second place, Commercial on third place, other sectors on fourth place and agriculture sector consumes very less than the remaining sectors.
Conclusion:
From the observation, more electricity is supplied to household sector and while very less is supplied to agriculture sector.
The pie chart given beLow shows the revenue of Kerala State Electricity Board, during 2011-2012.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 14
What all information do we get from this?
Answer:
From the observation, we can say that the revenue generated from Industrial sector is more and agriculture sector is very less.
Compare the two charts.
Answer:
By comparing the above two pie charts,
We can observe that more electricity is supplied to household while most of the revenue is generated from industrial sector.
Approximately, same revenue is generated from other groups.
The electricity consumed and revenue generated by agriculture sector is less than all the other sectors.

Time management Textbook Page No. 194

Given below is the pie-chart showing how Aravind, a class 7 student spends his time:
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 15
The circle is divided into 24 parts, each part indicating one hour.
What each colour shows is given below.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 16
Draw a bar graph showing these details.
Answer:
Given that circle is divided into 24 parts representing 5 different groups. By observing the above pie chart, we can calculate the number of parts covered under each sector.
Step 1: Number of parts covered under school are 6.
Step 2: Number of parts covered under sleep are 8.
Step 3: Number of parts covered under studies are 4.
Step 4: Number of parts covered under play/exercise are 3.
Step 5: Number of parts covered under other activities are 3.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_5

Change to circle

The bar graph below shows the number of girls in the three divisions of Class 7 in a school.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 19
Draw a pie chart of this.
Answer:
To draw a pie chart, we need to calculate the angles for each sector. To find the angles of each division we can use the below mentioned formula.
Formula:
Angle of sector = 360º×\(\frac{Given value}{Total sum of all the values}\)
Given data:
The vertical line or y-axes represents the number count of the girls in each division and horizontal line or x-axes represents the division for which it belongs to.
Division A is marked completely till 20, so 20 girls fall under it.
Similarly, 25 girls belong to division B and 15 girls belong to division.
Total number of girls will be 20+25+15=60
Solution:
Division A: Angle of sector for A will be 360º× \(\frac{20}{60}\)=120º.
Division B: Angle of sector for B will be 360º× \(\frac{25}{60}\)=150º.
Division C: Angle of sector for C will be 360º× \(\frac{15}{60}\)=90º.
Steps of Construction:
Step 1: Draw a circle with any convenient radius and mark the center as ‘O’.
Step 2: Next, mark a point A anywhere on the circumference of the circle and join OA.
Step 3: To represent division A, construct ∟AOB=120º.
Step 4: To represent division B, construct ∟BOC=150º.
Step 5: To represent division C, construct ∟COA=90º.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_4

Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts

Election Textbook Page No. 188

The picture below shown the votes each candidate got in a school election.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 1

• Who won the election?
Answer:
By observing the pie chart, we can say that Maria won the election as major part of the votes are covered under her sector.

• What all information do you get from this picture?
Answer:
From the picture, we can rank the candidates by observing the part covered under each of them.Maria ranks first and thus wons the election. Nazia is on the second place as the second largest number of votes are covered under her and Remya is on the third place, as she got less number of votes when compared to the other two.

House expenses

Look at this picture showing various expenses at Fathima’s home:
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 2
For what, is the most amount spent?
Answer:
By observing the pie chart, we can say that major amount was spent on other activities as the major part of the pie is covered under  it.
And the least?
Answer:
Least amount was spent on the education sector as the least part was covered under it.
For what all are the same amount spent?
Answer:
Same amount was spent on charity and food as both covers the equal part of pie.
How do we know that the same amount is spent on these?
Answer:
We can know the areas where same amount is spent by calculating the angle of sector which can be known by using the below formula:
360º×\(\frac{Given value}{Total sum of all the values}\)
After calculating the sector angles/values, if any of the sector angle is same as other than we can say that same amount was spent on those groups.
What more do we get from the pictures?
Answer:
We can get information such as,
Easy calculation of expenses made for different groups by using the above mentioned formula.
Easy understanding of major and minor expense made on groups by observing the pie chart.
We can get the order of the groups expenditure which is helpful to sort and plan the expenses accordingly.

A picture like this, which display information as parts of a circle is called pie chart.

Occupation

The pie chart below shows the various occupations of people in a panchayath.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 3

• What is the occupation of the most number of people?
Answer:
Labourers are the most number of people as the major part of circle is occupied by these group of people.

• Roughly, how many times the number of farmers is the number of labourers?
Answer:
By observing the areas covered under farmers and labourers, we can assume that the number of labourers are double the number of farmers.

• Roughly what fraction of the total is the number of factory workers?
Answer:
Let us make an assumption that the pie chart is divided into 6 equal sectors, labourers almost occupies 2 parts among the total parts.Therefore,we can assume that two-sixth of the complete circle is occupied by them.

• Arrange the occupations in the order of the number of people in each.
Answer:
By observing the pie chart, the increasing order of working people will be others,traders,framers ,government employees,factory workers and labourers.

Make some more questions relating to this picture.
Answer:
Which of the occupation has less number of people?
Which of the occupations has same number of people?
How can we say that the occupations are having the same number of people?

Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts

Agriculture Textbook Page No. 190

The pie chart below shows how the total farmland in a panchayat is used for various crops.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 6

Based on this figure, answer the following questions.

• For which is the least land used?
Answer:
By observing the figure, we can say that rice occupies least land area when compared to the remaining crops as the part of area covered under it is less than the remaining.

• For which is the most land used?
Answer:
By observing the figure, we can assume that the most of the land is used for other crops and vegetables as the major portion of the circle is covered under it.

• Roughly, what fraction of the total land is used for vegetables?
Answer:
Let us make an assumption that the circle is divided into four major sectors. First sector represents the rubber crop, while second sector represents Plantain, Tapioca, Coconut and Rice.Third sector represents vegetables and fourth other crops.
Therefore, we can assume that one-fourth of the land is utilized for vegetables.

Let’s draw!

It was decided to have a vegetable garden in the school. Amaranth in half the plot and beans and brinjal in equal
parts of the other half. Let’s draw a pie chart showing this;
First draw a circle.
Half the plot is for amaranth.

How do we show this in our picture?
Answer:
360º×\(\frac{1}{2}\)=180º, which is half of the complete circle.
How do we mark half the circle?
Answer:
Draw a circle with any convenient radius and mark the center as ‘O’.
To mark half of the circle, construct ∟AOB=180º.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 7
Now to show the parts for beans and brinjal one the circle must be halved again.
Can’t you do it?
Answer:
Given that:
Beans and brinjal in equal parts of the other half which is 180º×\(\frac{1}{2}\)=90º
Also colour each part to tell these apart.
Answer:
To mark beans and brinjal in the circle, construct ∟BOC=90º AND ∟COA=90º.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_6

Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts

Travel math Textbook Page No. 191

There are 40 students in class 7 A of a school. 20 of them come in the school bus, 5 of them walk and 5 ride bicycles
to school.
Let’s draw a pie chart showing these.
What fraction of children take the school bus?
Answer:
Fraction of students who take school bus will be \(\frac{No.of students who prefer school bus}{Total students}\)=\(\frac{20}{40}\).So, \(\frac{1}{2}\) of total students take school bus.
This we can mark in a circle as before.
What fraction of children ride bicycles?
Answer:
The number of students who ride bicycles will be \(\frac{5}{40}\)=\(\frac{1}{8}\)
How do we mark \(\frac{1}{8}\) of the circle?
What angle should we draw for this?
\(\frac{1}{8}\) of 360° = 45°

Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 10
The remaining part of the circle shows the number of children who walk to school.
• What fraction of the circle is this?
Answer:
Given that 20 of them come to school by school bus,5 of them by bicycle and 5 of them by walk and the total number of students in 7A class are 40.
Solution:
Students who come by walk will be the difference of total students from students coming from other modes.So,the total number of students who come by walk will be 40-(20+5)=15
Fraction of the students who come by walk will be \(\frac{15}{40}\)=\(\frac{3}{8}\)
So, \(\frac{3}{8}\) of the total students come by walk.

• How much degrees is its angle?
Answer:
Angle of sector for walk will be 360º×\(\frac{3}{8}\)=135º.

School clubs

In a school, there are 100 students in Class 7 and each one is a member of some club.
The table shows the number of students in various clubs.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 11
We want to draw a pie chart of this.
What all fractions of the circle should we mark to show the number of members of various clubs?
There are loo students altogether.
Math club has 30 numbers.
To show this, we must mark \(\frac{30}{100}\) of the circle.
What angle should we draw to get this?
360° × \(\frac{30}{100}\) = 108°
Answer:
Similarly, what are the angles we have to draw to show the numbers in other clubs?
Science : 360° × \(\frac{20}{100}\) = 72°
Social Science : 360° × \(\frac{25}{100}\) = 90º
English : 360° × \(\frac{10}{100}\) = 36º
Vidhyarangam : 360° × \(\frac{15}{100}\) = 54º
Now we can draw the pie chart:
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 12

Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts

Grade chart Textbook Page No. 193

In Class 7 ofa school, 25% got A grade, 45% got B, 20% got C and the remaining D. Let’s draw a pie chart of this.
Let’s compute the fractions of circles to show those who got different grades:
25% got grade A.
We must use 25% of the circle to show this.
360 × \(\frac{25}{100}\) = 90°

40 % got B, and the angle is
= 360° × \(\frac{45}{100}\) = 162°
Can’t you compute the angles to be drawn to show those who got C and D grades, and draw the pie chart?
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 17
Collect such information about your class and other classes in your school and make pie charts. Display these in Maths lab.
Answer:
20% got C, the angle is 360° × \(\frac{20}{100}\)=72º.
Given that 25% got A grade, 45% got B, 20% got C.
Let us assume 100 as the total percentage, so the percentage D got will be 100-(25+45+20)=10%.
So, the angle will be 360° × \(\frac{10}{100}\)=36º

Now draw pie charts of the following

• The final match of the school cricket tournament was between Ramanujan House and C.V. Rarnan House. The details of the scores are given below. Draw a pie chart of each:
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts 18
Answer:
Given table has the data of the school cricket tournament between Ramanujan House and C.V. Rarnan House.
Formula used to find the angle of sector made by each player:
360º×\(\frac{Given value}{Total sum of all the values}\)

C.V.Raman House:
Solution:
Runs made by Jishnu will be 360º× \(\frac{56}{140}\)=144º
Runs made by Abhi will be 360º× \(\frac{35}{140}\)=90º
Runs made by Sachu will be 360º× \(\frac{7}{140}\)=18º
Runs made by Ajmal will be 360º× \(\frac{21}{140}\)=54º
Runs made by other players will be 360º× \(\frac{21}{140}\)=54º
Steps of Construction:
Step 1: Draw a circle with any convenient radius and mark the center as ‘O’.
Step 2: Next, mark a point A anywhere on the circumference of the circle and join OA.
Step 3: To represent the runs made by Jishnu, construct ∟AOB=144º.
Step 4: To represent the runs made by Abhi, construct ∟BOC=90º.
Step 5: To represent the runs made by Sachu, construct ∟COD=18º.
Step 6: To represent the runs made by Ajmal, construct ∟DOE=54º.
Step 7: To represent the runs made by other players, construct ∟EOA=54º.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_7

Ramanujan House:
Solution:
Runs made by Ananthu will be 360º× \(\frac{72}{144}\)=180º
Runs made by Thoufiq will be 360º× \(\frac{36}{144}\)=90º
Runs made by Abhilash will be 360º× \(\frac{18}{144}\)=45º
Runs made by other players will be 360º× \(\frac{18}{144}\)=45º
Steps of Construction:
Step 1: Draw a circle with any convenient radius and mark the center as ‘O’.
Step 2: Next, mark a point A anywhere on the circumference of the circle and join OA.
Step 3: To represent the runs made by Ananthu, construct ∟AOB=180º.
Step 4: To represent the runs made by Thoufiq, construct ∟BOC=90º.
Step 5: To represent the runs made by Abhilash, construct ∟COD=45º.
Step 6: To represent the runs made by other players, construct ∟DOA=45º.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_8

• There are 1600 books in the school library, classified as shown below.
Short story – 320
Poetry – 192
Novel – 384
Science – 544
Biography – 160
Draw a pie chart showing this
Answer:
Given data depicts different classifications of 1600 books in library.
Formula:
360º×\(\frac{Given value}{Total sum of all the values}\)
Solution:
Angle of sector for short story books will be 360º×\(\frac{320}{1600}\)=72º.
Angle of sector for poetry books will be 360º×\(\frac{192}{1600}\)=43.2º.
Angle of sector for novel books will be 360º×\(\frac{384}{1600}\)=86.4º.
Angle of sector for science books will be 360º×\(\frac{544}{1600}\)=122.4º.
Angle of sector for biography books will be 360º×\(\frac{160}{1600}\)=36º.
Steps of Construction:
Step 1: Draw a circle with any convenient radius and mark the center as ‘O’.
Step 2: Next, mark a point A anywhere on the circumference of the circle and join OA.
Step 3: To represent the short story classification, construct ∟AOB=72º.
Step 4: To represent the poetry classification, construct ∟BOC=43.2º.
Step 5: To represent the novel classification, construct ∟COD=86.4º.
Step 6: To represent the science classification, construct ∟DOE=122.4º.
Step 7: To represent the biography classification, construct ∟EOA=36º.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_9

In a survey, the number of students who like diffèrent kinds of books were found as follows.
Short story – 84
Poetry – 36
Novel – 48
Science – 60
Biography – 12
Answer:
Given data depicts the number of students who likes different kinds of books.The total number of students are 240.
Formula:
360º×\(\frac{Given value}{Total sum of all the values}\)
Solution:
Angle of sector for short story books will be 360º×\(\frac{84}{240}\)=126º.
Angle of sector for poetry books will be 360º×\(\frac{36}{240}\)=54º.
Angle of sector for novel books will be 360º×\(\frac{48}{240}\)=72º.
Angle of sector for science books will be 360º×\(\frac{60}{1600}\)=90º.
Angle of sector for biography books will be 360º×\(\frac{12}{1600}\)=18º.
Steps of Construction:
Step 1: Draw a circle with any convenient radius and mark the center as ‘O’.
Step 2: Next, mark a point A anywhere on the circumference of the circle and join OA.
Step 3: To represent the number of students reading short story books, construct ∟AOB=126º.
Step 4: To represent the number of students reading poetry books, construct ∟BOC=54º.
Step 5: To represent the number of students reading novel books, construct ∟COD=72º.
Step 6: To represent the number of students reading science books, construct ∟DOE=90º.
Step 7: To represent the number of students reading biography books, construct ∟EOA=18º.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_10
Draw a pie chart of this also. Compare the two pie-charts.
Are books bought according to the preference of the students?
Answer:
By comparing both the pie charts, we can say that the short story and poetry books are not according to the preference. Short story and poetry books are less while the number of students preferring those classification are more.
Number of students who prefer short story book are 126 while the available books are 72.
Number of students who prefer poetry book are 54 while the available books are approximately 43.

• Collect various pictographs, bar graph and pie charts from various periodicals. Analyse and compare these.
Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_12
The above bar graph represents the newspaper read by different people in a society.
From the graph, we can observe that
The number of people preferring Hindi newspaper are 30.
The number of people preferring Gujarathi newspaper are 40.
The number of people preferring Tamil newspaper are 10.
The number of people preferring English newspaper are 50.
The number of people preferring Malayalam newspaper are 50.
So, the total number of people reading all the newspaper are 180.
The same can be represented in pie chart as,
Hindi newspaper: 360º×\(\frac{30}{180}\)=60º
Gujarati newspaper: 360º×\(\frac{40}{180}\)=80º
Tamil newspaper: 360º×\(\frac{10}{180}\)=20º
English newspaper: 360º×\(\frac{50}{180}\)=100º
Malayalam newspaper: 360º×\(\frac{50}{180}\)=100º
So, the pie chart for the above table will be:
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_13
By comparing both the charts
Both the data representation has its own advantage and disadvantages. Bar graph helps in reading the numeric data more easily by observing the vertical bar. While, the pie chart represents the proportions of each category by observing the slices.

• Make a pie chart of the number of sùidents in various classes of your school, using computer
Answer:
Data collected:
Class 1: 55
Class 2: 45
Class 3: 60
Class 4: 57
Class 5: 63
Class 6: 50
Class 7: 80
Class 8: 60
Class 9: 80
Class 10: 50
Total number of students are 600.
Formula:
360º×\(\frac{Given value}{Total sum of all the values}\)
Solution:
Class 1: 360º×\(\frac{55}{600}\)=33º
Class 2: 360º×\(\frac{45}{600}\)=27º
Class 3: 360º×\(\frac{60}{600}\)=36º
Class 4: 360º×\(\frac{65}{600}\)=39º
Class 5: 360º×\(\frac{55}{600}\)=33º
Class 6: 360º×\(\frac{50}{600}\)=30º
Class 7: 360º×\(\frac{80}{600}\)=48º
Class 8: 360º×\(\frac{60}{600}\)=36º
Class 9: 360º×\(\frac{75}{600}\)=45º
Class 10: 360º×\(\frac{65}{600}\)=39º
Pie chart for the above calculated angles can be drawn as shown below.
Kerala Syllabus 7th Standard Maths Solutions Chapter 14 Pie Charts img_11

Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math

You can Download Speed Math Questions and Answers, Activity, Notes, Kerala Syllabus 7th Standard Maths Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math

Speed Math Text Book Questions and Answers

Olympics 2012 Textbook Page No. 90

Look at the table showing the first five in men’s 100 metres sprint in the London Olympics
Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math 1
How much time do you take to run 100 metres?
Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math 3
Answer:
The time taken by me can be calculated by the following formula:
Speed = Distance ÷ Time.

Explanation:
Time taken to run in 100 meters sprint in the London Olympics:
speed = distance ÷ time.

Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math

Rotating earth

Do we ever remain still? The Earth which carries us all, turns all the time, spinning on its own and going round the Sun. It spins at about 1700 km/h and revolves round the Sun at about 100000 km/h.

Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math 5

Value of time Textbook Page No. 93

The smallest unit of time we usually use is a second. Sometimes we need to use smaller units such as a microsecond or a nanosecond. A microsecond is a millionth of a second. A nanosecond is a thousandth of a microsecond.
Do you know by what part of a second did P.T. Usha lose an Olympic medal?
Answer:
By 1/100th part of a second P.T. Usha lose an Olympic medal.

Explanation:
At the Games, she clocked 56.81 s in the heats and 55.54 s in the semi-final, setting a new Commonwealth record as she entered the final. At the final, she came fourth, at 55.42 seconds, falling behind the eventual bronze medalist by 1/100th of a second.

Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math 6

Look at the average speeds of some animals:
Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math 7

Over speeding

How much does a car doing 90 km/h move in a minute?
\(\frac{90}{60}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\) km
And in one second?
1\(\frac{1}{2}\) kilometers means 1500 metres.
\(\frac{1500}{60}\) = \(\frac{75}{3}\) = 25 metres
So what if the driver is one second late in applying the brakes? The vehicle would have moved 25 metres ahead.
Answer:
If the driver is one second late in applying the brakes, the vehicle would have moved 25 metres ahead in a slower speed.

Explanation:
If the driver is one second late in applying the brakes,the vehicle would have moved 25 metres ahead.
=> This shows that force can make a moving object go slower.

Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math

Road accidents Textbook Page No. 96

We read about traffic accidents every day in the papers. The main causes are speeding and carelessness. How many lives are lost on our roads everyday!
There is a law that heavy vehicles should in stall a speed-lock”. With this device, they can not go beyond a certain speed.
If each of us obey the traffic rules, we can reduce the number of accidents.

Who’s the fastest? Textbook Page No. 90

“We have to find the fastest runner in the school. How do we do it?” The teacher asked.
“Let everyone run 100 metres”, Raji said.
Reghu had another idea: “How about everyone running for 1 minute?”
They all went to the ground.
And all ran 100 metres.
These were the best four:
Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math 2
Who won the race?
Is it easy to conduct a race as Reghu suggested?
Answer:
Shyam won the race.
No, its not easy to conduct a race as Reghu suggested as its limited by time and distance and strength of students is more.

Explanation:
According to the speed chart of the students. Shyam has won the race as time taken by him is comparatively less than other students.
Its not easy to conduct a race as Reghu suggested because time limitation is high n distance to run is longer and speed of the student has to be more.

Sports meet

Reghu and friends went in a bus to the sports meet at Kozhikode. They started at 7 in the morning and reached there at 10, travelling 150 kilometres. Would the speed have been the same during the entire trip?
Could have been 40 kilometres the first hour, 60 kilometres the next and 50 kilometres the last hour.
It’s in such instances that we calculated the average, remember?
Here, the distance travelled is 150 kilometres.
And how much time did it take?
So, we can say they travelled \(\frac{150}{3}\) = 50 kilometres in one hour, on average.
We can put it this way: the average speed of the bus was 50 kilometres per hour,
This we write 50 km/h.

Average speed Textbook Page No. 91

Celina and Beena went to Kozhikode for the State Arts Festival. Celina travelled in a jeep, covering 90 kilometres in 2 hours; Beena came in a car, travelling 150 kilometres in 3 hours. Which vehicle is faster?
The jeep travelled 90 kilometres, right?
And took 2 hours for the trip.
So what is its average speed?
\(\frac{90}{2}\) = 45 km/h
The average speed of the car can also be computed like this.
Which is greater?
Now try to solve these problems:

• Sudheer travelled in a train, covering 240 kilometres in 3 hours to reach Thiruvananthapuram. Ramesh came in another train, travelling 120 kilometres in 2 hours. Which train is faster? How much faster?
Answer:
Sudheer train travelled faster by 20 km/hr than Ramesh train.

Explanation:
Distance travelled by Sudheer in a train = 240 km.
Time taken by Sudheer = 3 hours.
Distance travelled by Ramesh in a train = 120 km.
Time taken by Ramesh = 2 hours.
Speed taken by Sudheer train = Distance travelled by Sudheer in a train ÷ Time taken by Sudheer
= 240 ÷ 3
= 80 km/ hr.
Speed taken by Ramesh train = Distance travelled by Ramesh in a train ÷ Time taken by Ramesh
= 120 ÷ 2
= 60 km/ hr.
Difference:
Speed taken by Sudheer train – Speed taken by Ramesh train
= 80 – 60
= 20 km/hr.

• A train took 4 hours and 30 minutes to cover 360 kilometers. What is its average speed?
Answer:
Speed of the train = 80 km/hr.

Explanation:
Time taken by a train = 4 hours and 30 minutes = 4\(\frac{1}{2}\)
Distance travelled = 360 km.
Speed of the train = Distance travelled ÷ Time taken by a train
= 360 ÷ 4\(\frac{1}{2}\)
=  360 ÷ [(4× 2) + 1) ÷ 2]
= 360 ÷ \(\frac{9}{2}\)
= (360 × 2) ÷ 9
= 720 ÷ 9
= 80 km/hr.

Let’s look at another problem.
If the average speed of a bus is 52 km/h, how much does it travel in 6 hours?
Since it travels 52 kilometres each hour on average, in 6 hours it travels
52 × 6 = 312 km
How long does it take to travel 520 kilometres at this speed?

• The table below gives some details of Joy’s trip. Fill in the missing entries:
Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math 4
Answer:
Missing entries details of Joy’s trip:
Kerala-State-Syllabus-7th-Standard-Maths-Solutions-Chapter-7-Speed-Math-Average speed Textbook Page No. 91

Explanation:
Speed of the train = Distance travelled ÷ Time taken by a train
=> 60 km/hr = Distance travelled ÷ 4 hrs
=> 60 × 4 = Distance travelled
=> 240 km = Distance travelled.

Speed of the car = Distance travelled ÷ Time taken by a car
=> Speed of the car = 120km ÷ 2 hrs
=> Speed of the car = 60 km/ hr.

Speed of the Aero plane = Distance travelled ÷ Time taken by a Aero plane
=> 840 km/hr = 5040 km ÷ Time taken by a Aero plane
=> Time taken by a Aero plane × 840 = 5040
=> Time taken by a Aero plane = 5040 ÷ 840
=>  Time taken by a Aero plane = 6 hrs.

• Shyama’s exam starts at 2 o’ Hoot To reach the place, she has to travel 50 kilometres by bus and 175 kilometres by train. Average speed of bus journey is 20 km/h and average speed of train journey is 50 km/h. To reach there 1 hour before the exam, when should Shyama leave home?
Answer:
Time Shyama leave home = 7 AM.

Explanation:
Time of Shyama’s exam starts = 2 pm.
Distance travelled by bus = 50 km.
Average speed of bus = 20 km/ hr.
Distance travelled by train = 175 km.
Average speed of train = 50 km/ hr.
To reach there 1 hour before the exam.
Time taken by bus she travels = Distance travelled by bus ÷ Average speed of bus
=> Time taken by bus she travels = 50 ÷ 20
=> Time taken by bus she travels = 2.5 hrs = 2 hour 50minutes.
Time taken by train she travels = Distance travelled by train ÷ Average speed of train
=> Time taken by train she travels = 175 ÷ 50
=> Time taken by train she travels = 3.5 hrs = 3 hour 50 minutes.
=> Time Shyama has to travel = Time taken by bus she travels  +  Time taken by train she travels
=> 3.50 + 2.50
=> 6 hrs.
Time Shyama leave home = 2pm – Time Shyama has to travel
=>2 pm – 6 hrs – 1 hr
=> 7 AM.

Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math

Saving time

A bus leaves Ernakulam at 6 in the morning and reaches Thiruvananthapuram at 12 noon, running 40 km/h on average. To reach one hour earlier, by how much should the average speed be increased?
What is the total distance covered?
If the journey time is to be reduced by 1 hour, what should be the time for the trip?
What should be the average speed to reach an hour earlier?
Answer:
The total distance covered = 240 km.
The average speed to reach an hour earlier is 48 km/hr.

Explanation:
A bus leaves Ernakulam at 6 in the morning and reaches Thiruvananthapuram at 12 noon.
Average speed of the bus = 40 km/h.
Time taken by bus to travel from Ernakulam to Thiruvananthapuram  = 12 noon – 6am = 6 hours.
Speed = Distance ÷ Time.
=> 40 = Distance ÷ 6
=> 40 × 6 = Distance
=> 240 km = Distance

If the journey time is to be reduced by 1 hour,
Time taken = Time taken by bus to travel from Ernakulam to Thiruvananthapuram  – 1 hr
=> 6 hr –  1hr
=> 5 hours.
Speed = Distance ÷ Time.
=> Speed = 240 ÷ 5
=> Speed = 48 km/hr.

Railway station Textbook Page No. 93

Abu boarded a bus at 7 in the morning. Usually the bus travels 30 km/h on average, to reach the railway station at 11. But since it was raining, the bus could run only 20 km/h on average. Abu got down from the bus at 9, caught a car and reached the station at 11 itself. What was the average speed of the car?
What is the distance to the railway station?
How far did the bus travel in the first 2 hours?
So how many kilometres did the car travel?
How much time did it take?
Answer:
Distance to the railway station = 120 km.
40 kms the car travelled.
Distance travelled in 2 hours by bus = 60 km.
It took 2hrs of time by car.

Explanation:
Speed of bus = 30 km/h
Time of Abu boarded the Bus = 7 AM.
Time to reach the railway station = 11 AM.
Time taken = 11 – 7 = 4 hrs.
Due to raining, the speed of bus = 20km/ hr.
Time Abu got down the bus = 9 AM.
Time Abu reached the station = 11AM.
Speed = Distance ÷ Time.
=> 30 = Distance ÷ 4
=> 30 × 4 = Distance
=> 120 kms = Distance.
Distance travelled in 2 hours:
Speed = Distance ÷ Time.
=> 30 = Distance ÷ 2
=> 30 × 2 = Distance
=> 60 kms = Distance.
Due to raining:
Speed = Distance ÷ Time.
=> 20 = Distance  ÷ 2
=> 20 × 2 = Distance
=> 40 km = Distance.

Average of speeds and average speed

A vehicle did the first 120 kilometres of a trip at an average speed of 30 km/h and the next 120 kilometres at 20 km/h. What is the average speed of the entire trip?
The average of the two speeds is
\(\frac{30+20}{2}\) = 25 km/h
Is this what we want?
What is the correct reasoning?

To compute the average speed, we must divide the total distance by the total time. Time to travel 120 kilometres at
an average speed of 30 km/h is
\(\frac{120}{30}\) = 4 hours
Time to travel 120 kilometres at an average speed of
\(\frac{120}{20}\) = 6 hours
Total time of travel 4 + 6 = 10 hours
Total distance travelled = 240 km
Average speed = 24 km/h

Train and bus Textbook Page No. 94

Rahim travelled 350 kilometres in a train and 150 kilometres in a bus. The average speed of the train was 70 km/h. The bus trip lasted 5 hours. What is the average speed of the entire trip?
Answer:
Average speed of the entire trip = 50km/hr.

Explanation:
Distance Rahim travelled by train  = 350 km.
Distance Rahim travelled by bus = 150 km.
Average speed of the train = 70 km/h.
Time Rahim travelled by bus = 5 hours.
Average Speed of the bus = Distance Rahim travelled by bus ÷ Time Rahim travelled by bus
=> Average Speed of the bus = 150 ÷ 5
=> Speed of the bus = 30 km/hr.
Average speed of the entire trip = (Average speed of the train + Average Speed of the bus) ÷ 2
= (70 + 30) ÷ 2
= 100 ÷ 2
= 50km/hr.

To Ratnagiri

Ratnagiri is 360 kilometres from Pavizhamala. Gopika’s family travelled in a car, at an average speed of 60 km/h. They managed only 40 km/h during the return trip. What is the average speed of the entire trip?
Suppose we take the distance as 180 kilometres in this problem.
Does it change the average speed for the entire journey?
Answer:
No, it does not change the average speed for the entire journey because speed is not dependent of distance travelled.

Explanation:
Distance from Pavizhamala to Ratnagiri = 360 km.
Average speed Gopika’s family travelled in a car = 60 km/h.
Average speed they managed during the return trip = 40 km/h.
Average speed of the entire trip = (Average speed Gopika’s family travelled in a car + Average speed they managed during the return trip) ÷ 2
= (60 + 40) ÷ 2
= 100 ÷ 2
= 50 km/hr.
Suppose, Distance travelled = 180 km.
Average speed Gopika’s family travelled in a car = 60 km/h.

Unknown distance

Babu went to Mananthavady by bus to see his friend. The average speed of the bus was 40 km/h. He came back in a car, at an average speed of 60 km/h. What is the average speed of the entire trip?
To find this, we have to divide the total distance by the total time. But we don’t know the distance.
We have seen in an earlier problem the average speed would be the same, whatever be the total distance.
Let’s take the distance one way as 120 kilometres. Total distance is then 240 kilometres
Time for the onward trip = \(\frac{120}{40}\) = 3 hours
Time for the return trip = 2 hours
So, the average speed of the entire trip = 48 km/h
What if we take the one way distance as 240 kilometres?
Can’t we find the time of this trip?
Answer:
The time of this trip = 6 hours.

Explanation:
Average speed of the bus = 40 km/h.
He came back in a car,
=>Average speed of car = 60 km/h.
Average speed of the entire trip = (Average speed of the bus + Average speed of car) ÷ 2
= (40 + 60) ÷ 2
=> 100 ÷ 2
=> 50 km/hr.
Distance travelled one trip = 240 km.
Average speed of the bus = 40 km/h.
Speed = Distance ÷ Time
=> 40 = 240 ÷ Time
=> Time = 240 ÷ 40
=> Time = 6 hours.

Johny’s journey Textbook Page No. 95

• Johny went to his uncle’s house, riding his bicycle at 15 km/h and came back at 10 km/h. What’s his average speed for the whole journey?
Answer:
Average speed for the whole journey = 12.5 km/ hr.

Explanation:
Speed of his bicycle = 15 km/h
Speed of his bicycle while coming back = 10 km/h
Average speed for the whole journey = (Speed of his bicycle + Speed of his bicycle while coming back) ÷ 2
= (15 + 10) ÷ 2
= 25 ÷ 2
= 12.5 km/ hr.

In seconds

A vehicle travels at an average speed of 72 km/h. How much will it move in one second?
One hour means 60 minutes; and a kilometre is 1000 metres.
Thus in 60 minutes, this vehicle will travel 72000 metres.
So, in one minute, it would move \(\frac{72000}{60}\) = 1200 metres.
Which means that in one second it would move \(\frac{1200}{60}\) = 20 meters.
We can say that the average speed of the car is 20 meters per second, written 20 m/s.
Compute the speed per hour of a vehicle doing 15 m/s.
Answer:
Speed = 54000 m/ hr or 54 km/hr.

Explanation:
A vehicle travels at an average speed = 72 km/h.
Conversion:
1 hour = 3600 seconds.
A vehicle moves per second = A vehicle travels at an average speed ÷ 3600
=> 72 ÷ 3600
=> 0.02 km/ second.
Speed per hour of a vehicle doing 15 m/s.
=> Speed = 15 m/s
=> 15 × 3600
=> 54000 m/ hr.
Conversion:
1 km = 1000 m.
=> 54000m ÷ 1000m = 54 km.

Now try these problems.

• A train travels at an average speed of 36 km/h. How far will it go in 3 minutes?
Answer:
1.8 km the train goes in 3 minutes.

Explanation:
Average speed of a train travels  = 36 km/h
Time = 3 minutes.
Speed = Distance ÷ Time
Conversion:
1 hour = 60 minutes
=> 36 ÷ 60 = 0.6 km/min
=> 0.6 × 3 minutes = 1.8 km/ 3 mins

• A 180 meters long train takes 9 seconds to pass a vertical pole. What is the average speed of the train in kilometer per hour?
Answer:
Speed of the train = 72 km/hour.

Explanation:
Distance of train = 180 meters.
Time taken to pass a vertical pole = 9 seconds.
Average speed of the train = Distance ÷ Time
=> Speed = 180 ÷ 9
=> Speed = 20 m/s.
Conversion:
1 hour = 3600.
=> Average speed of the train = Distance ÷ Time
=> 20 × 3600
=> 72000 m/hour.
Conversion:
1 km = 1000 m.
=> Speed = 72000 ÷ 1000
=> Speed = 72 km/hour.

Kerala Syllabus 7th Standard Maths Solutions Chapter 7 Speed Math

Let’s do it! Textbook Page No. 95

Question 1.
A car goes an average speed of 36 km/h for 15 minutes and 60 km/h for the next 15 minutes. How much distance did it cover?
Answer:
Total distance covered = 1,440 km.

Explanation:
Speed of a car = 36 km/h
Time of a car = 15 minutes
Speed = Distance ÷ Time
36 = Distance ÷ 15
=> 36 × 15 = Distance
=> 540 km = Distance.
Speed of a car = 60 km/h
Time of a car = 15 minutes
Speed = Distance ÷ Time
=> 60 = Distance ÷ 15
=> 60 × 15 = Distance
=> 900 km = Distance
Total distance covered = 540 + 900 = 1,440 km

Question 2.
Ramu and Salim travelled to Thiruvananthapuram from the same place, each in his own car. Ramu did the onward journey at an average speed of 120 km/h and the return at 50 km/h. Salim did both trips at 60 km/h. Who made the total journey in less time?
Answer:
Ramu made the total journey in less time.

Explanation:
Average speed of Ramu the onward journey = 120 km/h
Average speed of Ramu the return journey = 50 km/h
Average speed of Ramu of the journey = (Average speed of Ramu the onward journey + Average speed of Ramu the return journey) ÷ 2
=> (120 + 50) ÷ 2
=> 170 ÷ 2
=> 85 km/ hr.
Average speed of Salim the both journeys = 60 km/hr.

Question 3.
Two trains travelling along parallel tracks in the same direction do 50 km/h and 100 km/h on average. The faster train starts 2 hours later than the slower. After how many kilometers will the faster train catch up?
Answer:
After 200 km the faster train will catch up.

Explanation:
Average speed of both Trains = 50 km/h and 100 km/h
The faster train starts 2 hours later than the slower.
=> Let first train be slower = 60 km/ hr.
=> Second train be faster = 100 km/ hr.
Speed of the first train = Distance ÷  Time
=>Let time taken by slower train to where it caught by faster one = x hour
And time taken by faster one upto where it catch = (x – 2)
=> Distance = speed × time
thus we have x × 50 = (x – 2)×100
50x = 100x- 200
50x = 200
x= 4 hours
So after 4 hours faster train catch slower one
Number of km at which it catch = 4×50 =200 km.

Question 4.
A 125 metre long train travels at 90 km/h. How much time would it take to cross a 175 meter long bridge?
Answer:
12 seconds time would it take to cross a 175 meter long bridge.

Explanation:
Length of the train = 125 m.
Speed of the train = 90 km/h.
First convert 90 km/hr to m/hr.
Conversion:
1 km = 1000 m.
90 x 1000 = 90,000 m/hr.
Conversion:
1hour = 3600 seconds.
90,000 ÷ 3600 = 25m/sec.
=> the train is traveling at 25 m/sec
Distance = speed × time
=> time = Distance ÷ speed
=> time = 175/25
=> time =  7 seconds.
Time at which the front of the train will cross the 175 meters.
The train is 125 meters long, so at 25 m/sec,
=> it will take another 5 seconds (125/25) for the rear of the train to cross the 175 meter distance.
Total time taken = 7 seconds + 5 seconds
=> Total time taken = 12 seconds.

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

You can Download Squares and Right Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 7th Standard Maths Solutions Chapter 12 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

Squares and Right Triangles Text Book Questions and Answers

Another Way Textbook Page No. 158

There is another way to cut and rejoin two squares of the same size to one large square:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 1
Is there any relation between the side of the large square and some length related to the smaller squares?
Answer:
There is a parallel relationship between the side of the large square and some length related to the smaller squares.

Explanation:
The side of large square = diagonal be d.
Sides of the square be a.
=> Squares of diagonal = two squares of sides.
=> d² = a² × a²

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

How to draw?

When we cut out two squares of the same size and rejoin to make one large square, the length of a side of this large square is length of the diagonal of the small square. So, given a square, it is easy to draw a square of double the area.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 7
Using this idea, can you draw a square of area 50 square centimeters?
How about 32 square centimeters?

Answer:
We can draw a square of area 50 square centimeters by having the side measurement as 7.07 cms and 5.656 cm for 32 square centimeters.

Explanation:
Area of square = 50 square centimeters.
Area of square = side of the square × side of the square
=> 50 = S²
=> 50 Square root = S
=> 7.07 cm = side of the square.
Area of square = 32 square centimeters.
Area of square = side of the square × side of the square
=> 32 = S²
=> 32 Square root = S
=> 5.656 cm = Side of the square.

Parallel way  Textbook Page No. 160

There is another way to draw a square of double the area of a given square.
First draw a diagonal:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 8
Next draw lines parallel to this diagonal through the other two corners:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 9
Draw the other diagonal also and draw lines parallel to it as before:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 10

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

How to halve

To get a rectangle of half the area of a square, we need only cut horizontally or vertically along the middle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 15
Suppose we want a square itself of half the area?
Fold all corners of the square into the Centre:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 16
If we now unfold and cut along the creases, We get a square of half the area:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 17

Three squares Textbook Page No. 162
We can also cut and join three squares of the same size to make one large square
First cut two squares along the diagonal:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 18
Arrange the triangular pieces around the uncut square like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 19
Draw a square by joining the outer vertices of the triangles as shown below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 20
Cut out the four slender triangles putting out from the square and plug the gaps within:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 21

Five Squares

Place five squares of the same size like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 27
Draw a square by joining some of the corners as below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 28
Now cut out the triangles outside the square and fill the gaps within:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 29
Have you seen such a picture in some other lesson?

Answer:
No, like the above picture and cuttings its not found in any other lessons.

Explanation:
Picture of making five squares from a rectangle is not found in any lessons till now.

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

Another cut Textbook Page No. 164

There are other ways to cut two squares and rearrange to form a single large square.
Mark a side of the small square on one side of the larger square. Join this point with a corner of the square, as shown below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 30
Now place the squares side by side and draw a line as below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 31
Cut along these lines:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 32
Move the pieces below to the top as shown below, to make a square:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 33

Some history

Abu al-Wafa was a famous mathematician and astronomer who lived in Baghdad during the tenth century AD.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 36
One of his books is ‘On Those Parts of Geometry Needed by Craftsmen.” In it, he discusses various methods of joining small squares to make large squares and cutting large squares to make smaller squares.

One such discussion demonstrates that the method used by the craftsmen then to join three squares into one is not accurate and also gives the correct method. It is this method that is given under Three squares in this lesson.

Art and Geometry Textbook Page No. 166

Even before Abu al-Wafa’s time, mosques in Persia were decorated with ornamental tiles on the floors and walls. Abu al-Wafa describes various geometric methods to cut and reassemble such tiles of different sizes.

Many beautiful geometric patterns can be seen in such tiles. The picture below shows such a work on a wall in the famous Janab Abbasi Mosque in Iran, built during the 17th century.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 37

Shrinking square

Draw a square of side 5 centimetres and mark points 1 centimetre from the corners as below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 41
We get a square by joining these points:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 42
What is its area?
we need only subtract the areas of four right triangles from the area of the large square.
25 – 4 × \(\frac{1}{2}\) × 4 × 1 = 25 – 8 = 17 sq. cm
Suppose we join points 2 centimetres from the corners.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 43
What is the area of the small square?

Answer:
Area of the small square = 22 square cm.

Explanation:
Area of big square = Side of square × Side of square
= 5 × 5
= 25 square cm.
Area of right triangle = (Side × Height) \(\frac{1}{2}\)
= 2 × 3 × \(\frac{1}{2}\)
= 1 × 3 × \(\frac{1}{1}\)
= 3 square cm.
Area of the small square = Area of big square – Area of right triangle
= 25 – 3
= 22 square cm.

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

Cutting off Textbook Page No. 168

From a square of side 8 centimetres, if we cut off four right triangles as shown below, we get a square of area 34 square centimetres:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 44
What if we cut off triangles as in the picture below?
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 45
What is the area of the remaining square?
Can we cut out a square of area 50 square centimetres from the large square?
How about a square of area 44\(\frac{1}{2}\) square centimetres?

Answer:
Yes, we can cut out a square of area 50 square centimetres from the large square as big square area is 64 square cm.
Yes, area 44\(\frac{1}{2}\) square cm can also be cut from big square.
Explanation:
Side of the big square = 8 cm.
Area of big square = Side of the big square × Side of the big square
= 8 × 8
= 64 square cm.
Area of right triangle = (Side × Height) \(\frac{1}{2}\)
= 2 × 6 × \(\frac{1}{2}\)
= 1 × 6 × \(\frac{1}{1}\)
= 6 square cm.
Area of the small square = Area of big square – Area of right triangle
= 64 – 6
= 58 square cm.

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

Right triangles

We have seen two ways of drawing a square of area 34 square centimetres:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 51
In both pictures, a side of such a square is the longest side of a right triangle
What is the relation between the lengths of the perpendicular sides of the triangle and area of the square?

Answer:
There is a direct relationship between the lengths of the perpendicular sides of the triangle and area of the square as it helps to find the area of the small square.

Explanation:
Lengths of the perpendicular sides of the right angled triangle are the base and height of the triangle, which help to find the area of the square formed by the longest side of a right triangle.

Using GeoGebra, we can verify the relation between the areas of squares drawn on the three sides of a right triangle.
Draw a line AB and the perpendicular through A as below.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 52
Mark a point C on the perpendicular.
Now we can slide the line AC.
Using the Polygon tool, draw triangle ABC. Then using the Regular Polygon tool, we can draw squares on the sides AB, BC and AC, By selecting Area tool and clicking within the squares, we can see their areas.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 53
What is the relation between the areas? Drag the vertices of the triangle and check. Does the relation between areas change? To make the area of the largest square 25, what should be the sides of the smaller squares?
Suppose we want the largest square to have area 41?
Answer:
We cannot draw the largest square to have area 41 with the sides of 3cm and 4cm.

Explanation:
Area of the largest square = 25 square cm.
Side of largest square = 5 cm.
Side of smaller squares can be of 3cm and 4cm because 32 + 42 = 52

Pythagoras Textbook Page No. 171

We don’t know much about Pythagoras, a famous mathematician of ancient times. He was born around 570 BC in the island of Samos in Greece.
History says, in his youth he studied in Egypt and later came back to Greece to start a school. He taught that:
“The real nature of things could be understood only through Mathematics”
The picture shows a statue of Pythagoras at Samos, his place of birth.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 58

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

Indian Math

Certain geometric texts of ancient India are collectively termed sulvastura.
They are written by various authors at different periods estimated to be written around 500 BC.

In the Baudhayana sulvasutra, method of doubling a square is given:
By stretching a rope along the diagonal of a square, another square of double the area can be made.

In the Katyana sulvasutra, written around 200 BC, a more general construction is described:

By stretching a rope along the
diagonal of a rectangle, the sum
of the areas of square make by
the horiztontal and vertical sides.

The word sulva in Sanskrit means string or rope. The word sutra also means the same, but it also used for shortened statements of general principles.

Pythagoras relation Textbook Page No. 173

The square of the longest side of a right triangle is equal to the sum of the squares of the other two sides
Putting it in reverse, if the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then it is a right triangle.

That is, the property of having the square of one side equal to the sum of the squares of the other two, is peculiar to right triangles.
For example, since 32 + 42 = 52, a triangle with lengths of sides 3, 4, 5 is right. What if the lengths of sides are 6, 8, 10?

Different uses

We can use Pythagoras Theorem to make one large square from two smaller ones and to draw squares of specified areas.
We can also use it to construct perpendiculars and to verify perpendicularity.
For example, look at these lines:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 64
To check whether they are perpendicular to each other, mark a point on the horizontal line, 3 centimetres from the point where the lines meet; and a point on the vertical line, 4 centimetres up.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 65
If the distance between these points is 5 centimetres, then the lines are perpendicular; otherwise not.

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

Pythagorean triples Textbook Page No. 175

The sum of the squares of two natural numbers may not be the square of another natural number.
12 + 22 = 5
But we do have
32 + 42 = 25 = 52
52+ 122 = 169 = 132
82 + 152 = 289 = 172
and so on.
Any three natural numbers, with the sum of the squares of two of them equal to the square of the third, is called a Pythagorean triple.
Some examples of Pythagorean
3, 4, 5
5, 12, 13
8, 15, 17
Can you find some more?

Answer:
Some other examples of Pythagorean are (20, 21, 29) and (9, 40, 41).

Explanation:
Some examples of Pythagorean:
202+ 212 = 400 + 441 = 841 = 292
92+ 402 = 81 + 1600 = 1681 = 412

Doubling a square Textbook Page No. 158 

Cut out two squares of the same size from thick paper:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 2
These are to be cut and the pieces arranged to form one large square.
It can be done like this.
First cut along the diagonals’ of one square to make four triangles:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 3
Flip all these triangles outwards:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 70
Now place the uncut square in the vacant spot in the middle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 5

Why does the second square fit inside exactly?
Suppose the sides of the first two squares are all 2 centimetres long. What is the area of each of these?
So, what is the area of the final large square?
Now let’s do this with two squares of area 9 square centimetres each. To make a square of area 9 square centimetres, how much should be the length of the sides?
Cut out two such squares and cut one of these as above to make a single large square. What is its area?
How do you make a square of area 50 square centimetres?
What about a square of area 32 square centimetres?

Answer:
To make a square of 50 square cm, side of the square as to be 7.07 cm and 32 square cm the side of square to be 5.65 cm.

Explanation:
Area of each two squares = 9 square cm.
=> Side × Side = Area of square.
Side of the square = 3 cm.
Area of square = 50 square centimeters.
=> Side × Side = Area of square.
=> √50 = 7.07 cm.
Area of square = 32 square centimeters.
=> Side × Side = Area of square.
=> √32 = 5.65 cm.

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

Making larger

We have seen that by cutting a square along the diagonals and placing the pieces around another square of the same size, we can make a square of double the size.
Now let’s cut in a different way.
Cut out a square of side 5 centimetres.
Instead of joining opposite corners to draw diagonals, mark points 1 centimetre away from corners and join:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 6
Cut along these lines and flip the four pieces outwards:

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 11
Slide the left piece a little downwards, the right piece towards the left, and the top piece left and down to make a
large outer square with a small square hole inside as below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 12
What is the length of the sides of the inner square?
So, if we cut out a square of side 3 centimetres, we can fit it exactly into the hole.

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 13
It was the first square that was cut into four pieces. So the sum of the areas of these four pieces is equal to the area of the first square; that is 52 = 25 square centimetres.
What about the area of the second square placed inside?
So, the area of the final large square is
52 + 32 = 34 sq.cm.
Now cut out another square of side 5 centimetres again and join points marked 2 centimetres off the corners:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 14
If we cut along these lines and flip the four pieces outwards it would be like this:

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 22

Let’s slide the pieces as before to make a square with a hole:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 23

What should be the length of the sides of the square needed to plug this hole?
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 24
What is the area of the filled up large square?
52 + 12 = 26 sq.cm

Again cut out a square of 5 centimetres and join points 1.5 centimetres off the corners. Cut out along these lines and rearrange the pieces as above.
What should be the length of the sides of the square to be placed inside?
What is the area of the filled up large square?

Answer:
Area of the filled square = 36 square cm.

Explanation:
Side of the big square = 5 + 1.5 = 6.5 cm
Area of big square = Side of the big square × Side of the big square
= 6.5 × 6.5
= 42.25 square cm.
Side of the filled square = 1 + 1.5 = 2.5 cm.
Area of the filled square = (6.5 × 6.5) – (2.5 × 2.5)
= 42.25 – 6.25
= 36 square cm.

Two squares Textbook Page No. 163

We cut a square of side 5 centimetres in various ways and joined with another square to make larger squares of different sizes:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 26

To cut the first square into pieces, we mark points at the same distance from the corners. What is the relation between this distance and the length of the sides of the second square cut out to plug the hole?
Let’s look again at the way we made the large square.

When we mark points, we subtract the distance from the side of the first square. When we slide the pieces into position, we again subtract this distance. Thus, the side of the hole is the side of the original square reduced twice by the distance between the corners and the marks.

So, suppose we cut a square of side 8 centimetres by marking points 3 centimetres from the corners?
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 34
If we cut along these lines and rearrange the pieces as before, the side of the square to fill the hole is.
8 – (2 × 3) = 2 cm
What is the area of this large square?

Answer:
The area of this large square = 64 square cm.

Explanation:
Side of the big square = 8 cm.
Area of big square = Side of the big square × Side of the big square
= 8 × 8
= 64 square cm.

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

Another question

We want to cut a square of side 8 centimetres as before and fit a square of side 6 centimetres within to make a
larger square. How should we cut the first square?
Twice the distance of the points to be marked on the first square from its corners subtracted from 8 centimeters should be 6 centimetres.
So, twice this distance is
8 – 6 = 2 cm
So this distance is half of 2 centimetres; that is 1 centimetre.

Cut out a square of side 8 centimetres, cut into pieces like this and rearrange as before. Isn’t the sides of the inner square to 6 centimetres?
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 35
What is the area of the final large square?
82 + 62 = 100 sq. cm

• Cut out a square of side 7 centimetres and another of side 3 centimetres. Cut the larger square suitably and join with the smaller square to make one large square.
What is the area of this square?
Answer:
Area of the large square = 58 square cm.

Explanation:
Length of side of the square = 7 cm.
Length of side of the square = 3 cm.
Area of the large square = 72 + 32
=> 49 + 9
=> 58 square cm.

• Cut out a square of side 8 centimetres. Cut this suitably and join with a suitable square to make a large square of area 80 square centimetres.
Answer:
Area of the large square = 80 square cm.

Explanation:
Length of side of the square = 8cm.
Area of the square = 80 square cm.
Length of side of the square = 4cm.
Area of the large square = 82 + 42
=> 64 + 16
=> 80 square cm.

• A square of side 9 centimetres is to be cut and joined with another square to make a large square of area 117 square centimetres. What should be the side of the second square? At what distance from the corners of the first square should points be marked for cutting it?

Answer:
Side of other square = 6 cm.
3cm distance from the corners of the first square should points be marked for cutting it.

Explanation:
Area of the square = 117 square centimeters.
Side of one square = 9 cm.
Side of other square = ?? cm.
92 + 72 = 576 + 49 = 625 = 25 cm.
=> 81 + ??2 = 117
=> ??2 = 117 – 81
=> ??2 = 36.
=> ?? =  6 cm.

Drawing Squares Textbook Page No. 166

Remember how we joined together squares of sides 5 centimetres and 8 centimetres and made a large square?
First we find half the difference of the sides.
(5 – 3 ) ÷ 2 = 1
Then we mark points on the large square, 1 centimetre away from the corners. Cutting along the lines joining those points and rearranging the pieces with the small square inside, we get a square of area 25 + 9 = 34 square centimetres.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 38

If we just want to draw a square of this area, instead of actually making it, we need only draw a line equal in length to the side of the square. Let’s see how we do this.

Each of the four pieces got by cutting the square of side 5 centimetres have two sides of length 4 centimetres and I centimetre. By stacking the pieces, we can see that the other sides also are of equal lengths.

Now look at these pictures:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 39

The length of the sides of the final large square is equal to the length of the lines drawn to cut the first square.
So, we have an easy method to draw a side of a square of area 34 square centimetres.
First draw a square of side 5 centimetres, mark two points 1 centimetre away from two opposite corners and join
them:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 40
The square drawn with this line as a side has area 34 square centimetres.

Instead of shifting 1 centimetre from each of the two corners we can shift 2 × 1 = 2 centimetres from one corner.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 46
Suppose we want to join a square of side 5 centimetres and another of side 1 centimetre and draw square of area 25 + 1 = 26 square centimetres. For this, we mark two points (5 – 1) ÷ 2 = 2 centimeters away from two opposite corners of the larger square and join them:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 47
Then we draw a square with this line as a side.
We can draw without taking half of 5 – 1 = 4. Instead of marking 2 centimetres on two sides, we mark 4 centimetres on one side.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 48

Let’s have another look at the two squares we have drawn.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 49
In each figure, the side of the square is the longest side of a right triangle.
What about its area?
The sum of the areas of squares drawn on the other two sides of the triangle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 50

Now draw a right triangle and squares on all three sides on thick paper:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 54

In the medium size square, mark the point of intersection of diagonals; and through this point, mark lines parallel to the perpendicular sides of the largest square:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 55

Now cut along these lines to split the square iito four pieces. Cut out the smallest square also. Arrange all these inside the largest square as below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 56
What do we see from all this?

Answer:
The area of squares drawn out are equal to the area of the largest side of the triangle formed from cutting.

Explanation:
It is noticed that the area of squares drawn out are equal to the area of the largest side of the triangle formed from cutting.

The area of the square drawn on the largest side of a right triangle is equal to the sum of the areas of the squares drawn on the other two sides.
This is known as Pythagoras Theorem, in honour of the philosopher Pythagoras, who lived in Greece.
Using this, we can draw a square of area 10 square centimetres. We have
10 = 32 + 12
So according to Pythagoras Theorem, we need only draw a right triangle of perpendicular sides 3 centimetres and 1 centimetre and then draw a square on the longest side.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 57
How about a square of area 7 square centimeters?

Answer:
It cannot be drawn the side of the square of having area of 7 square centimeters.

Explanation:
Area of 7 square centimeters cannot be drawn. So according to Pythagoras Theorem, we need only draw a right triangle of perpendicular sides 3 centimeters and 1 centimeters and then draw a square on the longest side.

7 cannot be written as the Sum of two perfect squares
But we have
7 = 42 – 32
So by Pythagoras Theorem, we need only draw a right triangle of longest side 4 centimetres and another side 3 centimetres.
How do we draw it?
First draw a 3 centimetre long line and the perpendicular at one end:
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 59
Now use a compass to mark a point on the perpendicular, at a distance of 4 centimetres from the other end of the first line.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 60
The area of the square in the vertical side of this triangle is 42 – 32 = 7 square centimetres, according to Pythagoras Theorem.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 61

Can’t you now draw squares of areas given below?

• 20 square centimeters.
Answer:
Yes, it can be done for  area of the square = 20 square centimeters.

Explanation:
Area of the square = 20 square centimeters.
42 + 22 = 16 + 4 = 20.
Sides of the squares = 4cm , 2cm.

• 39 square centimeters
Answer:
No, it cannot be drawn a square of having 39 square centimeters.

Explanation:
Area of the square = 39 square centimeters.

• 40 square centimeters.
Answer:
Yes, it can be done a square of having 40 square centimeters.

Explanation:
Area of the square = 40 square centimeters.
62 + 22 = 36 + 4 = 40.
Sides of the squares = 6cm , 2cm.

• 65 square centimeters
Answer:
Yes, it can be drawn a square of having 65 square centimeters.

Explanation:
Area of the square = 65 square centimeters.
72 + 42 = 49 + 16 = 65.
Sides of the squares = 7cm , 4cm.

Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles

Squared relation Textbook Page No. 173

We can state Pythagoras Theorem as a relation between the lengths of the sides of a right triangle. The longest side of a right triangle is called its hypotenuse.

The square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.
For example, if the length of the perpendicular sides of a right triangle are 3 centimetres and 4 centimetres, then the square of the hypotenuse is
32 + 42 = 25
and so the length of the hypotenuse is 5 centimetres.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 62
Look at this problem: What is the length of the diagonal of the rectangle below?
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 63

The diagonal of the rectangle is the hypotenuse of a right triangle :
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 66
So, the square of the diagonal is
52 + 122 = 169
and the length of the diagonal is
\(\sqrt{169}\) = 13 m

• What is the length of the fourth side of the quadrilateral shown below?
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 67
Answer:
Length of the fourth side of the quadrilateral shown below = 25 cm.

Explanation:
Draw a line to side 24cm.
Kerala-State-Syllabus-7th-Standard-Maths-Solutions-Chapter-12-Squares-and-Right-Triangles-Squared relation Textbook Page No. 173

Length of the other side = 242 + 72 = 576 + 49 = 625 = 25 cm.

• What is the area of the triangle shown below?
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 68
Answer:
Area of the big triangle = 175 square cm.

Explanation:
Area of the small triangle = (Base × height) ÷ 2
Height of triangle = 252 – 72 = 625 – 49 = 576 = 24 cm.
Base of triangle = 7 cm.
Area of the 1 triangle = (Base × height) ÷ 2
= (7 × 24) ÷ 2
= 168 ÷ 2
= 84 square cm.

Area of the 2 triangle = (Base × height) ÷ 2
= (7 × 26) ÷ 2
= 182 ÷ 2
= 91 square cm.

Area of the  big triangle = Area of the 1 triangle + Area of the 2 triangle
= 84 + 91
= 175 square cm.

• The picture shows a plot of land in the form of a right angled triangle joined to a square.
Kerala Syllabus 7th Standard Maths Solutions Chapter 12 Squares and Right Triangles 69
The total area of the plot is 2200 square meters. What is its perimeter’?
Answer:
Perimeter of figure = 280 m.

Explanation:
Total area of the plot = 2200 square meters.
Side of the square = 40 m.
Perimeter of the square = 4 × side of the square
= 4 × 40
= 160 meters.
Base of triangle = 40 m.
Height of triangle = 40 m.
Perimeter of the triangle = side + side + side
= 40 + 40 + 40
= 80 + 40
= 120 m.
Perimeter of figure = Perimeter of the square  + Perimeter of the triangle
= 160 + 120
= 280 m.