Class 7

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 11 Squares and Right Triangles Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 11 Solutions Squares and Right Triangles

Class 7 Maths Chapter 11 Squares and Right Triangles Questions and Answers Kerala State Syllabus

Squares and Right Triangles Class 7 Questions and Answers Kerala Syllabus

Page 149

Question 1.
Now compute the lengths of the sides of squares with area given below Write the answers using square roots.
i) 49 square centimetres
ii) 169 square centimetres
iii) 400 square centimetres
iv) 225 square centimetres
v) 1.69 square centimetres
vi) 6 \(\frac{1}{4}\) square metres
Answer:
i) 49 square centimetres
Area = 49 square centimetres
49 = 7 × 7 = 7²
By reversing, \(\sqrt{49}\) = 7
So the length of the sides of square is 7 cm

ii) 169 square centimetres
Area =169 square centimetres
169 = 13 × 13 = 13²
By reversing, \(\sqrt{169}\) = 13
So the length of the sides of square is 13 cm

iii) 400 square centimetres Area = 400 square centimetres
400 = 20 × 20 = 20²
By reversing, \(\sqrt{400}\) = 20
So the length of the side of square is 20 cm

iv) 225 square centimetres
Area = 225 square centimetres
225 = 15 × 15 = 15²
By reversing, \(\sqrt{225}\) = 15
So the length of the side of square is 15 cm

v) 1.69 square centimetres Area =1.69 square centimetres
1.69 = 1.3 × 1.3 = (1.3)²
By reversing, \(\sqrt{169}\) = 1.3
So the length of the side of square is 1.3 cm

vi) 6 \(\frac{1}{4}\) square metres
By reversing, \(\sqrt{6.25}\) = 2.5
6.25 = 2.5 × 2.5 = (2.5)²
Area = 6\(\frac{1}{4}\) square metres = \(\frac{25}{4}\) = 6.25
So the length of the side of square is 2.5 m

Page 152

Question 1.
Now can’t you draw squares of areas as below:
i) 32 sq. cm
ii) 50 sq. cm
iii) 12. 5sq. cm
iv) 24\(\frac{1}{2}\) sq. cm
Answer:
i) 32 cm
Here we want to draw a square of area 32 sq. cm
So first we want to find half its area, ie \(\frac{32}{2}\) = 16 cm
Next we want to find square root of 16 ie, \(\sqrt{16}\) = 4 cm
So we have to draw a square of length 4 cm

Next we have to draw its diagonal and measure the length of its diagonal.

Then extend any side of square, and mark the length of the diagonal on the extended side of the square.

ii) 50 sq. cm
We want to draw a square of area 50 sq. cm
So first we want to find half its area, ie \(\frac{50}{2}\) = 25cm
Next we want to find square root of 25 ie, \(\sqrt{25}\) = 5 cm
So we have to draw a square of length 5 cm

Next we have to draw its diagonal and measure the length of its diagonal.

Then extend any side of square, and mark the length of the diagonal on the extended side of the square.

Thus we get a square of Area 50 sq. cm

iii) 12.5 sq. cm
First we want to find half its area , ie \(\frac{12.5}{2}\)= 6.25cm
Next we want to find square root of 6.25 ie, \(\sqrt{6.25}\) =2. 5 cm
So we have to draw a square of length 2.5 cm

Next we have to draw its diagonal and measure the length of its diagonal.

Then extend any side of square, and mark the length of the diagonal on the extended side of the square.

Thus we get a square of Area 12.5 sq. cm

iv) 24 \(\frac{1}{2}\)sq. cm
Here 24\(\frac{1}{2}\) = \(\frac{49}{2}\) = 24.5
First we want to find half its area ie \(\frac{24.5}{2}\) =12.25 cm
Next we want to find square root of 12.25 ie, \(\sqrt{12.25}\) = 3. 5 cm
So we have to draw a square of length 3.5 cm

Next we have to draw its diagonal and measure the length of its diagonal.

Then extend any side of square, and mark the length of the diagonal on the extended side of the square.

Thus, we get a square of area 24 \(\frac{1}{2}\)sq. cm

Page 157

Question 1.
Draw squares of areas given below
i) 17 square centimetres
ii) 18 square centimetres
iii) 19 square centimetres
Answer:
i) 17 square centimetres
We can split 17 as 17= 16 + 1 = 4² + 1²
So, by Pythagoras’ Theorem, if we draw a right triangle with perpendicular sides 4 and 1 cm, then the area of the square on the hypotenuse is 17 sq.cm.

ii) 18 square centimetres We can split 18 as 18 = 9 + 9 = 3² + 3²
So, by Pythagoras’ Theorem, if we draw a right triangle with perpendicular sides 3 and 3 cm, then the area of the square on the hypotenuse is 18 sq.cm.

iii) 19 square centimetres
We can split 19 as difference of two squares:
19 = 100 – 81 = 10² – 9²
So, if we draw a right triangle with a hypotenuse of 10 centimetres and another side of 9 centimetres then the area of the square on the third side would be 19 square centimetres.

Question 2.
In the picture below, dots are marked 1 centimetre apart, horizontally and vertically and some of these are joined to make squares:

Calculate their areas.
Answer:
Square 1:
Let’s extend the two sides of square which creates a triangle.

Now the sides of the triangle will be 3 cm and 2cm with respect to the size of the dots given.
So, The area of the square 1 = 3² + 2² = 9 + 4 = 13 square centimetres

Square 2:
Let’s extend the two sides of square which creates a triangle.

Now the sides of the triangle will be 1cm and 1cm with respect to the size of the dots given.
So, The area of the square 2 = 1² + 1² = 1 + 1 = 2 square centimetres

Square 3:
Let’s extend the two sides of square which creates a triangle.

Now the sides of the triangle will be 1cm and 2cm with respect to the size of the dots given.
So, The area of the square 3 = 1² + 2² = 1 + 4 = 5 square centimetres

Square 4:
Let’s extend the two sides of square which creates a triangle.

Now the sides of the triangle will be 4cm and 2cm with respect to the size of the dots given.
So, The area of the square 1 = 4² + 2² = 16 + 4 = 20 square centimetres

Page 159

Question 1.
The lengths of two sides of some right triangles are given below. Calculate the length of the third side.
i) Perpendicular sides 6 centimetres, 8 centimetres
ii) Perpendicular sides 9 centimetres, 12 centimetres
iii) Perpendicular sides 7 centimetres, 24 centimetres
iv) Perpendicular sides 14 centimetres, 48 centimetres
v) Hypotenuse 17 centimetres, another side 15 centimetres
vi) Hypotenuse 34 centimetres, another side 30 centimetres
Answer:
i) Perpendicular sides 6 centimetres, 8 centimetres
Length of the third side = \(\sqrt{6^2+8^2}\)
= \(\sqrt{36+64}\)
= \(\sqrt{100}\)
= 10 centimetres

ii) Perpendicular sides 9 centimetres, 12 centimetres
Length of the third side = \(\sqrt{9^2+12^2}\)
= \(\sqrt{81+144}\)
= \(\sqrt{225}\)
= 15 centimetres

iii) Perpendicular sides 7 centimetres, 24 centimetres
Length of the third side = \(\sqrt{7^2+24^2}\)
= \(\sqrt{49+576}\)
= \(\sqrt{625}\)
= 25 centimetres

iv) Perpendicular sides 14 centimetres, 48 centimetres
Length of the third side = \(\sqrt{14^2+48^2}\)
= \(\sqrt{196+2304}\)
= \(\sqrt{2500}\)
= 50 centimetres

v) Hypotenuse 17 centimetres, another side 15 centimetres
Length of the third side = \(\sqrt{17^2-15^2}\)
= \(\sqrt{289-225}\)
= \(\sqrt{64}\)
= 8 centimetres

vi) Hypotenuse 34 centimetres, another side 30 centimetres
Length of the third side = \(\sqrt{34^2-30^2}\)
= \(\sqrt{1156-900}\)
= \(\sqrt{256}\)
= 16 centimetres

Question 2.
Two pillars of heights 2 metres and 5 metres stand 4 metres apart:

What is the distance between their tops?
Answer:
We are given that two pillars of heights 2 metres and 5 metres stand 4 metres apart
We want to find the distance between their tops.
When we draw a line parallel to 4 metre in the figure, we got a right triangle whose sides are 4 metre, 3 metre (ie 5 – 2 =3)

So the third side is the hypotenuse of the right triangle
We know that hypotenuse2 = Base² + Altitude²
Here Hypotenuse² = 4² + 3²
= 16 + 9
= 25
Hypotenuse = \(\sqrt{25}\)
= 5 cm
So the distance between their tops = 5 cm

Question 3.
Find the length of the bottom side of the triangle drawn below:

Answer:
In the given figure there are 2 right triangles,
Here hypotenuse and altitudes of both the triangles are given,
So, Base² = hypotenuse² – altitude²
= 13² – 12²
= 169 – 144
= 25m
Base = \(\sqrt{25}\) = 5 m
So the base of both the triangle = 5m
So Base of the large triangle = 5 + 5 = 10 cm

Question 4.
In the pictures below, O is the centre of the circle and A, B, P, Q are points on the circle: Calculate the lengths of the lines AB and PQ.

Answer:
Join OB and OQ in the figure;
So we get two right triangles in the 1st circle
OB = 5 cm(radius)

Let C be the Point where O meets the line AB
So OC = 4 m
AO = 5 m
So AC² = AO² – OC²
= 5² – 4²
= 25 – 16
= 9
So AC = √9 = 3 cm
CB = 3 cm
Thus AB = AC + CB
= 3 + 3
= 6 cm

In the second figure,
Let R be the point where O meets PQ.
Here PO = 5 cm, OR = 3cm ,OQ = 5 cm
So PR² = PO² – OR²
= 5² – 3²
= 25 – 9
= 16
PR = \(\sqrt{16}\) = 4 cm
∴ RQ = 4 cm
So, PQ = 4 + 4 = 8 cm
Therefore AB = 6 cm and PQ = 8 cm

Class 7 Maths Chapter 11 Kerala Syllabus Squares and Right Triangles Questions and Answers

Question 1.
The area of a square is 196 sq.cm .Then find the length of the sides?
Answer:
It is given that area of the square = 196 Sq.cm
We know that area of the square = side × side
= 14 × 14
= 196 Sq.cm
length of the sides = \(\sqrt{196}\) = 14 cm
Next we want to find square root of 49 ie, \(\sqrt{49}\) = 7 cm
So we have to draw a square of length 7 cm

Question 2.
Draw a square of area 99 sq.cm.
Answer:
Here we want to draw a square of area 98 sq. cm
So first we want to find half its area, ie \(\frac{98}{2}\) = 49 cm

Next we want to find square root of 49 ¡e, \(\sqrt{49}\) = 7 cm
So we have to draw a square of length 7 cm

Next we have to draw its diagonal and measure the length of its diagonal.

Then extend any side of square, and mark the length of the diagonal on the extended side of the square.

Thus we get a square of Area 98sq. cm

Question 3.
Draw squares of areas 52 square centimetres and 21 square centimetres.
Answer:
For the area of square 52 square centimetres
We can split 52 as
52 = 36 + 16 = 6² + 4²
So, by Pythagoras’ Theorem, if we draw a right triangle with perpendicular sides 6 and 4 cm, then the area of the square on the hypotenuse is 52 sq.cm.

For the area of square 21 square centimetres
We can split 21 as difference of two squares:
21 = 25 – 4 = 5² – 2²
So, if we draw a right triangle with a hypotenuse of 5 centimetres and another side of 2 centimetres then the area of the square on the third side would be 21 square centimetres.

Question 4.
Find out the area of the square on the diagonal of given rectangle.

Answer:
First draw the diagonal.

Now the sides of the triangle will be 8 cm and 3 cm.
So, the area of the square on the diagonal = 8² + 3² = 64 + 9 = 73 square centimetres

Question 5.
The lengths of two sides of a right triangles are given below. Calculate the length of the third side. Hypotenuse 10 centimetres, another side 8 centimetres
Answer:
The length of the third side = \(\sqrt{10^2-8^2}\)
= \(\sqrt{100-64}\)
= \(\sqrt{36}\)
= 6 cm

Question 6.
Find the sides of the triangle in the figure.

Answer:
In the given figure the one perpendicular side is 4em,
This side is parallel to the side of the rectangle and hence it is 4 cm
And the other side is 8 – 5 = 3 cm
Therefore third side of the triangle = \(\sqrt{3^2+4^2}\)
= \(\sqrt{9+16}\)
= \(\sqrt{25}\)
= 5 cm

Class 7 Maths Chapter 11 Notes Kerala Syllabus Squares and Right Triangles

In this chapter we will explore two important shapes, Squares and Right Triangles. We’ll begin by exploring squares, We will learn how to calculate the area of a square and how to draw a square with double the area of another. Next, we’ll move on to right triangles. We will explore the Pythagoras theorem, which helps us find the lengths of the sides of right triangles. This is a very useful tool in both, math and real-life situations.

• In the first topic, Areas of Squares, we will explore the concept of squaring a number and the square of a number. Then, We will also look into the square root of a number.
• In the second topic, Doubling the Area, we will learn how to draw a square that has double the area of a given square.
• In the third topic, Rectangles and Squares, we will discuss the hypotenuse, perpendicular sides, and Pythagoras theorem.
• In the fourth topic, Line Math, we will examine how the Pythagorean theorem relates the sides of a right triangle and how to solve related problems.

Through this chapter, we will build a foundational understanding of squares and right triangles.

Areas Of Squares
The second power of a number is called the square of that number. And the operation of finding the square of a number is called squaring.
For example, (1.3)², (1\(\frac{1}{2}\))²
(1.3)² = (1.3) × (1.3)
= \(\frac{13}{10} \times \frac{13}{10}\)
= \(\frac{169}{100}\)
= 24

The area of a (geometrical) square is the (arithmetical) square of its sides.
Here we can also say that, to draw a square of area 25 square centimeters, what should be the length of the sides?
So to find the answer, we want to know, the square of which number is 25,
ie, which number multiplied by itself gives 25
We know that 5 × 5 = 25, so the length of the sides should be 5 cm.
In another way we can say it as :
5 is the square root of 25.
In shorthand notation, “5 squared us 25”
ie, 5² = 25
The reverse statement is 5 is the square root of 25. We √ use to write this in shorthand that is,
√25 = 5.
It is not easy to compute the square roots of numbers. For small natural numbers, we can check the squares one by one and find the square root .For slightly larger natural numbers, we can try to factorize them and compute the square root.
For example consider the number 196,where we can factorize 196 and writeas:
196 = 2 × 2 × 7 × 7 = 2² × 7²
But we know that product of powers is the power of the product ie,
xⁿyⁿ =(xy)ⁿ
So, 196 = 2² × 7² = (2 × 7)² = 14²
Writing this in reverse, we get \(\sqrt{196}\) = 14

Doubling The Area
How do you make a square of double the area of the given figure?

For this we want to measure the length of its sides. So that we can calculate its area as the square of this number. If we double this number and calculate its square root, we get the side of the square of double the area. Without any measurement or computation we can find the same in another way,
For that,
First cut out another square of the same size and cut both sides along their diagonals

Now arrange the four triangles as below:
So that we get this square,

We used all the pieces of two small squares to make this large square. So its area is double that of a small square
If each of the small squares is made up of two right triangles of the same size; the large square is made up of four such right triangles.
Thus we can conclude that, if we want to just draw a square with double the area of another, we need just draw the square with the diagonal as side:
Here the side of the large square is equal to the length to the diagonal of the smaller square.

If we want to see the squares separately, we just need to one side of the original square and mark the length of the diagonal on it.

Rectangles And Squares
Now let’s discuss how to draw the square on the diagonal of a rectangle with unequal sides?

Here, the area of this square also depends on the squares on the sides.

To understand this, draw a rectangle, and squares on its sides and a diagonal as in the above picture, on a thick sheet of paper. Then,
Draw the diagonals of the bottom square and mark the point where they intersect.
Then erase the diagonals and draw lines through the point marked, lines parallel to the sides of the largest square:

Now cut out the squares; also cut the grey square along the lines drawn inside it:

Arrange the pieces of the grey square and the whole white square within the black square as shown below:

Here the square on the diagonal to the rectangle can be exactly filled with squares on the sides of the rectangle. So, we can say that,
The area of the square on the diagonal of a rectangle is equal to the sum of the areas of the squares on the adjacent sides.

The longest side of a right triangle is called its hypotenuse.
So the above result can also be stated as:
The area of the square on the hypotenuse of a right triangle is equal to the sum of the areas of . the squares on the perpendicular sides.
This results is known as the Pythagoras’ Theorem.
For example:
If the sides of the rectangle are 7 cm and 4cm.
If the areas of the squares on sides of the rectangle are :
7² = 49 square centimetres
4² = 16 square centimetres
So, the area of the square on the diagonal = 49 + 16 = 65 sq.cm

Let see how to draw squares of specified areas using Phythagor’s theorem,
For example,
Draw a square of area 25 square centimetres.
For that first split 25 as
25 = 16 + 9 = 4² + 3²
So, by Pythagoras’ Theorem, if we draw a right triangle with perpendicular sides 4 and 3 cm, then the area of the square on the hypotenuse is 25 sq.cm.

What about a square of area 16 square centimetres?
Here 16 cannot be written as sum of squares of two natural numbers. So we cannot draw the square using this method.
So let’s consider another method,

According to Pythagoras’ Theorem, if the area of the square on the hypotenuse of a right triangle is subtracted from the square on another side, the area on the third side is obtained.

So, for this we can split 16 as difference of two squares:
16= 25 – 9 = 5² – 3²

So, if we draw a right triangle with a hypotenuse of 5 centimetres and another side of 3 centimetres then the area of the square on the third side would be 16 square centimetres

To draw such a triangle follow the steps given below:

• Draw a line 3 centimetres long and the perpendicular at one end.
• Draw a piece of the circle centered at the other end of the line, with radius 5 centimetres.

The point where this circle meets the perpendicular is the third vertex of the triangle.

Line Math
The area of a square is the square of the length of its side; so, Pythagoras’ Theorem can also be stated as a relation between the sides of a right triangle:
The square of the hypotenuse of a right triangle is the sum of the squares of its perpendicular sides.
For example,

If the perpendicular sides of a right triangle are 3 centimetres and 4 centimetres, then the square of its hypotenuse is

3² + 4² = 25
So, the hypotenuse of this triangle is 5 centimetres

• The area of a (geometrical) square is the (arithmetical) square of its sides.
• If we want to just draw a square with double the area of another, we need just draw the square with the diagonal as side.
• The longest side of a right triangle is called its hypotenuse.
• The area of the square on the hypotenuse of a right triangle is equal to the sum of the areas of the squares on the perpendicular sides. This results is known as the Pythagoras’ Theorem.
• The square of the hypotenuse of a right triangle is the sum of the squares of its perpendicular sides.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 12 Algebra Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 12 Solutions Algebra

Class 7 Maths Chapter 12 Algebra Questions and Answers Kerala State Syllabus

Algebra Class 7 Questions and Answers Kerala Syllabus

Page 166

Now try these problems:

Question 1.
Take some sets of three consecutive natural numbers and add all three.
i) Check if the sum has any relation with any one of the three numbers added.
ii) Explain why this relation holds for any three consecutive natural numbers.
iii) Write this relation, first in ordinary language, and then using algebra.
Answer:
Let’s take three examples for three consecutive natural numbers and add them together:
1. 21 + 22 + 23 = 66
2. 34 + 35 + 36= 105
3. 78 + 79 + 80 = 237

i) For the consecutive natural numbers 21, 22, 23.
21 + 22 + 23 = 66 and 3 × 22 = 66
For the consecutive natural numbers 34, 35, 36.
34 + 35 + 36= 105 and 3 × 35 = 105
For the consecutive natural numbers 78, 79, 80.
78 + 79 + 80 = 237 and
3 ×79 = 237
That is, the sum of three consecutive numbers is always three times the middle number.

ii) Consider the three consecutive numbers 35, 36, and 37. These can be expressed in relation to the middle term as 36 – 1, 36, and 36 + 1.
Thus, the sum of these three numbers is:
(36 – 1) + 36 + (36 + 1) = 108
On simplifying this expression, we get (36 – 1) + 36 + (36 + 1) = 3 × 36 = 108
This is because in the addition of the three numbers, the 1 and -1 cancel each other out, resulting in a sum that is three times the middle number,

iii) In ordinary language, we can say it as:
The sum of any three consecutive natural numbers is three times the middle number.
Algebraically, if the three consecutive numbers are x – 1, x, and x + 1, the sum is:
(x – 1) + x + (x + 1) = 3x, for any natural numbers x

Question 2.
For some sets of four consecutive natural numbers, the sum of the first and the last, and the sum of the middle two, are shown separately below:

i) Explain why such sums are equal for any four consecutive natural numbers.
ii) Write this relation using algebra.
Answer:
i) Consider the four consecutive natural numbers as 4, 5, 6 and 7

The sum of the first and last numbers is:
4 + 7 = 4 + (4 + 3) = (2 × 4) + 3 = 11
The sum of the middle two numbers is:
5 + 6 = (4 + 1) + (4 + 2) = (2 × 4) + 3 = 11
When all three numbers are expressed in terms of the first number, we observe that the sum of the first and the last numbers, as well as the sum of the middle two terms, is equal to the sum of twice the first number and three.

ii) Algebraically, if the four consecutive natural numbers are n, n + 1, n + 2, n + 3
Then, the sum of the first and last numbers is:
n + (n + 3) = 2n + 3

The sum of the middle two numbers is:
(n + 1) + (n + 2) = 2n + 3
That is,
n + (n + 3) = (n + 1) + (n + 2) = 2n + 3, for any natural number n

Question 3.
For four consecutive natural numbers, is there any relation between the sum of the first two numbers and the last two numbers? Explain the reason for this relation and write the relation using algebra. What about the sum of the first and the third numbers and the sum of the second and the fourth? .
Answer:
Consider the four consecutive natural numbers as 5,6,7,8.

The sum of the first two numbers is:
5 + 6 = 5 + (5 +1) = (2 × 5) + 1 = 11

The sum of the last two numbers is:
7 + 8 = (5 + 2) + (5 + 3) = (2 × 5) + 5 = 15

Taking the difference between these two sums, that is
[(2 × 5) + 5] – [(2 × 5) + 1] = 15 – 11 = 4

When all three numbers are expressed in terms of the first number, we observe that the sum of the first two numbers equals twice the first number plus 1, and the sum of the last two numbers equals twice the first number plus 5.

The difference between these sums is a constant value of 4.
Algebraically, if the four consecutive natural numbers are n, n + 1, n + 2, n + 3 then

The sum of the first two numbers is:
n + (n + 1) = 2n + 1

The sum of the last two numbers is:
(n + 2) + (n + 3) = 2n + 5

Taking the difference between these two sums:
(2n + 5) – (2n + 1) = 4

That is, the sum of the first two numbers and the sum of the last two numbers differ by 4.This can be expressed as:
[n + (n + 1)] — [(n + 2) + (n + 3)] = (2n + 5) — (2n + 1) = 4, for any natural number n.

Now, we will examine the sums of the first and third numbers and the second and fourth numbers using the four consecutive natural numbers 5,6,7,8
The sum of the first and third numbers is:
5 + 7 = 5 + (5 + 2) = (2 × 5) + 2 = 12

The sum of the second and fourth numbers is:
6 + 8 = (5 + 1) + (5 + 3) = (2 × 5) + 4 = 14

Taking the difference between these two sums:
[(2 × 5) +4] – [(2 × 5) + 2] = 14 – 12 = 2

When all three numbers are expressed in terms of the first number, we observe that the sum of the first and third number equals twice the first number plus 2, and the sum of the second and fourth numbers equals twice the first number plus 4.

The difference between these sums is a constant value of 2.
Algebraically, if the four consecutive natural numbers are n, n + 1, n + 2, n + 3 then

The sum of the first and third numbers is:
n + (n + 2) = 2n + 2

The sum of the second and fourth numbers is:
(n + 1) + (n + 3) = 2n + 4

Taking the difference between these two sums:
(2n + 4) – (2n + 2) = 2

That is, the sum of the first and third numbers and the sum of the second and fourth numbers differ by 2. This can be expressed as:
[(n + 1) + (n + 3)] — [n + (n + 2)] = (2n + 4) — (2n + 2) = 2, for any natural number n.

Page 171

Question 1.
The bottom row of a five-storey number tower (like the three-storey and four-storey towers we have discussed) is shown below:

i) Before actually writing down the other numbers, see if you can guess how many times 10 is the topmost number. Check whether your guess is correct by filling in the other numbers
Answer:

Yes, here the topmost number is 16 times of 10.

ii) Explain using algebra that if we start with any five numbers equally apart and make a five- storey tower like this, then the topmost number is a fixed multiple of the middle number.
Answer:
Let’s start with x and add y each time to get the next four numbers. First line:
x, x + y, x + 2y, x + 3y, x + 4 y

Second line:
x + (x + y) = 2x + y .
(x + y) + (x + 2 y) = 2x + 3 y
(.x + 2y) + (x + 3 y) = 2x + 5y
(x + 3 y) + (x + 4 y) = 2x + 7 y

Third line: .
(2x + y) + (2.x + 3 y) = 4x + 4y
(2x + 3 y) + (2x + 5y) = 4x + 8y
(2x + 5 y) + (2x + 7 y) = 4x + 12 y

Fourth line:
(4x + 4 y) + (4x + 8y) = 8x + 12y
(4x + 8 y) + (4x + 12 y) = 8x + 20 y

At the Top:
(8x + 12y) + (8x + 20y) = 16x + 32y

From this we can say that starting with five numbers equally apart, we can construct a tower with the top number being 16 times the middle number of the first line.

iii) Is there any method to determine this without writing all the numbers?
Answer:
We can also find the top number just after writing the second line.
Here the numbers in the second line is:
2.x + y, 2x + 3y, 2x + 5y, 2x + 7y

The sum of the two middle numbers is
(2x + 3y) + (2x + 5y) = 4x + 8 y

And four times of this is:
4 x (4x + 8y) = (4 x 4x) + (4 x 8y) = 16x + 32y
So, the top number of a four-story tower must be four times the sum of the two middle number of the first line.

Question 2.
We can make number towers starting with any set of numbers, not necessarily equally apart. For example, look at this tower:

(i) Fill in the empty cells of the tower below:

Answer:

ii) In a tower like this, keep the 10’s where they are and change the second number of the bottom row to some number other than 1 or 2 and compute the other numbers. Is the topmost number still 50?
Answer:

Yes, here also the topmost number is 50 itself,

iii) Explain the reason for this using algebra.
Answer:
We can explain this using algebra,
Let’s consider x as the second number of the bottom row. So the tower will look like this:

First line will be:
First number : 10
Second number : x
Third number :10 – x
Fourth number: 10

Second line:
First number : 10 + x = 10 + x
Second number: 10
Third number : (10 – x) + 10 = 20 – x

Third line:
First number : (10 + x) + 10 = 20 + x
Second number : 10 + (20 – x) = 30 – x

At the top:
(20 + x) + (30 – x) = 50
From this we can find that the sum of middle numbers of the bottom line is equal to 10.
So whatever the number in the place of x the topmost number will be 50 which means 5 times of 10.

iv) Fill in the empty cells of this tower:

Answer:

Question 3.
Write the numbers from 1 to 100 in rows and columns as in the first picture below. Then draw some 9-cell squares in it, as in the second pictWe. Mark the middle number of each such square also:

Find the following for each of the squares:
i) The relation between the middle number and the sum of the numbers on its left and right.
Answer:
First Square

Middle number = 14
Left number = 13
Sum of left and right = 13 +
Relation between the middle and sum = \(\frac{28}{2}\) = 14

Second Square

Middle number = 63
Left number = 62
Sum of left and right = 62 +
Relation between the middle and sum = \(\frac{126}{2}\) = 63

Third Square

Middle number = 78
Left number = 77
Sum of left and right = 77 + 79 = 156
Relation between the middle and sum = \(\frac{156}{2}\) = 78
Let’s explain this using algebra,
Let the middle number be x and its left and right numbers be x – 1 and x + 1 respectively,
The sum of number on the left and right of the middle number = (x – 1) + (x + 1) = 2x
Thus, relation between the middle and sum = \(\frac{2x}{2}\) = x
This means the middle number is equal to the average of sum of its left and right numbers.

ii) The relation between the middle number and the sum of the numbers on its top and bottom.
Answer:
First Square

Middle number = 14
Top number = 4
Bottom number = 24
Sum of top and bottom = 4 + 24 = 28
Relation between the middle and sum = \(\frac{28}{2}\) =14

Second Square

Middle number = 63
Top number = 53
Bottom number = 73
Sum of top and bottom = 53 + 73 = 126
Relation between the middle and sum = \(\frac{126}{2}\) = 63

Third Square

Middle number = 78
Top number = 68
Bottom number = 88
Sum of top and bottom = 68 + 88 = 156
Relation between the middle and sum = \(\frac{156}{2}\) =78

Let’s explain this using algebra,
Let the middle number be x and its top and bottom numbers be x – 10 and x + 10 respectively,
The sum of number on the top and bottom of the middle number = (x – 10) + (x + 10) = 2x
Thus relation between the middle and sum = \(\frac{(x-10)+(x+10)}{2}=\frac{2 x}{2}\) = x
That is the middle number is equal to the average of sum of its top and bottom numbers.

iii) The relation between the middle number and the sums of the pairs of numbers diagonally on its top and bottom.
Answer:
First Square

Middle number =14
Sum of first pair of diagonal numbers = 3 + 25 = 28
Sum of second pair of diagonal numbers = 5 + 23 = 28
Sums of both diagonal pairs = 28 + 28 = 56
Relation between the middle and sum = \(\frac{56}{4}\) = 14

Second Square

Middle number = 63
Sum of first pair of diagonal numbers = 52 + 74 = 126
Sum of second pair of diagonal numbers = 54 + 72 = 126
Sums of both diagonal pairs = 126 + 126 = 252
Relation between the middle and sum = \(\frac{252}{4}\)= 63

Third Square

Middle number = 78
Sum of first pair of diagonal numbers = 67 +89 = 156
Sum of second pair of diagonal numbers = 69 + 87 = 156
Sums of both diagonal pairs= 156 + 156 = 312
Relation between the middle and sum = \(\frac{312}{4}\) = 78

Let’s explain this using algebra,
Let take the first number as x.
Then we can write the rest of the number as

so, the sum of first pair of diagonal numbers = x + x + 22
= 2x + 22
sum of second pair of diagonal numbers =x + 20 + x + 2
= 2x + 22
Thus, the sum of both diagonal pairs = 2x + 22 + 2x + 22
= 4x + 44
= 4 (x + 11)
Taking on fourth of the sum,that’s
\(\frac{4(x+11)}{4}\) = x + 11
That is , the middle number in equal to the one fourth of the sum of it’s diagonal pairs.

iv) The relation between the middle number and sum of all the numbers in the square Explain all this using algebra.
Answer:
First Square

Middle number =14
Sum of all numbers in the square
= 3 + 4 + 5 + 13 + 14 + 15 + 23 + 24 + 25 = 126
Relation between the middle and sum = \(\frac{56}{4}\) = 14

Second Square

Middle number = 63
Sum of all numbers in the square
= 52 + 53 + 54 + 62 + 63 + 64 + 72 + 73 + 74 = 567
Relation between the middle and sum = \(\frac{252}{4}\) = 63

Third Square

Middle number = 78
Sum of all numbers in the square
= 67 + 68 + 69 + 77 + 78 + 79 + 87 + 88 + 89
= 702
Relation between the middle and sum = \(\frac{312}{4}\) = 78
Let’s explain this using algebra,
Let take the first number as x.
Then we can write the rest of the number as x x + 1 x + 2

Thus the sum of all number in the square
= x + x + 1 + x + 2 + x + 10 + x + 11 + x + 12 + x + 20 + x + 21 + x + 22
= 9x + 99
= 9(x + 11)
Taking 1/9 th of the sum = \(\frac{9(x+11)}{9}\) = x + 11
That in the sum of all number in equal to the 9 times the middle number

Question 4.
On the calendar of any month, draw 4-cell squares at various positions.

i) Explain using algebra, why the sum of all four numbers in such a square is a multiple of 4.
ii) Explain using algebra, the relation between this sum and the smallest number in the square.
Answer:
Let’s consider each square
Sum of all four numbers in first square =3 + 4 + 10 + 11 = 28, which is a multiple of 4.
Sum of all four numbers in second square =13 + 14 + 20 + 21 = 68, which is a multiple of 4.
Sum of all four numbers in third square =23 + 24 + 30 + 31 = 108, which is a multiple of 4.
Let’s explain reason for this using algebra,
Consider x as the first number in the square.
Then what will be the other numbers?
Then we can form any four number in the square of the calendar as:

So, the sum of these four number will be:
= x + (x + 1) + (x + 7) + (x + 8)
= 4x + 16
= 4(x + 4)
That is that the sum of four numbers in a 2×2 square on a calendar is always a multiple of 4,
Let the smalles number be x the sum of all the four number in a square = 4x + 16 = 4 (x + 4)
That is by adding 4 to the smallest number and then taking four times that sum gives the total of all numbers in the square.

Page 176

Now try these problems:

Question 1.
Add a number leaving remainder 1 on division by 3, and a number leaving remainder 2 on division by 3. Explain, using algebra, why the sum of any two such numbers is divisible by 3
Answer:
Consider the number 7, which leaves a remainder 1 on division by 3. Thus, we can write 7 as:
7 = (2 × 3) + 1
Consider the number 17, which leaves a remainder 2 on division by 3. Thus, we can write 17 as:
17 = (5 × 3) + 2
Adding 7 and 17 we get
7 + 17 = 24 = 3 × 4
That is, there sum is divisible by 3.
Using algebra, let’s see why the sum of any two numbers is divisible by 3.
For that, consider the two numbers as 3n + 1 (remainder 1 when divided by 3) and 3m + 2 (remainder 2 when divided by 3), where m and n are natural numbers.
Then their sum is,
(3m + 1) + (3n + 2) = 3 m + 3n + 3 = 3(m + n + 1)
When m + n + 1 = p, then
(3m + 1) + (3n + 2) = 3p
where p is a natural number.
That is, the sum of these two numbers leaves no reminder. Thus, we can say that the sum of a number leaves add a number leaving remainder 1 on division by 3, and a number leaving remainder 2 on division by 3 division by 3 will leaves no reminder.

Question 2.
The numbers 12, 23, 34,… are got by starting with 12 and adding 11 again and again.
i) What is the remainder if any such number is divided by 11?
ii) Write the general algebraic form of all these numbers
iii) Is 100 among these numbers? What about 1000?
Answer:
i) The reminder will be 1. Thus, the number can be written as:
12 = (1 × 11)+ 1
23 = (2 × 11) + 1
34 = (3 × 11) + 1

ii) Algebraically we can write in the form of 1 In + 1 where n is one of the numbers 0, 1, 2, 3….

iii) For the number 100 we can write it as:
100 = (9 × 11) + 1
For the number 1000 we can write it as:
1000 = (90 × 11) + 10

Question 3.
The numbers 21,32, 43… are got by starting with 21 and adding 11 again and again
i) What is the remainder if any such number is divided by 11?
ii) Write the general algebraic form of all these numbers.
iii) Is 100 among these numbers? What about 1000?
Answer:
i) The reminder will be 10. Thus, the number can be written as:
21 =(1 × 11) + 10
32 = (2 × 11)+ 10
43 = (3 × 11)+ 10

ii) Algebraically, we can write in the form of 11n + 10 where n is one of the numbers 0, 1, 2, 3….
iii) For the number 100 we can write it as:
100 = (9 × 1) + 1
For the number 1000 we can write it as:
1000 = (90 × 11) + 10

Page 180

Now try these problems:

Question 1.
Add any two-digit number and the number got by reversing the digits. Explain using algebra, why all such sums are multiples of 11.
Answer:
Consider three examples for any two-digit number and the number got by reversing the digits, then 23 + 32 = 55
35 + 53 = 88
47 + 74 = 121
Using algebra, let’s check this.
Take the larger two numbers as 10m + n.
On reversing the digit, we get the reversed number as 10n + m.
We can now determine the sum between two-digit number and its reverse as follows:
(10m + n) + (10n + m) = 10m + n + 10n + m
= 11m + 11n = 11(m + n)
Thus, we get the sum as a multiple of 11 where m + n is the sum of the digits of the numbers.

Question 2.
From a two-digit number, subtract the sum of the digits. Explain using algebra why all such differences are multiples of 9?
Answer:
Consider a two-digit number 53,
Where the sum of the digits = 5 + 3 = 8.
Now, subtract the sum of the digits from the two-digit number, that is: 53 – 8 = 45
Using algebra, let’s check this.
Take the two-digit number as 10a + b, where the sum of the digits is a + b.
When the two-digit number is subtracted from the sum of the digits, then
(10a + b) – (a + b) = 10a + b – a – b = 9a
That is, the difference is always a multiple of 9.

Question 3.
i) Write the algebraic form of all three-digit numbers.
ii) Take any three-digit number and the number got by reversing its digits. Subtract the smaller from the larger. Explain using algebra, why all such differences are multiples of 99.
iii) From a three-digit number, subtract the sum of the digits. Explain using algebra why all such differences are multiples of 9.
Answer:
i) Consider the three-digit number as xyz, where x is in the hundreds place, y is in the tens place, and z is in the one’s place. Thus, the value of the three-digit number can be expressed as:
100x + 10y + z

ii) Consider the three-digit number 531; on reversing the digits, we get 135.
Then subtract the smaller from the larger we get, 531 – 135 = 396
On dividing 396 ÷ 99 = 4,
so 396 is a multiple of 99.
Let’s explain this using algebra;
Take the three-digit number as:
100a + 10b + c
On reversing the digit, we can write it as:
100c + 10b + a
So, the difference between these two numbers is:
(100a + 10b + c) – (100c + 10b + a) = 100a – a + 10b – 10b + c – 100c
= 99a – 99c = 99(a – c)
The difference between any three-digit number and its reverse is always a multiple of 99 because the algebraic form of the difference is 99(a – c), which is clearly divisible by 99.

iii) Consider the three-digit number 352
Sum of digits: 3 + 5 + 2 = 10
Difference: 352 – 10 = 342
On dividing, 342 ÷ 9 = 38, so 342 is a multiple of 9.
Let’s explain this using algebra;
For that, consider the three-digit number
100a + 10b + c
The sum of the digits of this number is:
a + b + c
On subtracting the three-digit number from the sum of the digits of this number
(100a + 10b + c) – (a + b + c) = 100a + 10b + c – a – b – c
= (100a – a) + (10b – b) + (c – c)
= 99a + 9b
= 9(11a + b)

Intext Questions And Answers

Question 1.
What if we take the starting numbers as x — 2, x, x + 2 instead?
Answer:
First line is x – 2, x, x + 2

Second line:
(x — 2) + x = 2x – 2 x + (x + 2) = 2x + 2

At the top:
(2x – 2) + (2x + 2) = 4x

So, starting with three numbers equally apart, we can construct a tower with the top number being four times the middle number.
For example,

Now consider given four-story towers,

The first tower starts with the four consecutive numbers 6, 7, 8, 9.
While in second 9, 12, 15, 18, was obtained by repeatedly adding 3 to 9 from the beginning.

In both of this cases we can see that the top number is equal to four times the sum of the two middle numbers of the first line.

Question 2.
Make some more towers like this, starting with other number ?
Answer:

Let’s write this using algebra:
Start with x and add y each time to get the next three numbers.
First line:
x, x + y, x + 2y, x + 3y

Second line:
x + (x + y) = 2x + y
(x + y) + (x + 2y) = 2x + 3y
(x + 2y) + (x + 3y) = 2x + 5y

Third line:
(2x + y) + (2x + 3 y) = 4x + 4y
(2x + 3 y) + (2x + 5y) = 4x + 8y

At the top:
(4x + 4y) + (4x + 8y) = 8x + 12y

The sum of the two middle numbers is
(x + y) + (x + 2y) = 2x + 3y

And four times of this is: ‘
4 × (2x + 3y) = (4 × 2x) + (4 × 3y) = 8x + 12y

Thus we can conclude that the top number equal to the four times the sum of the two middle number of the first line.
Now let’s see how can be find the top number just after writing the second line. .
The numbers in the second line are ‘
2x + y, 2x + 3y, 2x + 5y
Let’s rewrite it like this:
2x + y
2x + 3y = (2x + y) + 2y
2x + 5y = (2x + 3y) + 2y
The three numbers are at an equal distance of 2y apart.
So here also, the top number of a three-story tower must be four times the middle number 2x + 3y.

Question 3.
In the same way, can you explain using algebra, why the sum of an even number and an odd number is an odd number?
Answer:Consider the even number as 2m and the odd number as 2n + 1, where m and n are either 0 or some natural numbers.
Then their sum is,
2m + (2n + 1) = 2m + 2n +1 = 2(m + n) + 1
When m + n = p, then
2m + (2n + 1) = 2p + 1
where p is a natural number.
Here, 2p is an even number, as it is two times a natural number. Adding 1 to an even number results in an odd number.
Therefore, we can conclude that the sum of an even number 2m and an odd number 2n + 1 is always an odd number.

Now we can explore this based on the remainders obtained when dividing by 3.

Set of numbers	Specialty	Algebraic form
0, 3, 6, 9,…	Remainder 0 on division by 3	3n (n = 0,1,2,3 …)
1, 4, 7, 10,…	Remainder 1 on division by 3	3n + 1 (n = 0,1,2,3 …)
2, 5, 8, 11,…	Remainder 2 on division by 3	3n + 2 (n = 0,1,2,3 …)

Class 7 Maths Chapter 12 Kerala Syllabus Algebra Questions and Answers

Question 1.
Find the algebraic expression for the numbers got by adding 10 repeatedly to 9.
Answer:
9 + 10 = 19
19+ 10 = 29
29 + 10 = 39
Thus, the numbers are 19, 29, 39….
Using algebra, we can express it as 10n + 9.

Question 2.
Take a set of five consecutive natural numbers and add all five.
i) Check if the sum has any relation with any one of the five numbers added.
ii) Explain why this relation holds for any five consecutive natural numbers.
iii) Write this relation, first in ordinary language and then using algebra.
Answer:
Let’s take an example of consecutive natural numbers and add them together:
34 + 35 + 36 + 37 + 38 = 180
i) For the consecutive natural numbers 34, 35, 36, 37, 38.
34 + 35 + 36 + 37 + 38 = 180 and 5 × 36 = 180
That is, the sum of five consecutive numbers is always five times the middle number.

ii) Consider the five consecutive numbers 34, 35, 36, 37 and 38. These can be expressed in relation to the middle term as 36 – 2, 36 – 1, 36, and 36 + 1, 36 + 2.
Thus, the sum of these five numbers is:
(36 – 2) + (36 – 1) + 36 + (36 + 1) + (36 + 2) = 180
On simplifying this expression, we get
(36 – 2) + (36 – 1) + 36 + (36 + 1) + (36 + 2) = 5 × 36 = 180
This is because, in the addition of the five numbers the 2, 1 and -1,-2 cancel each other out, resulting in a sum that is five times the middle number.

iii) In ordinary language, we can say it as:
The sum of any five consecutive natural numbers is five times the middle number. Algebraically, if the five consecutive numbers are x – 2, x — 1, x, and x + 1, x + 2 the sum is:
(x – 2) + (x – 1) + x + (x + 1) + (x + 2) = 5x, for any natural numbers x

Question 3.
The numbers 25,36, 47… are got by starting with 25 and adding 11 again and again
i) What is the remainder if any such number is divided by 11?
ii) Write the general algebraic form of all these numbers.
iii) Is 100 among these numbers? What about 1000?
Answer:
i) The remainder will be 3. Thus, the number can be written as:
25 = (2 × 11) + 3
36 = (3 × 11) + 3
47 = (4 × 11) + 3

ii) Algebraically we can write in the form of 11n + 3 where n is one of the numbers 0, 1, 2, 3….

iii) For the number 100 we can write it as: 100 = (9 × 11) + 1
For the number 1000 we can write it as:
1000 = (90 × 11) + 10

Question 4.
Consider a 5 storey pyramid start with numbers one and adding 2 repeatedly to 1 from the beginning the first line of the pyramid and so on. Explain the relations you find using algebra.
Answer:

Algebraic form of pyramid:

The middle number in the bottom row = x + 4
Top most number = 16(x + 4) = 16x + 64
That is the top most number is 16 times the middle number in the bottom row.

Question 5.

a) Three number pyramids are given here. Find the relation between the first and second pyramid and complete the third pyramid accordingly.
Answer:

b) Which will be the starting number to get 100 as the topmost number?
Answer:
Beginning with 10, the top most number be 100.

c) Find the relation between the topmost number and the first number of the bottom row. Write the algebraic form of it.
Answer:

The first number of the bottom can be taken as x.
Then the rest of the numbers became 2x, 3x, 4x respectively.
For the second row x + 2x = 3x 3x + 4x = 7x
Top most number is 3x + 7x = 10x
From this we can say that the top most number will be 10 times of the first number of the bottom row.

Class 7 Maths Chapter 12 Notes Kerala Syllabus Algebra

In this chapter, we will explore the exciting world of algebra, looking at different ways to express mathematical ideas about measurements and numbers more simply. Algebra serves as a powerful tool that allows us to represent relationships and patterns using symbols and letters, making complex ideas easier to understand.

• The sum of two consecutive natural numbers is equal to the sum of twice the smaller number and’ one. We can represent the statements in algebraic form as follows: x + (x + 1) = 2x + 1, for any natural number x.
• For any three consecutive natural numbers, the sum of the first and the last, is equal to twice the middle number. We can represent the statements in algebraic form as follows:
  (x – 1) + (x + 1) = 2x, for any natural number x.
• Even numbers are those numbers that are divisible by 2. That is, any even number can be written in the form 2n, where n is one of the numbers 0, 1, 2, 3,… .
• Odd numbers are those that leave a remainder 1 on division by 2. That is, any odd number can be written in the form 2n + 1, where n is one of the numbers 0,1,2,3,…
• When arranging all two-digit numbers in rows and columns, we can express the general algebraic form of any two-digit number as: 10 m + n (m = 1,2, …,9; n = 0,1,2, …,9)
• When the digits of a two-digit number are reversed and the smaller number is subtracted from the larger, the resulting difference is always a multiple of 9. This difference is given by m – n, where m and n are the digits of the original number.

Throughout this chapter, we will uncover algebraic forms for various sets of consecutive numbers, enhancing our understanding of how algebra can simplify and clarify numerical relationships.

Numbers And Algebra
Let’s look at more ways to write facts about measurements and numbers in shorthand using algebra.
First, let’s examine the case of two consecutive numbers. Take 156 and 157 as an example. Let’s check if adding 156 and 157 gives the same result as adding 1 to 2 times 156.
That is,
156 + 157 = 313
(2 × 156) + 1 = 313
Here, the number 157 in the first operation is not there in the second operation. If we can write 157 as:
156 + 1 = 157
Then,
156 + 157 = 156 + (156 + 1)
That is,
156 + (156 + 1) = (156 + 156) + 1
= (2 × 156) + 1

In general, we can write it as:
The sum of two consecutive natural numbers is equal to the sum of twice the smaller number and one.
Let’s examine how this can be expressed in algebraic terms.

Let x and y be two consecutive natural numbers, with x as the smaller number and y as the larger number. Using x we can denote the next natural number as x + 1. That is, add 1 to the x.
So, let’s get started like this:

• x is a natural number
• The next natural number is x + 1
• Adding these two gives the number x + (x + 1)

Now, let’s examine the number obtained by adding one to twice the smaller number:

• The smaller number is x
• Twice this is 2x
• 1 added to this is 2x + 1

Thus, we can express the statements above in algebraic form as:
x + (x + 1) = 2x + 1, for any natural number x
This result holds true not only for natural numbers but also for fractions.
That is, it is true for all numbers. The only thing is, instead of two consecutive numbers, we should say, a number and one more than the number.

Hence, we can say
The sum of a number and one more than it, is equal to sum of twice the number and one.
We already know that:
(x + y) + z = x + (y + z) for any numbers x,y,z .
In reverse, we can read as
x + (y + z) = (x + y) + z
Taking y as x and z as 1, we get
x + (x + 1) = (x + x) + 1
That is,
x + (x + 1) = 2x + 1

Now, let’s examine the case of three consecutive natural numbers. Take 54, 55, 56 as an example.
Let’s check if 54 + 56 and 2 × 55 give the same result.
That is,
54 + 56 = 110
2 × 55 = 110

Now we can link 54 and 56 to 55 as:
54 = 55 + 56 = 55 + 1

Thus, we can write
54 + 56 = (55 – 1) + (55 + 1)

Now evaluate the operation done in (55 – 1) + (55 + 1), one by one:

• Two 55’s are added
• 1 is added
• 1 is subtracted

That is,
54 + 56 = (55 – 1) + (55 + 1)
= (2 × 55) + 1 – 1
= 2 × 55

In general, we can write it as:
For any three consecutive natural numbers, the sum of the first and the last, is equal to twice the middle number.

Let’s examine how this can be expressed in algebraic terms.
Here, the operation is described using the middle number. Therefore, let’s represent it as x.

• Middle number is x
• The first number is 1 subtracted from x, that is x – 1
• The last number is 1 added to x, that is x + 1 So, the above result as an equation is
  (x – 1) + (x + 1) = 2x

As said earlier, the operations in (x – 1) + (x + 1) means adding two x’s, then adding 1, and then subtracting 1; in effect, just adding two x’s. And this means multiplying x by 2.
(x – 1) + (x + 1) = 2x, for any natural number x
We can also state it like this: ‘
x + z = 2y for any consecutive natural numbers x, y, z.

Multiple And Remainder
We know that for any two natural numbers, one number can be expressed as the sum of a multiple of the other number plus a remainder, using division.
Consider two natural numbers, 7 and 3 On dividing 7 by 3, we can write:
7 = (3 × 2) + 1
On dividing 3 by 7, we can write:
3 = (0 × 7) + 3

Let’s consider the numbers that can be divided by 2 without a remainder.
For example,
2 = 1 × 2
4 = 2 × 2
6 = 3 × 2

These numbers are called even numbers.
Since
0 = 0 × 2
That is, zero is also an even number.
In general,
Even numbers are those numbers which are divisible by 2.
We can express this algebraically as:
Any even number can be written in the form 2n, where n is one of the numbers 0, 1, 2, 3,…

Now let consider the numbers that leave a remainder when divided by 2.
For example,
1 = (0 × 2) + 1
3 = (1 × 2) + 1 ‘
5 = (2 × 2) + 1 .
These numbers are called odd numbers.
In general,
Odd numbers are those that leave a remainder 1 on division by 2.
We can express this algebraically as:
Any odd number can be written in the form 2n + 1, where n is one of the numbers 0, 1, 2, 3,…

Using algebra, let’s demonstrate that the sum of two even numbers is also an even number. Consider two even numbers 2m and 2n, where m and n are either 0 or some natural numbers.
Then their sum is,
2m + 2n = 2(m + n)
When m + n = p, then
2m + 2n = 2p
where p is either 0 or some natural number.
Therefore, 2p is an even number.

Using algebra, now let’s demonstrate that the sum of two odd numbers is also an even number. Consider two even numbers 2m + 1 and 2n + 1, where m and n are either 0 or some natural numbers. Then their sum is,
(2m + 1) + (2 n + 1) = 2m + 2n + 2
= 2 (m + n + 1)
When m + n + 1 = p, then
(2m + 1) + (2 n + 1) = 2 p
where p is a natural number.
Therefore, 2p is an even number.

Number Curiosities
Let’s write any three consecutive natural numbers.in a line:
3 4 5
Add the adjacent pair of numbers and write on the top:

Again add these two number and write it op the top:

Now we can create some more set of consecutive numbers and create tower like this:

Here, the number at the top is 4 times the middle number of the three we start with
using algebra we can explain the reason.
For that let’s start by taking the middle number as x.
What about the numbers on the left and right?
But we know the numbers before and after the middle number in three consecutive numbers are one less and one more.
So the bottom line will be x – 1, x, x + 1.
Now we can find the numbers on above that:

Then the number at the top will be:

Digits And Numbers
We can now arrange all two-digit numbers in rows and columns as follows:

The general algebraic expression for the first row can be written as:
10 + n (n = 0, 1,2,3 )
For the second row, it can be written as:
20 + n (n = 0, 1,2,3 )
Thus, the algebraic form of numbers in each row can be represented as:

where n represents the digit in the one’s place in each of these numbers.
If we write the digit in the ten’s place as m, then all these numbers can be written in the algebraic form 10m + n. Thus, the general algebraic form of all two-digit numbers is
10m+ n (m = 1, 2,…………..9 n = 0,1,2, …,9)

Now let’s look at a problem where we can take any two-digit number and the two-digit number got by reversing its digits, then subtract the smaller from the larger.
For example:
32 – 23 = 9
42 – 24-= 18

Using algebra, let’s check this.
Take the larger two numbers as 10m +n. On reversing the digit, we get the reversed number as lOn + m.
We can now determine the difference between a two-digit number and its reverse as follows:
(10m + n) – (10n + m) = (10m + n — 10n) – m
= (10m – 10n + n) – m
= (10m – 9 n) – m – 10m – m – 9n
= 9m  -9n
= 9(m – n)
Thus, we get the difference as a multiple of 9 where m – n is the difference of the digits of the numbers.

The sum of two consecutive natural numbers is equal to the sum of twice the smaller number and one. We can represent the statements in algebraic form as follows:
x + (x + 1) = 2x + 1, for any natural number x.

• For any three consecutive natural numbers, the sum of the first and the last is equal to twice the middle number. We can represent the statements in algebraic form as follows:
  (x – 1) + (x + 1) = 2x, for any natural number x.
• Even numbers are those numbers that are divisible by 2. That is, any even number can be written in the form 2n, where n is one of the numbers 0, 1, 2, 3,…
• Odd numbers are those that leave a remainder of 1 on division by 2. That is, any odd number can be written in the form 2n + 1, where n is one of the numbers 0, 1, 2, 3,…
• When arranging all two-digit numbers in rows and columns, we can express the general algebraic form of any two-digit number as:
  10m + n (m = 1, 2,…,9; n = 0, 1, 2,…,9)
• When the digits of a two-digit number are reversed and the smaller number is subtracted from the larger, the resulting difference is always a multiple of 9. This difference is given by m – n, where m and n are the digits of the original number.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 9 Number Relations Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 9 Solutions Number Relations

Class 7 Maths Chapter 9 Number Relations Questions and Answers Kerala State Syllabus

Number Relations Class 7 Questions and Answers Kerala Syllabus

Page 131

Now try these problems:

Question 1.
Find the number of factors of each number below:
i) 40
ii) 54
iii) 60
iv) 100
v) 210
Answer:
i) The factors of 40, expressed as powers of prime numbers, are:,
40 = 2³ × 5 = 8 × 5
Therefore, number of factors = (3 + 1)(1 + 1) = 4 × 2 = 8

ii) The factors of 54, expressed as powers of prime numbers, are:
54 = 3³ × 2 ,
Therefore, number of factors = (3 + 1)(1 + 1) = 4 × 2 = 8

iii) The factors of 60, expressed as powers of prime numbers, are:
60 = 2² × 3 × 5
Therefore, number of factors = (2 + 1) (1 + 1) (1 + 1) = 3 × 2 × 2 = 12

iv) The factors of 100, expressed as powers of prime numbers, are:
100 = 5² × 2²
Therefore, number of factors = (2 + 1)(2 + 1) = 3 × 3 = 9

v) The factors of 210, expressed as powers of prime numbers, are:
210 = 7 × 5 × 3 × 2
Therefore, number of factors = (1 + 1) (1 + 1) (1 + 1) (1 + 1) = 2 × 2 × 2 × 2 = 16

Question 2.
From the number of factors of a number, we can deduce some peculiarities of the number. The table below lists these for number of factors up to 5. Extend it to number of factors up to 10

Answer:

6	Fifth power of a prime	p5(p, a prime)
	Product of two primes	pq2 or p2q(p, q primes)
7	Sixth power of a prime	p6(p, a prime)
8	Seventh power of a prime	p7 (p, a prime)
	Product of three primes	pqr (p, q, r primes)
	Product of two primes	p3q or pq3 (p, q primes)
9	Eight power of a prime	p8 (p, a prime)
	Product of two primes	p2q2 (p, q primes)
10	Ninth power of a prime	p9(p, a prime)
	Product of two primes	pq4 or p4q (p,q primes)

Page 134

Question 1.
For each pair of numbers given below, find the largest common factor and all other common factors:
i) 45, 75
ii) 225, 275
iii) 360, 300
iv) 210, 504
v) 336, 588
Answer:
i) Factors of 45 = 3² × 5
Factors of 75 = 3 × 5²
Common primes: 3 and 5
Take the lowest powers: 3¹ × 5¹ = 15
Largest common factor: 15
Other common factors: 1,3,5

ii) Factors of 225 = 3² × 5²
Factors of 275 = 5² × 11
Common prime: 5
Take the lowest power: 2 = 25
Largest common factor: 25
Other common factors: 1,5

iii) Factors of 360 = 2³ × 3² × 5
Factors of 300 = 2² × 3 × 5²
Common primes: 2, 3, and 5
Take the lowest powers: 2², 3¹, 5¹
Largest common factor = 2² × 3¹ × 5¹ = 4 × 3 × 5 = 60
Other common factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

iv) Factors of 210 = 2 × 3 × 5 × 7
Factors of 504 = 2³ × 3² × 7
Common primes: 2, 3, and 5
Take the lowest powers: 2¹, 3¹, 5¹
Largest common factor = 2¹ × 3¹ × 5¹ = 2 × 3 × 5 = 30
Other common factors: 1, 2, 3, 5, 6, 10, 15, 30

v) Factors of 336 = 2⁴ × 3 × 7
Factors of 558 = 2² × 3 × 7²
Common primes: 2, 3, and 7
Take the lowest powers: 2², 3¹, 7¹
Largest common factor = 2² × 3¹ × 7¹ = 4 × 3 × 7 = 84
Other common factors: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84

Question 2.
i) What is the largest common factor of two different prime numbers?
ii) Can the largest common factor of two composite numbers be 1?
iii) If two numbers are divided by their largest common factor, what would be the largest common factor of the quotients?
Answer:
i) The largest common factor of two different prime numbers is 1, since prime numbers have no common factors other than 1.
ii) Yes, the largest common factor of two composite numbers can be 1 if they are co prime, meaning they have no common factors other than 1.
iii) If two numbers are divided by their largest common factor, the largest common factor of the quotients will be I. This is because the largest common factor is the greatest number that divides both original numbers, and dividing by it eliminates any common factors from the quotients.

Class 7 Maths Chapter 9 Kerala Syllabus Number Relations Questions and Answers

Question 1.
Find the number of factors of each number below:
i) 36
ii) 84
iii) 144
Answer:
i) The factors of 36, expressed as powers of prime numbers, are:
36 = 2² × 3²
Therefore, number of factors = (2 + 1) (2 +1) = 3 × 3 = 9

ii) The factors of 84, expressed as powers of prime numbers, are:
84 = 2² × 3¹ × 7¹
Therefore, the number of factors = (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12

iii) The factors of 144, expressed as powers of prime numbers, are:
144 = 2⁴ × 3²
Therefore, the number of factors = (4 + 1)(2 + 1) = 5 × 3 = 15

Question 2.
For each pair of numbers given below, find the largest common factor:
i) 48, 180
ii) 90, 150
iii) 84, 126
Answer:
i) Prime factors of 48 = 2⁴ × 3¹
Prime factors of 180 = 2² × 3² × 5¹
Common primes: 2 and 3
Take the lowest powers: 2², 3¹
Largest Common Factor = 2² × 3¹ = 4 × 3 = 12

ii) Prime factors of 90 = 2¹ × 3² × 5¹
Prime factors of 150 = 2¹ × 3¹ × 5²
Common primes: 2,3, and 5
Take the lowest powers: 2¹, 3¹ , 5¹
Largest Common Factor = 2¹ × 3¹ × 5¹ = 2 × 3 × 5 = 30

iii) Prime factors of 84 = 2² × 3¹ × 7¹
Prime factors of 126 = 2¹ × 3² × 7¹
Common primes: 2, 3, and 7
Take the lowest powers: 2¹, 3¹, 7¹
Largest Common Factor = 2¹ × 3¹ × 7¹ = 2 × 3 × 7 = 42

Class 7 Maths Chapter 9 Notes Kerala Syllabus Number Relations

In this chapter, we will explore important concepts related to numbers and their factors. You’ll learn how to determine the number of factors for prime numbers and how to identify common factors between two numbers.

In short, we can explain as,

• To find the number of factors of the product of powers of two prime numbers, we have to add one to each exponent and multiply these numbers.
• To find the common factors of two numbers, first write the factors of each number as the product of prime numbers raised to their powers. Then, identify the common prime numbers and take the lowest power of each. These will be the common factors.
• These fundamental ideas are essential for understanding the relationships between numbers and will help you solve problems with greater confidence.

Number Of Factors

Prime numbers are natural numbers that can only be divided by 1 and the number itself. Examples include 2, 3, 5, 7, and 11.
Each of these numbers has exactly two factors.,
Now, let’s examine how the number of factors changes as the power of a prime number varies.

In general,
The number of factors of a power of any prime number is one more than the exponent.
Algebraically we can say,
If p is a prime number and n is a natural number, then the number of factors of pⁿ is n + 1.

Now, let’s explore how the number of factors changes when a prime number is multiplied by another number.

For 3 × 5 = 15
Factors of 3 = 1, 3
Factors of 5 = 1, 5
Factors of 15 = 1, 3, 5, 15
Thus, the number of factors = 4
We can tabulate it like this:

1	3	(Factors of 3)
5	15	(Factors of 3 multiplied by 5)

Thus, the number of factors = 4

For 3² × 5 = 45
Factors of 3² as the power of prime numbers = 1, 3, 3² that is 1, 3, 9
Factors when each of them multiplied by 5 are:

Thus, the number of factors = 3 + 3 = 3 × 2 = 6

For 3² × 5² = 225

Thus, the number of factors = 3 × 3 = 9

For 3³ × 5³

The factors are:

Thus, the number of factors = 4 × 4 = 16

In general,
The number of factors of the product of powers of two prime numbers is got by adding one to each exponent and multiplying these numbers.
Algebraically we can say,
If p and q are two different primes and m, n are any two natural numbers, then the number of factors of p^m qⁿ is (m + 1)(n + 1).

Now, let’s consider the situation with three different prime numbers.
Examine the expression 3³ × 5² × 11
First, tabulate the factors of 3³ × 5²

Here, the number of factors for 3³ × 5² = 4 × 3 = 12
Then, tabulate the factors of 3³ × 5² × 11

Thus, all together 3³ × 5² × 11 = 12 × 2 = 24

In general,
To compute the number of factors of a number, we write it a product of powers of different prime numbers, and find the product of the numbers got by adding one to each exponent.

Common Factors
Let’s understand what a common factor of two numbers means:
For that consider the two numbers 180 and 270.
Factors of 180 = 2² × 3² × 5
Factors of 270 = 2 × 3³ × 5
Here, the prime common for both numbers are 2, 3, 5
The smaller power of these primes in the factorizations is 2, 32, 5
Here, the common factors are the factors of 2 × 3² × 5
We can tabulate it as follows:

In this case, the largest common factor of 180 and 270 is 90.
In short,
To find the common factors of two numbers, first write each number as the product of prime numbers raised to their powers. Then, identify the common prime numbers and take the lowest power of each. These will be the common factors.

• The number of factors of a power of any prime number is one more than the exponent.
• If p is a prime number and n is a natural number, then the number of factors of pn is n + 1.
• The number of factors of the product of powers of two prime numbers is got by adding one to each exponent and multiplying these numbers.
• If p and q are two different primes and m, n are any two natural numbers, then the number of factors of pmqn is (m + 1)(n + 1).
• To find the common factors of two numbers, first write each number as the product of prime numbers raised to their powers. Then, identify the common prime numbers and take the lowest power of each. These will be the common factors.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 8 Repeated Multiplication Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 8 Solutions Repeated Multiplication

Class 7 Maths Chapter 8 Repeated Multiplication Questions and Answers Kerala State Syllabus

Repeated Multiplication Class 7 Questions and Answers Kerala Syllabus

Page 112

Question 1.
Write each number below either as a power of a single prime or as a product of powers of different primes:
i) 125
ii) 72
iii) 100
iv) 250
v) 3600
vi) 10800
Answer:
i) 125
125 = 5 × 5 × 5 = 5³

ii) 72
72 = 2 × 2 × 2 × 3 × 3
= 2³ × 3²

iii) 100
100 = 2 × 2 × 5 × 5
= 2² × 5²

iv) 250
250 = 2 × 5 × 5 × 5
= 2 × 5³

v) 3600
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 2⁴ × 3² × 5²

(vi) 10800
10800 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5
= 2⁴ × 3³ × 5²

Page 115

Question 1.
Calculate the powers below as fractions:
(i) \(\left(\frac{2}{3}\right)^2\)
Answer:
\(\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)=\frac{4}{9}\)

(ii) \(\left(1 \frac{1}{2}\right)^2\)
Answer:
\(\left(1 \frac{1}{2}\right)=\left(\frac{3}{2}\right)\)
= \(\left(\frac{3}{2}\right)\left(\frac{3}{2}\right)=\left(\frac{9}{4}\right\)

(iii) \(\left(\frac{2}{5}\right)^3\)
Answer:
\(\left(\frac{2}{5}\right)\left(\frac{2}{5}\right)\left(\frac{2}{5}\right)\)
= \(\frac{8}{125}\)

(iv) \(\left(2 \frac{1}{2}\right)^3\)
Answer:
\(\left(2 \frac{1}{2}\right)=\left(\frac{5}{2}\right)\)
\(\left(\frac{5}{2}\right)\left(\frac{5}{2}\right)\left(\frac{5}{2}\right)=\frac{125}{8}\)

Question 2.
Calculate the powers below in decimal form:
(i) (0.5)²
(ii) (1.5)²
(iii) (0.1)³
(iv) (0.01)³
Answer:
i) (0.5)²
= 0.5 × 0.5
= 0.25

ii) (1.5)²
= (1.5)(1.5)
= 2.25

iii) (0.1)³
= (0.1)(0.1)(0.1)
= 0.001

iv) (0.01)³
= (0.01)(0.01)(0.01)
= 0.000001

Question 3.
Using 15³ = 3375 calculate the powers below:
i) (1.5)³
ii) (0.15)³
iii) (0.015)³
Answer:
i) (1.5)³ = 1.5 × 1.5 × 1.5 = 3.375
ii) (0.15)³ = 0.15 × 0.15 × 0.15 = 0.003375
iii) (0.015)³ = 0.015 × 0.015 × 0.015 = 0.000003375

Page 118

Question 1.
Write each product below as the product of powers of different primes:
i) 72 × 162
ii) 225 × 135
iii) 105 × 175
iv) 25 × 45 × 75
Answer:
i) 72 × 162
72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²
162 = 3 × 3 × 3 × 3 × 2 = 3⁴ × 2
72 × 162 = (2³ × 3²)(3⁴ × 2)
= 2⁴ × 3⁶

ii) 225 × 135
225 = 3 × 3 × 5 × 5 = 3² × 5²
135 = 3 × 3 × 3 × 5 = 3³ × 5¹
225 × 135 = (3² × 5²)( 3³ × 5¹)
= 3⁵ × 5³

iii) 105 × 175
105 = 3 × 5 × 7 = 3¹ × 5¹ × 7¹
175 = 5 × 5 × 7 = 5² × 7¹
105 × 175 = (3¹ × 5¹ × 7¹)(5² × 7¹)
= 3 × 5³ × 7²

iv) 25 × 45 × 75
25 = 5¹ × 5¹
45 = 3² × 5¹
75 =3¹ × 5²
25 × 45 × 75 = (5¹ × 5¹)(3² × 5¹)(3¹ × 5²)
= 5⁵ × 3³

Question 2.
Write the product of the numbers from 1 to 15 as the product of powers of different primes.
Answer:
1 (no primes)
2 = 2¹
3 = 3¹
4 = 2²
5 = 5¹
6 = 2¹ × 3¹
7 = 7¹
8 = 2³
9 = 3²
10 = 2¹ × 5¹
11 = 11¹
12 = 2² × 3¹
13 = 1³¹
14 = 2¹ × 7¹
15 = 3¹ × 5¹
∴ The product of the numbers from 1 to 15 as the product of powers of different primes are
1 × 2¹¹ × 3⁶ × 5³ × 7² × 11¹ × 13¹

Question 3.
Consider the numbers from 1 to 25
i) Which of them are divisible by 2, but not by 4?
ii) Which of them are divisible by 4, but not by 8?
iii) Which of them are divisible by 8, but not by 16?
iv) Which of them are divisible by 16?
v) What is the highest power of 2 that divides the product of the numbers from 1 to 25 without remainder?
Answer:
i) The numbers from 1 to 25 that are divisible by 2 but not by 4 are:
2, 6, 10, 14, 18, 22

ii) The numbers from 1 to 25 that are divisible by 4 but not by 8 are:
4, 12, 20

iii) The numbers from 1 to 25 that are divisible by 8 but not by 16 are:
8, 24

iv) The numbers from 1 to 25 that are divisible by 16 are:
16

v) Numbers that are divisible by 2 from 1 to 25 are:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24

Thus, the total numbers that can be divisible by 2 are 12.
Number that are divisible by 4 from 1 to 25 are:
4, 8, 12, 16, 20, 24

Thus, the total number numbers that can be divisible by 4 are 6.
Number that are divisible by 8 from 1 to 25 are:
8, 16, 24

Thus, the total number numbers that can be divisible by 8 are 3.
Number that are divisible by 16 from 1 to 25 are:
16

Thus, the total number numbers that can be divisible by 16 are 1.
From this, the highest power of 2 = 12 + 6 + 3 + 1 = 22
So, the highest power of 2 that divides the product of number from 1 to 25 is 2²².

Question 4.
Consider the product of the numbers from 1 to 25
i) What is the highest power of 5 which divides this product without remainder?
ii) And the highest power of 10 dividing this product without remainder?
iii) How many zeros does this product end with?
Answer:
i) The numbers divisible by 5 are 5, 10, 15, 20, 25 that is 5 number.
The number divisible by 25 is 25 that is 1 number.
Thus, the total contribution from multiple of 5 = 5 + 1 = 6
Thus, the highest power of 5 that divides the product between 1 to 25 is 56.

ii) Since 10 = 2 × 5. We need to count how many times both 2 and 5 appear as factors in the product.
Thus, the smallest count between the 2 and 5 will give us the highest powers of 10.
The highest power of 2, which divides the product of numbers from 1 to 25 is 22 and
the highest power of 5, which divides the product of numbers from 1 to 25 is 6.
Thus, the highest power of 10 = smallest count of the highest power of 2 and 5
Therefore, the highest power of 10 dividing the product from 1 to 25 is 106.

iii) Since we found that 106 divides the product of numbers from 1 to 25.
This means that the product ends with 6 zeros.

Page 122

Question 1.
Calculate the following quotients:
i) 512 ÷ 64
ii) 3125 ÷ 125
iii) 243 ÷ 27
iv) 1125 ÷ 45
Answer:
i) 512 ÷ 64
512 = 29 = 26 × 23
64 = 26 = 23 × 23
512 ÷ 64 = ( 2⁶ × 2³) ÷ (2³ × 2³)
= 2⁶ ÷ 2³
= 2^6-3
= 2³
= 8

ii) 3125 ÷ 125
3125 = 5⁵ = 5² × 5³
125 = 53 = 5² × 5¹
3125 ÷ 125 = (5² × 5³) ÷ (5² × 5¹)
= 5³ ÷ 5¹
= 5^3-1
= 5²
=25

iii) 243 ÷ 27
243 = 3⁵ = 3² × 3³
27 = 3³ = 3² × 3¹
243 ÷ 27 = (3² × 3³) – (3² × 3¹)
= 3³ – 3¹
= 3^3-1
= 3²
= 9

iv) 1125 ÷ 45
1125 = 5³ × 3²
45 = 5¹ × 3²
1125 ÷ 45 = (53 × 32) ÷ (51 × 32)
= 5^3-1
= 5²
= 25

Question 2.
i) Write half of 2¹⁰ as a power of 2.
ii) Write one-third of 3¹² as a power of 3.
Answer:
i) Half of 2¹⁰ = 2¹⁰ ÷ 2¹
= 2^10-1
= 2⁹

ii) One-third of 3¹² = 3¹² ÷ 3¹
= 3¹¹

There is another way for doing this type of problems, and it can be stated as a general principle, using algebra:
\(\frac{x^m}{x^n}=\frac{1}{x^{n-m}}\), for all natural numbers x ≠ 0 and for all natural numbers m < n
For example, let’s factorize the numerator and denominator, and calculate 64 ÷ 512.

Page 123

Question 1.
Can’t you simplify the fractions below like this?
i) \(\frac{27}{243}\)
Answer:

ii) \(\frac{125}{3125}\)
Answer:

iii) \(\frac{48}{64}\)
Answer:

iv) \(\frac{54}{81}\)
Answer:

Page 126 

Question 1.
Calculate the products below in head:
i) 5² × 4²
ii) 5³ × 6³
iii) 25³ × 4³
iv) 125² × 8²
Answer:
i) 5² × 4² = (5 × 4)²
= 20²
= 400

ii) 5³ × 6³ = (5 × 6)³
= 30³
= 27000

iii) 25³ × 4³ = (25 × 4)³
= (100)³
= 1000000

iv) 125² × 8² = (125 × 8)²
= (1000)²
= 1000000

Question 2.
Write each number below as a product of powers of different primes;
i) 15²
ii) 30³
iii) 12² × 21²
iv) 12² × 21³
Answer:
i) 15² = (3 × 5)³
= 3² × 5²

ii) 30³ = (2 × 3 × 5)³
= 2³ × 3³ × 5³

iii) 12² × 21²
12² = (2 × 2 × 3)²
= 2² × 2² × 3²

21² = (7 × 3)²
= 7² × 3²

12² × 21² = 2² × 2² × 3² × 7² × 3²
= 2⁴ × 3⁴ × 7²

iv) 12² × 21³
12² = (2 × 2 × 3)² = 2² × 2² × 3²
21³ = (7 × 3)³ = 7³ × 3³
12² × 21³ = 2² × 2² × 3² × 7³ × 3³
= 2⁴ × 3⁵ × 7³

Class 7 Maths Chapter 8 Kerala Syllabus Repeated Multiplication Questions and Answers

Question 1.
Write each number below either as a power of a single prime or as a product of powers of different primes.
i) 3125
ii) 200
iii) 1600
Answer:
i) 3125 = 5 × 5 × 5 × 5 × 5 = 5s
ii) 200 = 2 × 2 × 2 × 5 × 5 = 2³3 × 5²
iii) 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2⁶ × 5²

Question 2.
Calculate the powers below as fractions:
i) \(\left(\frac{3}{2}\right)^3\)
Answer:
\(\left(\frac{3}{2}\right)^3=\left(\frac{3}{2}\right) \times\left(\frac{3}{2}\right) \times\left(\frac{3}{2}\right)\)
= \(\left(\frac{27}{8}\right)\)

ii) \(\left(\frac{3}{5}\right)^2\)
Answer:
\(\left(\frac{3}{5}\right)^2=\left(\frac{3}{5}\right) \times\left(\frac{3}{5}\right)\)
= \(\left(\frac{9}{25}\right)\)

iii) \(\left(2 \frac{3}{2}\right)^2\)
Answer:
\(\left(2 \frac{3}{2}\right)^2=\left(\frac{7}{2}\right)^2\)
= \(\left(\frac{49}{4}\right)\)

Question 3.
Write each product below as the product of powers of different primes:
i) 75 × 45
ii) 96 × 144
iii) 72 × 175
Answer:
i) 75 × 45
75 = 3 × 5²
45 = 3² × 5
5 × 45 = (3 × 5²) × (3² × 5)
= 3³ × 5³

ii) 96 × 144
96 = 2⁵ × 3¹
144 = 122 = (2² × 3)² = 2⁴ × 3²
96 × 144 = (2⁵ × 3¹) × (2⁴ × 3²)
= 2⁹ × 3³

iii) 72 × 175
7² = 8 × 9 = 2³ × 3²
17⁵ = 25 × 7 = 5² × 7¹
7² × 17⁵ = (2³ × 3²) × (5² × 7¹)

Question 4.
Calculate the following quotients:
(i) \(\frac{1440}{120}\)
(ii) \(\frac{729}{27}\)
Answer:
i) 1440 = 12² × 10
= (2² × 3)² × (2¹ × 5¹)
= 2⁵ × 3²2 × 5¹
120 = 12 × 10 = (22 × 3) × (2 × 5)
120 = 23 × 31 × 51
= \(\frac{1440}{120}=\frac{2^5 \times 3^2 \times 5^1}{2^3 \times 3^1 \times 5^1}\)
= 2^5-3 x 3^2-1 x 5^1-1
= 2² × 3¹ × 5⁰
=4 × 3 × 1
= 12

(ii) \(\frac{729}{27}\)
729 = 3⁶
27 = 3³
\(\frac{729}{27}=\frac{3^6}{3^3}\) = 3^6-3 = 3³ = 27

Question 5.
Write each number below as a product of powers of different primes:
i) 28
ii) 45²
iii) 18² × 30²
iv) 20³ × 27¹
Answer:
i) 28 = 2² × 7¹
ii) 45² = (3² × 5¹)² = 3⁴ × 5²
iii) 18² × 30² = (2¹ × 3²)² × (2¹ × 3¹ × 5¹)² = 2²⁺² × 3⁴⁺² × 5² = 2⁴ × 3⁶ × 5²
iv) 20³ × 27¹ = (2² × 5¹)³ × (3³)
= 2⁶ × 5³ × 3³

Class 7 Maths Chapter 8 Notes Kerala Syllabus Repeated Multiplication

In this chapter, we will explore the concept of repeated multiplication, which is a fundamental idea in mathematics. Repeated multiplication occurs when we multiply a number by itself multiple times. In this chapter we discuss about Factors, Power of Fractions, Product of Powers, Quotient of Powers, Multiples And Powers. So,

• In the first topic ‘Factors’, we will discuss about exponents, powers and its related problems.
• And in the second topic ‘Power of fractions’, we discuss about how powers of fractions increases and decreases.
• In the third topic, we discuss about product of powers, here we discuss products and its powers.
• In the fourth topic, we discuss about Quotient of Powers, here we studied how we deal with powers in division.
• And in the last topic, Multiples and Powers, here we discuss about how we use powers in multiplication.

Understanding repeated multiplication helps us simplify complex calculations and is essential for learning about powers and roots.

Factors
Numbers can be split into products of prime numbers.
For example,128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
We can write this product in a shortened form as 27 (“read as, two to the seventh power”)
27 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Therefore, Repeated multiplication of the same number is in this form Similarly, we write repeated addition as multiplication.
For example,
5 + 5 = 2 × 5
5 × 5 = 5²
5 + 5 + 5 = 3 × 5
5 × 5 × 5 = 5³
5 + 5 + 5 + 5 = 4 × 5
5 × 5 × 5 × 5 = 5⁴
The operation of multiplying a number by itself repeatedly is called exponentiation.
The number showing how many are multiplied together is called exponent.
We write the exponent in a smaller size, to the right and slightly above the number multiplied.
The number got by repeatedly multiplying a number by itself are called powers.

For example,
3 × 3 × 3 = 3³ Third power of three
5 × 5 × 5 × 5 = 5⁴ Fourth power of five
7 × 7= 7² Second power of seven

We can consider, any number as the first power of itself.
How do we split 576 as a product of primes?
576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
= 2⁶ × 3³
Any natural number greater than 1 can be written either as the power of a prime or as the product of powers of different primes.

Powers of Fractions
We know that area of a square of side 4 meters is 4 × 4 as 4²
Then what is the area of a square of side \(\frac{1}{4}\) meter?
i.e, \(\frac{1}{4} \times \frac{1}{4}=\left(\frac{1}{4}\right)^2\)
\(\frac{1}{4} \times \frac{1}{4}=\frac{1}{4 \times 4}\)
\(\left(\frac{1}{4}\right)^2=\frac{1}{4^2}\)

Similarly, what is the area of a square of sides 0.33 metres?
(0.33)² = 0.33 × 0.33
= 0.1089 square metres

In general, the area of a square with sides of length x is x².
Find the volume of a cube with lengths of edges 0.5 metre
(0.5)³ = 0.5 × 0.5 × 0.5
= 0.125 cubic metre

Using algebra, the volume of a cube with the length of the edges x is x³.
Then, Find the volume of a cube of edges 2\(\frac{1}{4}\) metres?
(2\(\frac{1}{4}\)) = \(\left(\frac{9}{4}\right)^3\)
= \(\frac{9}{4} \times \frac{9}{4} \times \frac{9}{4}\)
= \(\frac{9 \times 9 \times 9}{4 \times 4 \times 4}\)
= \(\frac{729}{64}\)

We can split this into quotient and reminder
\(\frac{729}{64}\) = 11\(\frac{25}{64}\)

Or use a calculator to compute
\(\frac{729}{64}\) = 11.390625

1 metre = 100 centimetres
= 10² centimetres

1 cubic metres = 100 × 100 × 100 cubic centimetres
= 10⁶ cubic centimetres

11.390625 cubic metres = 11. 390625 × 10⁶ cubic centimetres
= 11390625 cubic centimetres

Powers of 2
2² = 2 × 2 = 4
2³ = 4 × 2 = 8
2⁴ = 8 × 2 = 16
2⁵ = 16 × 2 = 32
By knowing the powers of 2, we can easily compute the powers of \(\frac{1}{2}\).

Each multiplication by 2 doubles, while each multiplication by \(\frac{1}{2}\) halves.
In general, Powers of numbers greater than 1 steadily increase. Powers of numbers greater than 0 and less than 1 steadily decrease. All powers of 1 remain 1 and all powers of 0 remain 0.

Multiples And Powers
Let’s see how we can add 2 times 4 and 2 times 6.

We know,
2 times 4 = 4 + 4
2 times 6 = 6 + 6
Adding this we get,
(2 times 4) + (2 times 6) = (4 + 6) + (4 + 6)
= 10 + 10 = 2 times 10

In short, we can write,
(2 × 4) + (2 × 6) = 2 × (4 + 6)

Using power, we can do the same as:
2^nd power of 4 = 4 × 4
2^nd power of 6 = 6 × 6

Multiplying,
(2^nd power of 4) × (2^nd power of 6)
= (4 × 6) × (4 × 6)
= 2nd power of (4 + 6)

That is,
4² × 6² = (4 × 6)²
In general,
The product of the same powers of two numbers is equal to the same power of the product of these numbers

And by using algebra,
xⁿyⁿ = (xy)ⁿ for all numbers x, y and for all natural numbers

Now let’s check what is 5 times 2 times 7?
We can calculate like this:
5 times 2 is 5 × 2 = 10
10 times 7 = 10 × 7 = 70

Also, we can calculate like this:
2 times 7 = 2 × 7 = 14
5 times 14 is 5 × 14 = 70

Let’s see how the second computation works:
2 times 7 = 7 + 7 = 14
5 times 14 is (7 + 7) (7+ 7) (7+ 7) (7+ 7) (7+ 7) = 10 times 7

Thus, we can write the computation as:
5 × (2 × 7) = (5 × 2) × 7

Using power, we can do the same as:
That is 5th power of 7 = 7 × 7
5th power of (7 × 7) = (7 × 7)(7 × 7)(7 × 7)(7 × 7)(7 × 7)
= 10th power of 7

In short using exponent we can write this as,
(7 × 7)⁵ = (7²)⁵ = 7¹⁰
Thus, we get the relation, that is,
In computing a power of a power of a nmnber the exponents should be multiplied
This can be written using algebra as an equation.
(x^m)ⁿ = x^mn for all numbers x and for all natural numbers m and n

Products Of Powers
Let’s check how we can find 3² × 3⁴
We know that
3² = 3 × 3
3⁴ = 3 × 3 × 3

Multiplying

Thus 3² × 3⁴ = 3²⁺⁴ = 3⁶
In general, we can say:
In multiplying two powers of a number, the exponents should be added

Using algebra, we can write it as follows:
x^m × xⁿ = x^m+n for all numbers x and all natural numbers m and n

We should note two things here:
(i) The product of two powers of the same number is a power of that number
(ii) The exponent of the product is the sum of the exponents of the numbers multiplied
For example, compute 5² × 5³ × 54⁴
5² × 5³ × 5⁴ = (5² × 5³) × 54⁴
= 5⁵ × 5⁴
= 5⁹

This method is also used for prime factorization of a product.
For example consider, 96 × 144
96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3
144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3²

and then compute the product as
96 × 144 = (2⁵ × 3) × (2⁴ × 3²)
= (2⁵ × 2⁴) × (3 × 3²)
= 2⁹ × 3³

Quotient Of Powers
Now let’s see how we can find the quotient using the powers.
For that consider an example, 288 ÷ 36
First factorize the numbers 288 and 36
288 = 2⁵ × 3²
36 = 2² × 3²

We can remove common factors. So that we get,
(2⁵ × 3²) ÷ (2² × 3²) = 2⁵ ÷ 2²

Now we can do the division easily
2⁵ ÷ 2² = 2^5-2 = 2³

Write all these steps together we get:
288 ÷ 36 = (2⁵ × 3²) ÷ (2² × 3²)
= 2⁵ ÷ 2²
= 2³
= 8
As a general principle we can state this as:

In dividing the larger power of a non-zero number by a smaller power of the same number, the exponents should be subtracted
And we can state this using algebra like this:
\(\frac{x^m}{x^n}\) = x^m-n, for all numbers x ≠ 0 and for all natural numbers m > n

• The operation of multiplying a number by itself repeatedly is called exponentiation.
• The number showing how many are multiplied together is called exponent.
• The number got by repeatedly multiplying a number by itself are called powers.
• Any natural number greater than 1 can be written either as the power of a prime or as the product of powers of different primes.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 6 Ratio Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 6 Solutions Ratio

Class 7 Maths Chapter 6 Ratio Questions and Answers Kerala State Syllabus

Ratio Class 7 Questions and Answers Kerala Syllabus

Page 87

Question 1.
Write down the ratio of the height to width of each of the following rectangles using the smallest possible natural numbers.
(i) Height 8 centimetres, width 10 centimetres
(ii) Height 8 metres; width 12 metres
(iii) Height 20 centimetres, width 1 metre
(iv) Height 40 centimetres; width 1 metre
(v) Height 1;5 centimetres; width 2 centimetres
Answer:
(i) Height 8 centimetres, width 10 centimetres.
The ratio of the height to width = 8 : 10 = 4 : 5

(ii) Height 8 metres; width 12 metres.
The ratio of the height to width = 8 : 12 = 2 : 3

(iii) Height 20 centimetres, width 1 metre.
i.e, Height 20 centimetres, width 100 centimetres.
The ratio of the height to width = 20 : 100 = 1:5

(iv) Height 40 centimetres; width 1 metre.
i.e, Height 40 centimetres, width 100 centimetres.
The ratio of the height to width = 40 : 100 = 2:5

(v) Height 1.5 centimetres; width 2 centimetres.
The ratio of the height to width – 1.5 : 2 = 3 : 4

Question 2.
In the table below, the height, width and their ratio of some rectangles are given, but only two of each. Can you calculate the third and complete the table.

Answer:

Page 89

Question 1.
Amina has 105 rupees with her and Mercy has 175 rupees. What is the ratio of the smaller amount to the larger?
Answer:
Amount with Amina =105 rupees
Amount with Mercy = 175 rupees
The ratio of the smaller amount to the larger = 105 : 175
= 3:5

Question 2.
96 women and 144 men attended a meeting. Find the ratio of the number of women to the number of men.
Answer:
Number of women = 96
Number of men =144
The ratio of the number of women to the number of men = 96 : 144
= 2:3

Question 3.
Of two pencils, the shorter is 4.5 centimetres long and the longer 7.5 centimetres. What is the ratio of the length of the longer to the shorter?
Answer:
Length of the longer pencil = 7.5 centimetres
Length of the shorter pencil = 4.5 centimetres
The ratio pf the length of the longer to the shorter = 7.5 : 4.5
= 5:3

Question 4.
When a rope was used to measure the sides of a rectangle, the width was \(\frac{1}{4}\) of the rope and the height was \(\frac{1}{3}\) of the rope. What is the ratio of the height to the width?
Answer:
Width = \(\frac{1}{4}\) of the rope
Height = \(\frac{1}{3}\) of the rope

The ratio of the height to the width = \(\frac{1}{3}: \frac{1}{4}\)
\(\frac{1}{4}=\frac{1 \times 3}{4 \times 3}=\frac{3}{12}\)
\(\frac{1}{3}=\frac{1 \times 4}{3 \times 4}=\frac{4}{12}\)

∴ Ratio = \(\frac{4}{12}: \frac{3}{12}\) = 4 : 3

Question 3.
3\(\frac{1}{2}\) glasses of water would fill a large bottle and 2\(\frac{1}{2}\) glasses of water could fill a smaller bottle. What is the ratio of the capacities of the larger bottle to the smaller bottle?
Answer:
Capacity of the large bottle = 3\(\frac{1}{2}\) glasses of water
= \(\frac{7}{2}\) glasses of water.

The capacity of the small bottle = 2\(\frac{1}{2}\) glasses of water
= \(\frac{5}{2}\) glasses of water

The ratio of the capacities of the larger bottle to the smaller bottle = \(\frac{7}{2}: \frac{5}{2}\)
= 7:5

Page 90,91

Question 1.
To make dosa, we need to take 2 cups of black gram for every 6 cups of rice. For 9 cups of rice, how many cups of black gram shall be taken?
Answer:
2 cups of black gram for every 6 cups of rice.
Ratio of black gram to rice = 2:6 = 1: 3
For 9 cups of rice,
Black gram : 9 = 1: 3
\(\frac{\text { Black gram }}{9}=\frac{1}{3}\)
Black gram = \(\frac{1}{3}\) × 9 = 3 cups
So, 3 cups of black gram should be taken for 9 cups of rice.

Question 2.
To plaster the walls of a house, cement and sand are mixed in the ratio 1: 5.45 sacks of cement were bought. How many sacks of sand are needed?
Answer:
The ratio of cement to sand = 1: 5 ,
45 sacks of cement were bought,
45: sand = 1 :5
\(\frac{45}{\text { sand }}=\frac{1}{5}\)
sand = 45 × \(\frac{1}{5}\) = 225 sacks
So, 225 sacks of sand are needed for 45 sacks of cement.

Question 3.
12 litres of paint was mixed with 8 litres of turpentine while painting the house. How many litres of turpentine should be mixed with 15 litres of paint?
Answer:
12 litres of paint was mixed with 8 litres of turpentine.
∴ The ratio of”turpentine to paint = 8: 12 = 2: 3
For 15 litres of paint,
Turpentine: 15 = 2:3
\(\frac{\text { Turpentine }}{15}=\frac{2}{3}\)
Turpentine = \(\frac{2}{3}\) × 15 = 10 litres.
So, 10 litres of turpentine should be mixed with 15 litres of paint.

Question 4.
In a ward of a panchayat, the women and men are in the ratio 11 : 10. There are 1793 women in the ward. How many men are there in the ward? What is the total number of women and men?
Answer:
The ratio of women to men = 11: 10.
There are 1793 women in the ward.
\(\frac{1793}{\mathrm{men}}=\frac{11}{10}\)
∴ Men = 1793 × \(\frac{10}{11}\) = 1630
Total number of women and men = 1793 + 1630 = 3423

Page 92

Question 1.
Suhara and Sita started a business. Suhara invested 40000 rupees and Sita 50000 rupees. They made a profit of 9000 rupees. It was divided in the ratio of their
Answer:
Amount invested by Suhara = 40000 rupees
Amount invested by Sita = 50000 rupees
Total profit = 9000 rupees
Ratio of investment of Suhara to Sita = 40000: 50000 = 4: 5
So, Suhara got \(\frac{4}{9}\) of the total profit and Sita got \(\frac{5}{9}\) of the total profit.
∴ Profit earned by Suhara = 9000 × \(\frac{4}{9}\) = 4000 rupees
Profit earned by Sita = 9000 × \(\frac{5}{9}\) = 5000 rupees.

Question 2.
Ramesan and John took up a contract for a work. Ramesan worked the first 6 days and John 7 days. They got 6500 rupees. They divided it in the ratio of the number of days each worked. How much did each get?
Answer:
Number of days Ramesan worked = 6 days
Number of days John worked = 7 days
Total amount they got = 6500 rupees
The ratio of the number of days worked by Ramesan to John = 6:7
So, amount Ramesan get is \(\frac{6}{13}\) of the toatal amount and John get \(\frac{7}{13}\) of the total amount.
Amount Ramesan get = 6500 × \(\frac{6}{13}\) = 3000 rupees
Amount John get = 6500 × \(\frac{7}{13}\) = 3500 rupees

Question 3.
When Ramu and Raju divided a sum of money in the ratio 3 : 2, Ramu got 480 rupees.
(i) How much did Raju get?
(ii) What was the sum that was divided?
Answer:
(i) The ratio of amount got by Ramu to Raju = 3:2
480 : Raju = 3: 2
\(\frac{480}{\text { Raju }}=\frac{3}{2}\)
∴ Raju = 480 × \(\frac{2}{3}\) = 320 rupees.
So, the amount Raju get = 320 rupees.

(ii) Total amount = 480 + 320 = 800 rupees.

Question 4.
Draw a line AB, 9 centimetres long. Mark a point P on it. The lengths of AP and PB should J>e in the ratio 1:2. How far away from A should we mark P ? Calculate and mark it.
Answer:
Length of AB = 9 cm .
AP : PB = 1:2
So, the length of AP is \(\frac{1}{3}\) of AB and that of PB is \(\frac{2}{3}\) of AB.,
∴ AP = 9 × \(\frac{1}{3}\) = 3 cm
PB = 9 × \(\frac{2}{3}\) = 6 cm
So, we have to mark P at a distance of 3 cm from A

Question 5.
Draw a line 15 centimetres long. Mark a point on it that divides the line in the ratio 2 : 3. Calculate the length and mark the point.
Answer:
Let AB = 15 cm and P be the point thet divide AB in the ratio 2:3.
i.e., AP: PB = 2: 3
So, the length of AP is j of AB and that of PB is \(\frac{3}{5}\) of AB.
∴ AP = 15 × \(\frac{2}{5}\) = 6 cm
PB = 15 × \(\frac{3}{5}\) = 9 cm

Question 6.
Draw a rectangle of perimeter 30 centimetres and sides of length in the ratio 1:2.
(i) With the same perimeter draw two more rectangles with sides in the ratio 2 : 3 and 3 : 7.
(ii) Calculate the areas of the three rectangles. Which rectangle has the greatest area?
Answer:
Perimeter of the rectangle = 30 cm
2 (length + breadth) =30
Length+breadth =15
The ratio of sides =1:2
∴ Length = \(\frac{2}{3}\) × 15 = 10 cm
Breadth = \(\frac{1}{3}\) × 15 = 5 cm

(i) The ratio of sides = 2:3
∴ Length = \(\frac{3}{5}\) × 15 = 9 cm
Breadth = \(\frac{2}{5}\) × 15 = 6 cm

The ratio of sides = 3:7
∴ Length = \(\frac{7}{10}\) × 15 = 10.5 cm
Breadth = \(\frac{3}{10}\) × 15 = 4.5 cm

(ii) Area of first rectangle = 10 × 5 = 50 cm²
Area of second rectangle = 9 × 6 = 54 cm²
Area of third rectangle = 10.5 × 4.5 = 47.25cm²
So, second rectangle has the greatest area.

Intext Questions And Answers

Question 1.
The walls of Aji’s house needs painting. First, 25 litres of green paint and 20 litres of white paint were mixed. To get the same shade of green, how many litres of green should be mixed with 16 litres of white?
Ans:
The ratio of green to white = 25: 20 = 5: 4
That is, 5 litres of green for every 4 litres of white.
4 times 4 litres is 16 litres.
So, 4 times 5 litres, 20 litres of green should be mixed with 16 litres of white.
In terms of ratio,
Green: White = 5: 4
For 16 litres of white,
Green: 16 = 5 : 4
\(\frac{\text { Green }}{16}=\frac{5}{4}\)
∴ Green = \(\frac{5}{4}\) × 16 = 20 litres

Question 2.
The school needs a vegetable garden. A rectangular plot is to be roped off for this. The length of the rope is 32 metres. They decided to have width and length in the ratio 3:5. What should be the width and length?
Answer:
Length of the rope = 32 metres.
So, perimeter of rectangular garden = 32 metres
2 (length + width) = 32
Length + width = 16
The ratio of width to length = 3: 5
∴ Length = 16 × \(\frac{3}{8}\) = 6 metres
Breadth = 16 × \(\frac{5}{8}\) =10 metres

Class 7 Maths Chapter 6 Kerala Syllabus Ratio Questions and Answers

Question 1.
Angles of a linear pair are in the ratio 4: 5. What is the measure of each angle?
Answer:
Ratio of angles in the linear pair = 4: 5
Sum of angles = 180
In 180, \(\frac{4}{9}\) is one angle and – is the other angle.
So, one angle = 180 × \(\frac{4}{9}\) = 80°
Other angle = 180 × \(\frac{5}{9}\) = 100°

Question 2.
Sita and Soby divided some money in the ratio 1: 2 and sita got 400 rupees. What is the total amount they divided?
Answer:
Ratio in which money divided =1:2
Amount Sita got = 400 rupees
400: Sita =1: 2
\(\frac{400}{\text { Sita }}=\frac{1}{2}\)
∴ Sita = 400 × \(\frac{2}{1}\) = 800 rupees
So, total amount = 400 + 800 = 1200 rupees.

Question 3.
Ramesh’s father divided his saving as follows:

• \(\frac{2}{7}\) of his savings to Ramesh
• \(\frac{5}{7}\) of his savings to his mother.

Find the ratio of this division.
Answer:
Ratio = \(\frac{2}{7}: \frac{5}{7}\) = 2: 5

Question 4.
What does it mean to say that the width to length ratio of a rectangle is 1:1? What sort of rectangle is it?
Answer:
The width to length ratio of a rectangle is 1:1,
Which means, length = width
When the sides of a rectangle are equal, it will be a square.

Question 5.
Santha decided to give her salary to her children Ravi and Shinu in the ratio 3:2. If Ravi gets Rs.4500, then.
a) Find the amount Shinu gets?
b) How much salary does Santha get?
Answer:
(a) Ratio of amount got by Ravi to Shinu = 3:2
Amount Ravi got = 4500 rupees
4500: Shinu = 3:2
\(\frac{4500}{\text { Shinu }}=\frac{3}{2}\)
∴ Shinu = 4500 × \(\frac{2}{3}\) = 3000 rupees.

(b) Santha’s salary = 4500 + 3000 = 7500 rupees

Question 6.
Raju has an amount of 120 rupees and Mary has 180 rupees.
A) What is the ratio of the amounts Mary and Raju have?
(a) 3:2
(b) 2:3
(c) 6:5
(d) 5:9

B) Mother gave 60 rupees more to Mary. How much more money Raju needed to make the same ratio?

C) If they divide 800 rupees in the same ratio, how much amount did each get?
Answer:
A) Ratio of the amount Mary and Raju have = 180 : 120 = 3:2
B) Mother gave 60 rupees more to Mary.
Now the amount with Mary = 180 + 60 = 240 rupees.
240 : Raju = 3: 2
\(\frac{240}{\text { Raju }}=\frac{3}{2}\)
∴ Raju = 240 × \(\frac{2}{3}\) = 160
So, more money Raju needed = 160 – 120 = 40 rupees

C) Amount Raju got = 800 × \(\frac{2}{5}\) = 320 rupees
Amount Mary got = 800 × \(\frac{3}{5}\)
= 480 rupees.

Question 7.
Last year the ratio of the female teachers and male teachers of Ramapuram UP School was 6:1.
a) If the number of male teachers is 6, what is the number of female teachers? Instead of the female teachers who got transfer from the school, male teachers joined there. Now the ratio of the female teachers and male teachers is 11:10.
b) How many female teachers got transfer this year?
c) This year some female teachers and male teachers will be 1:1. If so, how many female teachers will retire this year?
Answer:
a) Ratio of female teachers to male teachers = 6:1
Number of male teachers = 6
Number of female teachers = 6 × 6 = 36

b) Total number of teachers = 36 + 6 = 42
Ratio = 11: 10
Number of female teachers = 42 × \(\frac{11}{21}\) = 22
So, number of female teachers transferred = 36 – 22 = 14

c) Number of male teachers = 42 × \(\frac{10}{21}\) = 20
So, inorder to become ratio 1:1, number of male teachers = number of female teachers
∴ The number of female teachers retire this year = 22 – 20 = 2

Practice Questions

Question 1.
Express the width and length in ratios.
(i) Width = 3 cm Length= 9 cm
(ii) Width = 6 cm Length= 14 cm
Answer:
(i) 1:3
(ii) 3:7

Question 2.
The perimeter of a rectangular garden is 28 metres. The ratio of length and width is given by 3:4. Find its length and width.
Answer:
6 cm, 8 cm

Question 3.
The capacity of small tank is 500 litres and big tank is 1500 litres.
(a) Find the ratio of the capacities of the small tank and the big tank.
(b) Small tank is fully filles with water and big tank is half filled. Then, find the ratio of the capacities of the small tank and the big tank.
(c) 1500 litres of water is distributed to the houses of Appu and Muthu in the ratio 3:2. How much water will each house get?
Answer:
(a) 1:3
(b) 2:3
(c) 900 litres, 600 litres

Question 4.
24 litres of curd was mixed with 96 litres of water to make buttermilk.
(a) What is the ratio of water and curd used?
(b) How many litres of water is needed to mix with 96 litres of curd to make buttermilk in the same ratio?
(c) How many litres of curd is needed to make 600 litres of buttermilk?
Answer:
(a) 4:1
(b) 384 litres
(c) 120 litres

Question 5.
Anu and Manu divided ah amount in the ratio 3:2. If Anu got 100 rupees more, what is the total amount they divide?
Answer:
500 rupees

Question 6.

(a) What part of the circle is shaded in the picture?
(b) What part of the circle is unshaded?
(c) What is the ratio of shaded to unshaded?
Answer:
(a) \(\frac{1}{4}\)
(b) \(\frac{3}{4}\)
(c) 1:3

Question 7.
For making idlis, rice and Urad are to be taken in the ratio 2:1. In 9 cups of such a mxiture of rice and urad, how may cup of rice and urad are taken?
Answer:
6 cups,3 cups

Question 8.
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?
Answer:
8, 12

Question 9.
A rectangular piece of land is to be marked on the school ground for a vegetable garden.
Hari and Mary started making 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 length and width of the rectangle? .
Answer:
7.5 m , 4.5 m

Question 10.
When measuring the length of two buses with one rope the length of the first bus is \(\frac{2}{3}\) of the length of the rope and the length of the second bus was \(\frac{3}{5}\) of the length of the rope. What is the ratio between the length of the buses.
Answer:
10:9

Class 7 Maths Chapter 6 Notes Kerala Syllabus Ratio

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. We can use ratio to compare parts of a whole also

• Generally, when we state ratios, we avoid fractions and decimals. In other words, the smallest possible counting number is used to express ratios. By multiplying or dividing the ratio by the same number, we can make it in terms of the smallest possible counting number.
• We can use ratios to express any two measures, not just lengths, as multiples and parts.
• In any mixture, components are in fixed ratio. So, by knowing the ratio one can find the quantity of one component if other is given.
• If we are given the ratio in which a quantity is divided, we can find how much each part is using this ratio.

Rectangle Problems
Consider the following rectangle.

In both the rectangles, the width is 3 times the height. Or we can say that the height is ^ times the width.
So, the ratio of width to height is 3 : 1 and the ratio of height to width is 1 : 3 (The ratio of width to height = 6 : 2 = 3 : 1 or 4.5 : 1.5 = 3 : 1)
If we extend the height and width by the same multiple or shorten them by the same fraction, the ratio remains the same.

Other Measures
We can use ratios to express any two measures (not just lengths) as multiples and parts. For example, .
Suppose we have a 15 litre bucket and a 25 litre bucket.
The small bucket holds \(\frac{15}{25}=\frac{3}{5}\) of what large bucket holds.
Or
we can say that,
The ratio of capacity of small bucket to large bucket = 15 : 25 = 3:5

• If we extend the height and width by the same multiple or shorten them by the same fraction, the ratio remains the same.
• We can use ratios to express any two measures as multiples and parts.
• In a mixture, the components are mixed in fixed ratio.
• We can use ratio to compare parts of a whole also.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 7 Shorthand Math Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 7 Solutions Shorthand Math

Class 7 Maths Chapter 7 Shorthand Math Questions and Answers Kerala State Syllabus

Shorthand Math Class 7 Questions and Answers Kerala Syllabus

Page 96

Question 1.
Write the following statements using the language of algebra.
1) Zero added to any number gives the same number.
2) Zero subtracted from any number gives the same number.
3) Any number subtracted from the same number gives zero.
4) Any number multiplied by zero gives zero.
5) Any number divided by the same number gives 1.
6) Twice a number added to the number makes three times the number.
7) Twice a number subtracted from thrice the number gives the number.
8) A number added to a number, and then the added number subtracted gives the original number.
Answer:
Let n be the number.
1) n + 0 = n
2) n – 0 = n
3) n – n = 0
4) n × 0 = 0
6) n + 2n = 3n
7) 3n – 2n = n
8) Let m be the other number, n + m – m = n

Page 98

Question 1.
Do the following problems mentally:
(i) 49 + 125 + 75
(ii) 3\(\frac{1}{2}\) + 8\(\frac{3}{4}\) + \(\frac{1}{4}\)
(iii) 15.5 + 0.25 + 0.75
(iv) 38 + 27
(v) 136 + 64
Answer:
(i) 49 + 125 + 75 = 49 + 100 + 25 + 75
= 49 + 100+ 100
= 49 + 200
= 249

(ii) 3\(\frac{1}{2}\) + 8\(\frac{3}{4}\) + \(\frac{1}{4}\)
= \(\frac{7}{2}+\frac{36}{4}\)
= \(\frac{7}{2}+\frac{18}{2}\)
= \(\frac{25}{2}\)
= 12\(\frac{1}{2}\)

(iii) 15.5 + 0.25 + 0.75 = 15.5 + 1.0
= 16.5

(iv) 38 + 27 = 38 + 2 + 25
= 40 + 25
= 65

(v) 136 + 64 = 136 + 4 + 60
= 140 + 60
= 200

Question 2.
Do the following proble mentally:
(i) (135 – 73) – 27
(ii) (37 – 1\(\frac{1}{2}\)) – \(\frac{1}{2}\)
(iii) (298 – 4.5) – 3.5
(iv) 78 – 29
(v) 140 – 51
Answer:
(i) (135 – 73) – 27 = 135 – (73 + 27)
= 135 – 100
= 35

(ii) (37 – 1\(\frac{1}{2}\)) – \(\frac{1}{2}\) = (37 – 1.5) – 0.5
= 37 – (1.5 + 0.5)
= 37 – 2
= 35

(iii) (298 – 4.5) – 3.5
= 298 – (4.5 + 3.5)
= 298 – 8
= 290

(iv) 78 – 29 = 78 – (30 – 1)
= 78 – 30 + 1
= 48 + 1
= 49

(v) 140 – 51 = 140 – (50 + 1)
= 140 – 50 – 1
= 90 – 1
= 89

Page 99

Do the following problems mentally:
(i) (136 + 29) – 19
(ii) (3\(\frac{1}{2}\) + 5\(\frac{3}{4}\)) – 2\(\frac{1}{4}\)
(iii) (298 + 14.5) – 12.5
(iv) 23 + (35 – 18)
(v) 65 + 98
Answer:
(i) (136 + 29) – 19 = 136 + (29 – 19)
= 136 + 10
= 146

(ii) (3\(\frac{1}{2}\) + 5\(\frac{3}{4}\)) – 2\(\frac{1}{2}\) = (3.5 + 5.75) – 2.25
= 3.5+ (5.75 – 2.25)
= 3.5 + 3.5
= 7

(iii) (298 + 14.5) – 12.5 = 298 + (14.5 – 12.5)
= 298 + 2
= 300

(iv) 23 + (35 – 18) = (23 + 35) – 18
= 58 – 18
= 40

(v) 65 + 98 = 65 + (100 – 2)
= (65 + 100) – 2
= 165 – 2
= 163

Page 101

Do the following problems mentally:
(i) (135 – 73) + 23
(ii) (38 – 8\(\frac{1}{2}\)) + \(\frac{1}{2}\)
(iii) (19 – 6.5) + 2.5
(iv) 135 – (35 – 18)
(v) 240 – (40 – 13)
Answer:
(i) (135 -73) + 23 = 135 -(73 – 23)
= 135 – 50
= 85

(ii) (38 – 8\(\frac{1}{2}\)) + \(\frac{1}{2}\) = (38 – 8.5) + 0.5
= 38 – (8.5 – 0.5)
= 38 – 8
= 30

(iii) (19 – 6.5) + 2.5 = 19 – (6.5 – 2.5)
= 19 – 4
= 15

(iv) 135 – (35 – 18) = (135 – 35) + 18
= 100 + 18
= 118

(v) 240 – (40 – 13) = (240 – 40) + 13
= 200 + 13
= 213

Page 102

Do the following problems mentally:
(i) 103 × 15
(ii) 98 × 25
(iii) (63 × 12) + (37 × 12)
(iv) (65 × 11) – (55 × 11)
(v) (15 × \(\frac{3}{4}\)) + (5 × \(\frac{3}{4}\))
(vi) (5\(\frac{1}{2}\) × 23) – (4\(\frac{1}{2}\) × 23)
Answer:
(i) 103 × 15 = (100+ 3) × 15
= (100 × 15) +(3 × 15)
= 1500 + 45
=1545

(ii) 98 × 25 = (100 – 2) × 25
= (100 × 25)-(2 × 25)
= 2500 – 50
= 2450

(iii) (63 × 12) + (37 × 12)
= (63 + 37) × 12
= 100 × 12
= 1200

(iv) (65 × 11) – (55 × 11)
= (65 – 55) × 11
= 10 × 11
= 110

(v) (15 × \(\frac{3}{4}\)) + (5 × \(\frac{3}{4}\))
= \(\frac{3}{4}\) × (15 + 5)
= \(\frac{3}{2}\) × 15
= \(\frac{45}{2}\)

(vi) (5\(\frac{1}{2}\) × 23) – (4\(\frac{1}{2}\) × 23)
= 23 × (5\(\frac{1}{2}\) + 4\(\frac{1}{2}\))
= 23 × (5.5 + 4.5)
= 23 × 10
= 230

Class 7 Maths Chapter 7 Kerala Syllabus Shorthand Math Questions and Answers

Question 1.
Write the following statements using the language of algebra.
i. When 15 is added to a number, then it is equal to three times that number.
ii. If 25 is added to three times a number, the result is 70.
iii. One third of a number when added to 1 gives 15
Answer:
i. n + 15 = 3n
ii. 3n + 25 = 70
iii. 1 + \(\frac{n}{3}\) = 15

Question 2.
Find 24 + 16 + 34
Answer:
24 + 16 + 34 = 24 + (16 + 34)
= 24 + 50
= 74

Question 3.
Find 79 – 52 – 18
Answer:
79 – 52 – 18 = 79 – (52 + 18)
= 79 – 70
= 9

Question 4.
Find 3 × 13 + 3 × 7
Answer:
3 × 13 + 3 × 7 = 3(13 + 7)
= 3 × 20
= 60

Question 5.
Find 7 × 48
Answer:
7 × 48 = 7(50 – 2)
= 350 – 14
= 336

Class 7 Maths Chapter 7 Notes Kerala Syllabus Shorthand Math

Algebra is one of the main branches of mathematics. In algebra, we make use of letters to solve mathematics. This chapter introduces some basic concepts of algebra. Following are the main topics discussed in this chapter.

Numbers and letters

• The method of stating relations between measures or numbers using letters is called algebra.
• In algebra, we write products without multiplication signs.
• In products with numbers and letters, we write the number first in algebra
• We can use any letter to denote the number or the measure in algebra
• We write division as a fraction in algebra
• We need to use more than one letter when we talk about several numbers in algebra

One by one and altogether

• While adding three numbers in a row, we can use the following result.
  (x + y) + z = x + (y + z), for any three numbers x, y, z.
• While subtracting three numbers in a row, we can use the following result.
  (x – y) – z — x – (y + z) for any three numbers x, y, z.

Addition and Subtraction
Adding a larger number to the given number and then subtracting a smaller number gives the same result as adding the difference between, the larger number and the smaller number to the given number. That is,
(x + y) – z = x + (y – z), for any three numbers x, y, z with y > z

Subtracting a larger number from the given number and then adding a smaller number gives the same result as subtracting the difference between the larger number and the smaller number from the given number. That is,
(x – y) + z = x – (y – z), for any x, y,-z with y > z
We must use brackets to show the order of operations clearly.

Addition and subtraction and then multiplication

• Multiplying a sum by the given number gives the same result as multiplying each number in the
  sum separately by the given number and then adding them. That is,
  (x + y)z = xz + yz, for any three numbers x, y and z
• Multiplying a difference by a number gives the same result as multiplying each number in the difference and then subtracting them. That is,
  (x – y)z = xz – yz, for any three numbers x, y, z

Numbers And Letters
The method of stating relations between measures or numbers using letters is called algebra. Eg:
Consider the following fact: “A number added to itself is twice the number.”
In the language of math, we write it as follows; a number + the same number = 2 × number Let’s denote the number by n.
Then, n + n = 2n for any number n, which is the algebraic representation of the given fact.

Some features of algebra:

• Write products without multiplication sign.
  Eg:
  3 × n can be written as 3n.
• In products with numbers and letters, write the number first.
  Eg;
  5 × m is written as 5m, not m5.
• We can use any letter to denote the number or the measure.
  Eg:

Consider the fact; “A number added to itself is twice the number.”

• If we denote the number as n, the algebraic form is n + n = 2n.
• If we denote the number as x, the algebraic form is x + x = 2x.

We write division as a fraction.
Eg:
Consider the fact: “Five times a number divided by five gives that number.”
If we denote the number as n, the algebraic form is \(\frac{5n}{5}\) = n.

We need to use more than one letter when we talk about several numbers.
Eg:
Consider the fact: If we add a number to another number and then subtract the original number, we get the added number.
Let n be the original number and m be the added number. Then the fact becomes; n + m – n = m

The letters that we use to represent numbers or measures in algebra are generally know as variables

One By One And Altogether
Adding two numbers one after another to a number, or adding their sum, gives the same result. Algebraically, (x + y) + z = x + (y + z), for any three numbers x, y, z.
Eg:
Consider the numbers 1, 2 and 3.
(1 + 2) + 3 = 3 + 3 = 6
1 + (2 + 3) = 1 + 5 = 6

There are situations where adding the sum of these numbers is easier than adding one after another. .
Eg:
15 + 28 + 2 =15 + 30
= 45

There are situations where adding one after another is easier than adding the sum.
Eg:
25 + 18 = 25 + 5 + 13
= 30 + 13
= 43

Subtracting two numbers one after the other from a number or subtracting the sum of these two numbers from the first number both give the same result. Algebraically,
(x – y) – z = x – (y + z) for any three numbers x, y, z.
Eg:
Consider the numbers 10, 7 and 2.
(10 – 7) – 2 = 3 – 2 = 1
10 – (7 + 2) = 10 – 9 = 1

There are situations where subtracting the sum is easier than subtracting the numbers one after the other.
Eg:
35 – 17 – 3 = 35 – (17 + 3)
= 35 – 20
= 15

There are situations where subtracting the numbers one after the other is easier than subtracting the sum.
Eg:
500 – 201 = 500 – 200 – 1
= 300 – 1
= 299

Addition And Subtraction
(x + y) – z = x + (y – z), for any three numbers x, y, z with y > z
Eg:
Consider the numbers 5, 4 and 3.
(5 + 4) – 3 = 9 – 3 = 6
5 + (4 – 3) = 5 + 1 = 6
Therefore, (5 + 4) – 3 = 5 + (4 – 3)

Sometimes, it is more convenient to apply it in the reverse.
Eg:
25 + 99 = 25 + (100 – 1)
= (25 + 100) – 1
= 125 – 1
= 124

(x – y) + z = x – (y – z), for any x, y, z with y > z Eg:
Consider the numbers 10, 7 and 4.
(10 – 7) + 4 = 3 + 4 = 7
10 – (7 – 4) = 10 – 3 = 7
Therefore, (10 – 7) + 4 = 10 – (7 – 4)

We must use brackets to show the order of operations clearly.

Explanation:
Adding 7 and 4 and then adding 2 gives 13.
Adding 4 and 2 first and then adding this to 7 also gives 13.
That is, (7 + 4) + 2 = 7 + (4 + 2)

So, we may write this sum as 7 + 4 + 2 without brackets.
But if we subtract 4 from 7 and then subtract 2, we get 1.

That is, (7 – 4) – 2 = 1. Whereas if we
subtract 2 from 4 and then subtract it from 7, we get 5.

That is 7 – (4 – 2) = 5.
So, if we just write 7 – 4 – 2, the answer will be different depending on which operation we do first.
So, we must use brackets to show the order of operations clearly.

Addition And Subtraction And Then Multiplication

(x + y)z = xz + yz, for any three numbers x, y and z Eg:
Consider the numbers 1, 2, 3.
(1 + 2) × 3 = 3 × 3 = 9
(1 × 3) + (2 × 3) = 3 + 6 = 9
Therefore, (1 + 2) × 3 = (1 × 3) + (2 × 3)

Reading these in reverse is also useful in some problems. That is, xz + yz = (x + y)z
Eg:
32 + 56 =(4 × 8)+ (7 × 8)
= 8 × (4 + 7)
= 8 × 11
= 88

(x – y)z = xz – yz, for any three numbers x, y, z
Eg:
Consider the numbers 7, 5 and 3. .
(7 – 5) × 3 = 2 × 3 = 6
(7 × 3) – (5 × 3) = 21 – 15 = 6
Therefore, (7 – 5) × 3 = (7 × 3) – (5 × 3)

Reading these in reverse is also useful in some problems, That is, xz – yz = (x – y)z
Eg:
(\(\frac{1}{2}\) × 35) – (\(\frac{1}{2}\) × 15) = \(\frac{1}{2}\) (35 – 15)
= \(\frac{1}{2}\) × 20
= 10

The method of stating relations between measures or numbers using letters is called algebra.

Features of algebra:

• Write products without multiplication sign.
• In products with numbers and letters, write the number first.
• We can use any letter to denote the number or the measure.
• We write division as a fraction.
• We need to use more than one letter when we talk about several numbers.

Adding two numbers one after another to a number, or adding their sum, gives the same result.
Algebraically, (x + y) + z = x + (y + z), for any three numbers x, y, z.

• There are situations where adding the sum of these numbers is easier than adding one after another.
• There are situations where adding one after another is easier than adding the sum.

Subtracting two numbers one after the other from a number or subtracting the sum of these two numbers from the first number both give the same result. Algebraically,
(x – y) – z = x – (y + z) for any three numbers x, y, z. ,

There are situations where subtracting the numbers one after the other is easier than subtracting the sum.
(x + y) – zfc x + (y – z), for any three numbers x, y, z with y > z

Sometimes, it is more convenient to apply it in the reverse. That is, x + (y – z) = (x + y) – z

• We must use brackets to show the order of operations clearly.
• (x – y) + z = x – (y – z), for any x, y, z with y > z
• (x + y)z = xz + yz, for any three numbers x, y, z

Reading these in reverse is also useful in some problems. That is,
xz + yz = (x + y)z ,

• (x – y)z = xz – yz, for any three numbers x, y, z
• Reading these in reverse is also useful in some problems. That is, xz – yz = (x – y)z

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 1 Parallel Lines Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 1 Solutions Parallel Lines

Class 7 Maths Chapter 1 Parallel Lines Questions and Answers Kerala State Syllabus

Parallel Lines Class 7 Questions and Answers Kerala Syllabus

Page 17

Question 1.
Draw the parallelogram below with the given measures.

Calculate the other three angles.
Answer:
Draw a horizontal line segment AB that is 5 cm long. This will be one side of the parallelogram.

At point A, use a protractor to measure and mark an angle of 60 degrees from line AB.

From point A, draw a line segment AC that is 3 cm long, making sure it forms the 60° angle with AB.

From point B, draw a line segment BD that is 3 cm long, parallel to AC. You can use a ruler and a set square to ensure parallelism.
From point C, draw a line segment CD that is 5 cm long, parallel to AB. The points C and D should connect to complete the parallelogram.

Now finding the remaining angles,
We know that, ∠A + ∠C = 180°
∠C = 180° – 60° = 120°
Similarly, ∠A + ∠B = 180°
∠B = 180° – 60° = 120°
And, ∠B + ∠D = 180°
∠D = 180° – 120° = 60°

Question 2.
The top and bottom blue lines in the figure are parallel. Find the angle between the green lines.

Answer:
Draw a horizontal line parallel to the other two parallel lines

From the figure, we get,
∠EAB = ∠ABD = 40°
∠DBC – ∠BGF = 50°
:. ∠ABC = ∠ABD + ∠DBC
= 40° + 50° = 90°

Question 3.
In the figure, the pair of lines slanted to the left are parallel; and also the pair of lines slanted to the right.
Draw this figure:

Answer:
Draw a line segment AB measuring 2 cm.

Draw a perpendicular line from point A.

Measure 20° on the protector, draw two lines on both sides of the perpendicular line, and mark C and D.

Using a set square, draw a line parallel to line AC starting from point B

Using a set square, draw a line parallel to line AD starting from point B and mark the intersection point as E.

Page 21

Question 1.
Draw the triangle with the given measures.

Answer:
First off all we have to find the third angle of the given triangle
⇒ Third angle = 180° – (60° + 40°) = 180° – 100° = 80°
Now we can start drawing the triangle, for that
Draw a horizontal line segment that is 5 cm long.

At left end of the line, use a protractor to measure and mark an angle of 40° from the line

At right end of the line, use a protractor to measure and mark an angle of 40°from the line

Construct line segments forming angles of 40° and 80°, extending each until the lines intersect

Question 2.
The figure shows a triangle drawn in a rectangle. Calculate the angles of the triangle.

Answer:
Consider the figure

Since, ABCD is rectangle ∠A, ∠B, ∠C, ∠D equals to 90°
Now, it is given that one part of ∠A is 40°, so the remaining part of A is 50°
Similarly, one part of ∠D is 25°, so the remaining part of ∠D is 65°
It implies that two angles of the triangle are 50° and 65°.
Therefore, the third angle of the triangle is = 180° – (50° + 65°)
= 180° – (105°)
= 75°

So, the angles of triangles are 50°, 65°, 75o

Question 3.
The top and bottom lines in the figure are parallel. Calculate the third angle of the bottom triangle and all angles of the top triangle.

Answer:

Consider the above figure,
Given that AB and CD are parallel lines
The two given angles of the bottom triangle are 35° and 45°
Therefore, third angle of bottom triangle = 180° – (35° +45°)
= 180° – (80°)
= 100°
In bottom triangle, ∠B and in top triangle, ∠C are Alternate angles
Therefore, ∠A and ∠D are equal. ∠D of top triangle is = 35°
Similarly, in bottom triangle, ∠B and in top triangle, ∠C are Alternate angles
Therefore, B and C are equal, ∠C of top triangle is = 45°
So, the third angle of top triangle = 180° – (35° + 45o)
= 180° – (80°)
= 100°
All angles of bottom triangle = 35°, 45°,
All angles of top triangle = 35°, 45°, 100°.

Question 4.
The left and right sides of the large triangle are parallel to the left and right sides of the small triangle. Calculate the other two angles of the large triangle and all angles of the small triangle.

Answer:

Consider the above figure,
Given that AB and CD are parallel, and AC and DE are also parallel,
It’s given that the middle angle is 70°.
Considering the parallel lines AB and CD, AC is a slanting line intersecting the parallel lines.
Therefore, the middle angle and ∠A of the larger triangle are alternate angles,
Therefore, these two angles are equal
i.e., ∠A = 70°
Now, we can calculate the third angle of the larger triangle
i.e., ∠B = 180° – (70° +60°)
= 180° – (130°) = 50°

All three angles of larger triangle are 50°, 60°, 70°
Similarly, considering the parallel lines AC and DE, CD is a slanting line intersecting the parallel lines.
Considering the parallel lines AC and DE, CD is slanting line through the parallel lines.
Here, middle angle and the ∠D of smaller triangle are alternate angles,
Therefore, these two angles are equal
i.e., ∠D = 70°
Now, ZC of larger triangle, middle angle.and ∠C of the smaller triangle form a straight line.
Therefore, ∠C of smaller triangle is = 180° − (60° + 70°)
= 180° – 130°
= 50°
Now we can find the third angle of smaller triangle (∠E) = 180° – (50° + 70°)
= 180° – (120°)
= 60°
All three angle of smaller triangle are 50°, 60°, 70°

Question 5.
A triangle is drawn inside a parallelogram.
Calculate the angles of the triangle.
Answer:

Consider the above figure,
∠D and ∠B are opposite angles of parallelogram.
Therefore, D = ∠B = 110°
But, one part of ∠D is 60o, so remaining part of ∠D = 110° – 60° = 50°
Now, considering the angles A and <D, their sum is equals to 180°
i.e., ∠A + ∠D = 180°
∠A = 180° – ∠D
= 180° – 110° = 70°
But, one part of ∠A is 30°, so remaining part of ∠A = 70° – 30° = 40°
It means, we get two angles of the triangle which are 50° and 40°
So, third angle of the triangle = 180° – (50° + 40°)
= 180° – (90°)
= 90°

Hence, all three angles of the triangles are 50°, 40°, 90°

Intext Questions and Answers

Question 1.
Find the unknown angle in the following figures?

Answer:
i. By extending the parallel lines and slanting line we get the figure below

From the figure, it is clear that two small angles are formed when the slanting line intersects the top and bottom parallel lines. Since these angles are equal in measure.
Therefore, the unknown angle is 45°

ii. By extending the parallel lines and slanting line we get the figure below

From the figure, it is clear that two small angles are formed when the slanting line intersects the top and bottom parallel lines. Since these angles are equal in measure.
Therefore, the unknown angle is 40°.

iii. By drawing a vertical line, as shown in the figure below, we can observe that this vertical line is parallel to the other parallel lines and bisects the middle angle. Therefore, the angles on the left side of the vertical line are 30°

Similarly, since the angle on the right side is also 30°, then the unknown angle will also be 30o.

When two parallel lines are cut by a third line (called a slanting line), different pairs of angles are formed.

Question 2.
One angle of a triangle is 72°. The other two angles are of equal measure. What are their measures?
Answer:
Sum of two other angles = 180° – 72° = 108°
Since the other two angles are equal, each angle is = \(\frac{108^0}{2}\) = 54

Question 3.
When a line crosses another line, how many angles are formed between them ?

Answer:
When a line crosses another line, then 4 angles are formed between them.

When two lines cross perpendicular, then all angles are 90°.

Consider the figure below,

Here, 4 angles are 21, 22, 23, 24, where 21 and 22 are small angles, and 23 and 24 are large angles Then we can say that,
The two small angles are of the same measure.
i.e., ∠1 = ∠2

The two large angles are of the same measure.
i.e., ∠3 = ∠4

The sum of a small angle and a large angle is 180°.
i.e., ∠1 + ∠4 = 180° and ∠3 + ∠4 = 180°.

Consider the figure below, which shows two parallel lines intersected by another line, forming eight angles.

Then we can say that,
∠1 = ∠5
∠2 = ∠6
∠3 = ∠7
∠4 = ∠8
For example,
In the given figure, one of the angles is given as 50°

Then, the remaining angles are shown below

A line intersects two parallel lines at angles of the same measure

Question 4.
Can you calculate the other seven angles which the parallel lines make with the slanting line?

Answer:

Class 7 Maths Chapter 1 Kerala Syllabus Parallel Lines Questions and Answers

Question 1.
Draw the parallelogram below with the given measures.

Answer:
Draw a horizontal line segment AB that is 6 cm long. This will be one side of the parallelogram.

At point A, use a protractor to measure and mark an angle of 50° from line AB.

From point A, draw a line segment AC that is 4 cm long, making sure it forms the 50° angle with AB.

From point B, draw a line segment BD that is 4 cm long, parallel to AC. You can use a ruler and a set square to ensure parallelism.

From point C, draw a line segment CD that is 6 cm long, parallel to AB. The points C and D should connect to complete the parallelogram.

Question 2.
The top and bottom lines in the figure are parallel. Find the unknown angle shown in the figure.

Answer:
Draw a horizontal line parallel to the other two parallel lines

From the figure, we get,
∠EAB = ∠ABD = 30°
∠DBC = ∠BCF = 60°
∴ ∠ABC = ∠ABD + ∠DBC
= 30° + 60° = 90°

Question 3.
Two vertical lines in the figure are parallel. Find the unknown angle shown in the figure.

Answer:
Consider the figure below

Given that AB and CD are parallel line
Draw a line PQ parallel to both AB and CD
∠A and ∠AQP are alternate angles
Therefore, ∠A = ∠AQP = 30°

Similarly,
∠C and ∠CQP are alternate angles
Therefore, ∠C = ∠CQP = 40°
But, ∠Q = ∠AQP + ∠CQP
= 30° + 40° = 70°

Question 4.
The figure shows a triangle drawn in a rectangle. Calculate the angles of the triangle.

Answer:
Consider the figure below,

Since, ABCD is rectangle ∠P, ∠Q, ∠R, ∠S equals to 90°
Now, it is given that one part of ∠Q is 50°, so the remaining part of∠A is 40°
Similarly, one part of ∠R is 30°, so the remaining part of ∠R is 60°
It implies that two angles of the triangle are 40° and 60o.
Therefore, the third angle of the triangle is = 180° – (40° + 60°)
= 180° — (100°) = 80°
So, the angles of triangles are 40°, 60°, 80°

Question 5.
A triangle is drawn inside a parallelogram. Calculate the angles of the triangle.

Answer:
Consider the figure below

∠D and ∠B are opposite angles of parallelogram.
Therefore, ∠D = ∠B = 105°
But, one part of ∠D is 50°, so remaining part of ∠D = 105° – 50° = 55°
Now, considering the angles ∠A and ∠D, their sum is equals to 180°
i.e., ∠A + ∠D = 180°
∠A = 180° – ∠D
= 180° – 105° = 75o
But, one part of ∠A is 20°, so remaining part of ∠A = 75° – 20o = 55°
It means, we get two angles of the triangle which are 55° and 55°
So, third angle of the triangle = 180° – (55° + 55°)
= 180° – (110°) = 70°
Hence, all three angles of the triangles are 55°, 55°, 70°.

Practice Questions

Question 1.
Draw the parallelogram below with the given measures.

Calculate the other three angles.
Answer:
The other three angles of Parallelogram 60°, 60°, 120°

Question 2.
Given that PQ and QR are parallel lines, find the angle shown in the figure?

Answer:
90°

Question 3.
In the figure, the pair of lines slanted to the left are parallel; and also the pair of lines slanted to the right.

Draw this figure:
Answer:
third angle of the bottom triangle = 90°

Question 4.
The top and bottom lines in the figure are parallel.

Calculate the third angle of the bottom triangle and all angles of the top triangle.
Answer:
all angles of the top triangle = 90°, 40o, 50°,

Question 5.
The left and right lines in the figure are parallel.

Calculate the third angle of the larger triangle and all angles of the smaller triangle.
Answer:
Third angle of the larger triangle = 80°

Question 6.
A triangle is drawn inside a parallelogram.

Calculate the angles of the triangle.
Answer:
All angles of smaller triangle = 80°, 55°, 45°

Question 7.
The figure shows a triangle drawn in a rectangle.

Calculate the angles of the triangle.
Answer:
All angles of triangle = 70°, 70°, 40°

Question 8.
Draw the triangle with the given measures.

Answer:
90°, 55°, 35° 70°.

Class 7 Maths Chapter 1 Notes Kerala Syllabus Parallel Lines

Understanding geometry begins with recognising the relationships between lines and angles. In this chapter, we delve into the fascinating world of parallel lines and the angles they create when intersected by other lines. This exploration will lay the foundation for more advanced geometric concepts and develop your ability to visualise and solve geometric problems.

Lines and Angles
We start by examining lines and the angles formed when they intersect. A special focus is given to pairs of angles that are formed on the same side of the slanting line, both above and below the parallel lines. These pairs of angles are equal in measure, helping us understand the special properties of angles created when parallel lines are intersected by a slanting line.

Matching Angles
Next, we explore the idea of matching angles. This section introduces alternate angles and co-interior angles more explicitly. Alternate angles are pairs of angles that lie on opposite sides of the transversal but inside the two lines. These angles are crucial in identifying and proving the parallel nature of lines. Co-interior angles are also revisited here, reinforcing their properties and significance.

Triangle Sum
Finally, we turn our attention to triangles, a fundamental shape in geometry. One of the most important properties of triangles is the sum of their interior angles. You will learn and prove that the total sum of all three angles in any triangle is always 180°. This section will include various exercises to solidify your understanding and application of this essential geometric principle.

By the end of this chapter, you will have a solid understanding of how parallel lines interact with transversals, the relationships between the resulting angles, and the fundamental properties of triangles. These concepts are not only crucial for your current studies but also form the bedrock of many advanced topics in geometry.

Matching Angles
Of the angles made when two parallel lines are cut by a slanting line,

• • the small angles are of the same measure.
  • the large angles are of the same measure.
  • a small angle and a large angle add up to 180°
• If the intersecting line is perpendicular to one of the parallel lines, it would be perpendicular to the other line too, and all angles would be right angles.

Interior Angles:
These are the angles on the inside of the parallel lines.
When you look at the pairs of interior angles on the same side of the slanting line, they are called co-interior angles.
The sum of each pair of co-interior angles is always 180°.

Exterior Angles:
These are the angles on the outside of the parallel lines.
Similarly, when you look at the pairs of exterior angles on the same side of the slanting line, they are called co-exterior angles.
The sum of each pair of co-exterior angles is also 180°.

Alternate Interior Angles:
Alternate interior angles are pairs of angles that lie between the two parallel lines and on opposite sides of the slanting line.
These angles are same measure

Alternate Exterior Angles:
Alternate exterior angles are pairs of angles that lie outside the two parallel lines and on opposite sides of the slanting line.
Like alternate interior angles, alternate exterior angles are also equal.

Corresponding Angles:
These angles are in the same relative position at each intersection where a slanting line crosses the parallel lines.
Each pair of corresponding angles has the same measure.

Find out the other angles in the given parallelogram?

Answer:
First, consider the given angle (55° angle). To determine other angles, observe the intersections made by the left side with the top and bottom parallel lines.
The 55° angle and the angle above it form a pair of angles that add up to 180°. Therefore, the top angle is:
180° – 55° = 125°
Next, look at the angle to the right of the marked 55° angle.
To find this angle, examine the angles formed by the left and right parallel sides with the bottom line. The 55° angle and the angle to its right also form a pair of angles that add up to 180°.
Thus, this angle is also 125°, as calculated earlier.

Triangle Sum
The sum of all angles of a triangle is 180°

• A line intersects two parallel lines at angles of the same measure
• When two parallel lines are cut by a slanting line
  • the small angles are of the same measure.
  • the large angles are of the same measure.
  • a small angle and a large angle add up to 180°.
• If the intersecting line is perpendicular to one of the parallel lines, would be perpendicular to the other line too, and all angles would be right angles.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 4 Reciprocals Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 4 Solutions Reciprocals

Class 7 Maths Chapter 4 Reciprocals Questions and Answers Kerala State Syllabus

Reciprocals Class 7 Questions and Answers Kerala Syllabus

Page 64

Question 1.
Suma has 16 rupees with her. Safeer has 4 rupees.
(i) What part of Suma’s money does Safeer have?
(ii) How many times Safeer’s money does Suma have?
Answer:
Suma’s amount = 16
Safeer’s amount = 4
(i) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{4}{16}\)
= \(\frac{1}{4}\)
So, Safeer has of Suma’s money.

(ii) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{16}{4}\)
= 4
So, Suma has 4 times that of Safeer’s money.

Question 2.
A large bag contains 9 kilograms of sugar. A small bag contains 6 kilograms.
(i) The weight of sugar in the heavier bag is how much times that in the lighter bag?
(ii) The weight of sugar in the lighter bag is what part of that in the heavier bag?
Answer:
(i) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{9}{6}\)
= \(\frac{3}{2}\)
So, the weight of sugar in the heavier bag is \(\frac{3}{2}\) times that in the lighter bag.

(ii) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{6}{9}\)
= \(\frac{2}{3}\)
So, the weight of sugar in the lighter bag is \(\frac{2}{3}\) part of that in the heavier bag.

Question 3.
The weight of an iron block is 6 kilograms. The weight of another block is 26 kilograms.
(i) The weight of the lighter block is what fraction of that of the heavier block?
(ii) The weight of the heavier block is how much times that of the lighter block?
Answer:
(i) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{6}{26}\)
= \(\frac{3}{13}\)
So, the weight of the lighter block is \(\frac{3}{13}\) fraction of that of the heavier block.

(ii) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{26}{6}\)
= \(\frac{13}{3}\)
So, the weight of the heavier block is \(\frac{13}{3}\) times that of the lighter block.

Question 4.
The length of a ribbon is 2 times the length of a smaller ribbon. What part of the length of the large ribbon is the length of the small ribbon?
Answer:
Times = \(\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)
2\(\frac{2}{3}=\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)
\(\frac{8}{3}=\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)

8 × length of the smaller ribbon = 3 × length of the large ribbon
\(\frac{\text { length of the smaller ribbon }}{\text { length of the large ribbon }}=\frac{3}{8}\)
Part = \(\frac{3}{8}\)
So, length of the smaller ribbon is \(\frac{3}{8}\) part of the length of the large ribbon.

Page 67

Question 1.
27 students of a class got A plus in Maths. They form of the entire class. How many students are there in the class?
Answer:
\(\frac{3}{4}\) × entire class = 27
entire class = 27 ÷ \(\frac{3}{4}\)
= 27 × \(\frac{4}{3}\)
= 9 × 4
= 36
Thus, there are 36 students in the class.

Question 2.
\(\frac{2}{3}\) of a bottle was filled with \(\frac{1}{2}\) litre of water. How many litres of water will the bottle hold?
Answer:
\(\frac{2}{3}\) x entire bottle = \(\frac{1}{2}\) litres
entire bottle = \(\frac{1}{2} \div \frac{2}{3}\)
= \(\frac{1}{2} \times \frac{3}{2}\)
= \(\frac{3}{4}\)
Thus, the bottle will hold – litres of water.

Question 3.
\(\frac{3}{4}\) of a vessel holds 1\(\frac{1}{2}\) litres of water. What is the capacity of the vessel in litres if it is completely filled with water?
Answer:
\(\frac{3}{4}\) × entire vessel = 1\(\frac{1}{2}\) = \(\frac{3}{2}\) litres
entire vessel = \(\frac{3}{2} \div \frac{3}{4}\)
= \(\frac{3}{2} \times \frac{4}{3}\)
= \(\frac{4}{2}\)
= 2
Thus, the capacity of the vessel if it is completely filled with water is 2 litres.

Question 4.
Two of the three ribbons of the same length and half the third ribbon were placed end to end. It came to 1 metre. What is the length of a ribbon in centimetres?
Answer:
1 metre = 100 cm
2.5 ribbons = 100 cm
\(\frac{5}{2}\) × length of a ribbon = 100 cm
length of a ribbon = 100 ÷ \(\frac{5}{2}\)
= 100 × \(\frac{2}{5}\)
= 20 × 2
= 40 cm

Page 68

Question 1.
A 16 metres long rod is cut into pieces of length \(\frac{2}{3}\) metre. How many such pieces will be there?
Answer:
Total length of the rod = 16 m
Length of a piece = \(\frac{2}{3}\) m
Number of pieces = Total length of the rod ÷ Length of a piece
= 16 ÷ \(\frac{2}{3}\)
= 16 × \(\frac{3}{2}\)
= 8 × 3
= 24.

Question 2.
How many \(\frac{3}{4}\) litre bottles are needed to fill 5\(\frac{1}{4}\) litres of water?
Answer:
Total water = 5\(\frac{1}{4}=\frac{21}{4}\) litres
Amount of water in a bottle = \(\) litres
Number of bottles = Total water ÷ Amount of water in a bottle
= \(\frac{21}{4} \times \frac{4}{3}\)
= \(\frac{21}{4} \times \frac{4}{3}\)
= \(\frac{21}{3}\)
= 7

Question 3.
13\(\frac{1}{2}\) kilograms of sugar is to be packed into bags with 2\(\frac{1}{4}\) kilograms sugar each. How many bags are needed?
Answer:
Total sugar = 13\(\frac{1}{2}\) = \(\frac{27}{2}\) kg
Amount of sugar in one bag = 2\(\frac{1}{4}=\frac{9}{4}\)kg
Number of bags = Total sugar Amount of sugar in one bag
= \(\frac{27}{2} \div \frac{9}{4}\)
= \(\frac{27}{2} \times \frac{4}{9}\)
= 3 × 2
= 6.

Question 4.
The area of a rectangle is 22\(\frac{1}{2}\) square centimetres, and one side is 3\(\frac{3}{4}\) centimetres long. What is the length of the other side?
Answer:
Area of a rectangle = 22\(\frac{1}{2}\) = \(\frac{45}{2}\)sq.cm
length of one side = 3\(\frac{3}{4}=\frac{15}{4}\) cm
length of the other side = area of a rectangle ÷ length of one side
= \(\frac{45}{2} \div \frac{15}{4}\)
= \(\frac{45}{2} \times \frac{4}{15}\)
= 3 × 2
= 6 cm

Question 5.
How many pieces, each of length 2\(\frac{1}{2}\) metres, can be cut off from a rope of length 11\(\frac{1}{2}\) metres? How many metres of rope will be left?
Answer:
Total length of the rope = 11\(\frac{1}{2}\) = \(\frac{23}{2}\) m
Length of a piece = 2\(\frac{1}{2}=\frac{5}{2}\)m
Number of pieces = Total length of the rope ÷ Length of a piece
= \(\frac{23}{2} \div \frac{5}{2}\)
= \(\frac{23}{2} \times \frac{2}{5}\)
= \(\frac{23}{5}\)
= 4\(\frac{3}{5}\)

So, we can cut 4 pieces of length 2\(\frac{1}{2}\) m.
4 × 2\(\frac{1}{2}\)
= 4 × \(\frac{5}{2}\)
= 2 × 5
= 10 m
Length of the remaining rope =11\(\frac{1}{2}\) – 10 = \(\frac{23}{2}\) – 10 = \(\frac{3}{2}\)m

Page 68

Question 1.
A rod is 36 metres long. How many pieces each of length 2\(\frac{1}{2}\) metres can be cut off from it? What is the length of the rod left?
Appu did the problem this way.
36 ÷ 2\(\frac{1}{2}\) = 36 ÷ \(\frac{5}{2}\)
= 36 ÷ \(\frac{2}{5}\)
= \(\frac{72}{5}\)
When we divide 72 by 5, the quotient is 14 and remainder is 2. So we get 14 pieces. The remaining rod is of length 2 metres.
Ammu used another idea.
2 pieces, each of length 2\(\frac{1}{2}\) metres makes 5 metres.
7 × 5 = 35
So 7 × 2 = 14 pieces can be cut off. The remainder is 36 – 35 = 1 metre. Whose answer is right?
Solution:
In Appu’s case, he find that 14 pieces of length 2\(\frac{1}{2}\) metres can be cut off from it.
This 14 pieces together forms;
14 × 2\(\frac{1}{2}\) = 14 × \(\frac{5}{2}\)
= 7 × 5
= 35 m
So, the remaining length = 36 – 35 = 1 m.
Thus, Ammu’s answer is the right one.

Class 7 Maths Chapter 4 Kerala Syllabus Reciprocals Questions and Answers

Question 1.
Find the reciprocal of \(\frac{3}{7}\)
Answer:
reciprocal of \(\frac{3}{7}=\frac{7}{3}\)

Question 2.
Simplify \(\frac{3}{4} \div \frac{5}{6}\)
Answer:
\(\frac{3}{4} \div \frac{5}{6}=\frac{3}{4} \times \frac{6}{5}=\frac{18}{20}=\frac{9}{10}\)

Question 3.
A recipe requires cup of sugar to make 12 cookies. How much sugar is needed to make 36 cookies?
12 കുക്കീസ് ഉണ്ടാക്കുവാനായി \(\frac{3}{4}\) കപ്പ് പഞ്ചസാര വേണം. അങ്ങനെയങ്കിൽ 36 കുക്കീസ് ഉണ്ടാക്കുവാൻ എത്ര കപ്പ് പഞ്ചസാര വേണം?
Answer:
\(\frac{3}{4}\) cup of sugar gives 12 cookies.
12 × 3 gives 36.
So, for 36 cookies we need \(\frac{3}{4}\) × 3 = \(\frac{9}{4}\) cups of sugar.

Question 4.
If of the cake is eaten, what fraction of the whole cake remains?
ഒരു കേക്കിന്റെ \(\frac{2}{5}\) ഭാഗം കഴിച്ചെങ്കിൽ ബാക്കി എത്ര ഭാഗം ഉണ്ട്?
Answer:
The whole cake is \(\frac{5}{5}\) = 1
Remaining cake = \(\frac{5}{5}-\frac{2}{5}=\frac{3}{5}\)
Thus, \(\frac{3}{5}\) part of the whole cake remains.

Question 5.
A ribbon is cut into 12 equal parts. What fraction of the ribbon is 3 parts?
ഒരു റിബൺ 12 തുല്യ ഭാഗങ്ങളായി മുറിച്ചു. അതിലെ 3 ഭാഗം ആകെയുള്ള റിബണിന്റെ എത്ര ഭാഗമാണ്?
Answer:
Part (or fraction) = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{3}{12}\)
= \(\frac{1}{4}\)

Practice Questions

Question 1.
Solve \(\frac{4}{5} \div \frac{5}{4}\)
Answer:
\(\frac{16}{25}\)

Question 2.
Calculate \(\frac{3}{7} \div \frac{7}{3}\)
Answer:
1

Question 3.
A company produces \(\frac{2}{5}\) of its daily output in the morning and \(\frac{3}{8}\) in the afternoon. What fraction of the total daily output is produced in the afternoon?
Answer:
\(\frac{15}{31}\)

Question 4.
Calculate the reciprocal of \(\frac{2}{5}\). Which is larger?
Answer:
\(\frac{5}{2}\), reciprocal is the largest

Question 5.
When a tank is \(\frac{1}{4}\) full, it contains 80 litres of water. What will it contain when it is \(\frac{3}{8}\) full?
Answer:
120 litres

Class 7 Maths Chapter 4 Notes Kerala Syllabus Reciprocals

We are already familiar with a number of mathematical operations, such as addition, subtraction, multiplication and division. This chapter introduces a new type of mathematical operation named “reciprocal”. Following are the topics discussed in this chapter.

Times and parts
If we are given a large number and a small number, we can say the large number is how many times the small number. Always,
Times = \(\frac{\text { large number }}{\text { small number }}\)

If we are given a large number and a small number, we can say the small number is what part of the large number. Always,
Part = \(\frac{\text { small number }}{\text { large number }}\)

Topsy – Turvy
A fraction obtained by interchanging the numerator and denominator of the given fraction is called the reciprocal of the given fraction.
If we multiply a natural number by its reciprocal, we will get one as the result.
Eg:
The reciprocal of \(\frac{2}{3}\) is \(\frac{3}{2}\)

If we multiply a fraction by its reciprocal, we will get one as the result.
Eg:
Consider the fraction \(\frac{3}{4}\). Its reciprocal is \(\frac{4}{3}\). Now, \(\frac{3}{4} \times \frac{4}{3}\) = 1

Zero does not have a reciprocal because any number multiplied by zero is zero, not 1.

Fraction division
Division of natural numbers is the same as multiplication by the reciprocal.
Dividing a fraction (or a number) by another fraction is exactly same as multiplying the first fraction (or number) by the reciprocal of the second fraction.
Eg:
\(\frac{1}{2} \div \frac{3}{4}=\frac{1}{2} \times \frac{4}{3}=\frac{4}{6}=\frac{2}{3}\)
8 ÷ \(\frac{2}{5}\) = 8 × \(\frac{5}{2}\) = 4 × 5 = 20

If we are given a large number and a small number, we can say;
the large number is how many times the small number
the small number is what part of the large number large number
Times = \(\frac{\text { large number }}{\text { small number }}\)
Part = \(\frac{\text { small number }}{\text { large number }}\)

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 2 Fractions Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 2 Solutions Fractions

Class 7 Maths Chapter 2 Fractions Questions and Answers Kerala State Syllabus

Fractions Class 7 Questions and Answers Kerala Syllabus

Page 24

Do the following problems mentally. Write each as how many times and also as a product.

Question 1.
Each piece of a pumpkin weighs a quarter kilogram. What is the weight of two pieces together? What is the weight of four such pieces? Six pieces?
Answer:
Two times quarter kilogram is half kilogram.
Four times quarter kilogram is one kilogram.
Six times quarter kilogram is one and a half kilogram.
Mathematically,
Each piece of a pumpkin weighs \(\frac{1}{4}\) kilogram.
The weight of two pieces is two times \(\frac{1}{4}\), which is \(\frac{1}{2}\) kilogram.
The weight of four pieces is 4 times \(\frac{1}{4}\) which is 1 kilogram.
The weight of six pieces is 6 times \(\frac{1}{4}\) which is 1\(\frac{1}{2}\) kilogram.

Using products,
The weight of two pieces = 2 × \(\frac{1}{4}=\frac{2}{4}\) = \(\frac{1}{2}\) kilogram
The weight of four pieces= 4 × \(\frac{1}{4}=\frac{4}{4}\) = 1 kilogram
The weight of six pieces = 6 × \(\frac{1}{4}\) = 1\(\frac{1}{2}\) kilogram.

Question 2.
We can fill a cup with one-third of a litre of milk. How much milk is needed to fill two cups? Four cups?
Answer:
Two times one-third of a litre is two-third of a litre.
Four times one-third of a litre is four-third of a litre. Mathematically,
Each cup contains one-third of a litre of milk.
So, milk needed to fill two cups is 2 times \(\frac{1}{3}\) which is \(\frac{2}{3}\) litre.
Milk needed to fill four cups is 4 times \(\frac{1}{3}\), which is 1\(\frac{1}{3}\) litre.
Using products,
Milk needed to fill two cups = 2 × \(\frac{1}{3}=\frac{2}{3}\) litre
Milk needed to fill four cups = 4 × \(\frac{1}{3}\) = 1\(\frac{1}{3}\) litre.

Question 3.
What is the total length of four pieces of ribbons, each of length three fourths of a metre? What about five pieces?
Answer:
Four times three fourth of a metre is three metre.
Five times three fourth of a metre is three and three-fourth of a metre.
Mathematically,
Length of each piece is three-fourth of a metre.
So, total length of four pieces of ribbon is 4 times \(\frac{3}{4}\), which is 3 metre.
Total length of five pieces of ribbon is 5 times \(\frac{3}{4}\), which is 3\(\frac{3}{4}\) metre.
Using products,
Total length of four pieces of ribbon = 4 × \(\frac{3}{4}\) = 3 metre
Total length of five pieces of ribbon = 5 × \(\frac{1}{2}\) = 3\(\frac{3}{4}\) metre

Question 4.
It takes hour to walk around a play ground once.
(i) How much time does it take to walk 4 times around at this speed?
(ii) What about 7 times?
Answer:
(i) 4 times \(\frac{1}{4}\) is 1 hour.
Using products,
Time taken to walk 4 time around = 4 × \(\frac{1}{4}\) = 1 hour

(ii) 7 times \(\frac{1}{4}\) is 1\(\frac{3}{4}\) hour.
Using products,
Time taken to walk 7 times around = 7 × \(\frac{1}{4}\) = 1\(\frac{3}{4}\)hour.

Page – 25,26

Question 1.
The weight of an iron block is kilogram.
(i) What is the total weight of such 15 blocks?
(ii) 16 blocks?
Answer:
(i) Weight of one block = \(\frac{1}{4}\) kilogram.
Weight of 15 blocks = 15 × \(\frac{1}{4}=\frac{15 \times 1}{4}\)
= \(\frac{15}{4}=\frac{12+3}{4}\)
= \(\frac{12}{4}+\frac{3}{4}\)
= 3\(\frac{3}{4}\)

(ii) Weight of 16 blocks = 16 × \(\frac{1}{4}=\frac{16}{4}\) = 4 kilogram.

Question 2.
Some 2 metre long rods are cut into 5 pieces of equal length.
(i) What is the length of each piece?
(ii) What is the total length of 4 pieces?
(iii) of 10 pieces?
Answer:
(i) Total length of the rod = 2 metre.
Length of each piece = \(\frac{2}{5}\) metre

(ii) Total length of 4 pieces = 4 × \(\frac{2}{5}=\frac{4 \times 2}{5}=\frac{8}{5}\)
= \(\frac{5+3}{5}=\frac{5}{5}+\frac{3}{5}\)
= 1\(\frac{1}{5}\) metre

Question 3.
5 litres of milk is filled in 6 bottles of the same size.
(i) How many litres of milk does each bottle hold?
(ii) How many litres in 3 bottles together?
(iii) In 4 bottles?
Answer:
Total milk = 5 litre
Number of bottles = 6
(i) Milk in each bottle = \(\frac{5}{6}\) litre

(ii) Milk in 3 bottles together = 3 × \(\frac{5}{6}=\frac{3 \times 5}{6}=\frac{15}{6}\)
= \(\frac{12+3}{6}=\frac{12}{6}+\frac{3}{6}\)
= 2\(\frac{1}{2}\) litre.

(iii) Milk in 4 bottles = 4 × \(\frac{5}{6}=\frac{4 \times 5}{6}=\frac{20}{6}\)
= \(\frac{18+2}{6}=\frac{18}{6}+\frac{2}{6}\)
= 3\(\frac{1}{3}\) litre

Page – 28

Do these problems in head. Then write each as a part and also as a product of numbers.

Question 1.
Nine litres of milk is divided equally among three children. How many litres will each get? What if there are four children?
Answer:
Each will get a third of nine litres, that is three litres.
As a part,
Each will get \(\frac{1}{3}\) of 9, which is 3 litres.

As a product,
Each will get \(\frac{1}{3}\) × 9 = \(\frac{9}{3}\) = 3 litres.
If there are four children,
Each will get a fourth of nine litres. A fourth of 8 litre is 2 litre and then a fourth of the remaining one litre. So, two and one-fourth of a litre.

As a part,
Each will get \(\frac{1}{4}\) of 9, which is 2\(\frac{1}{4}\) litres.

As a product,
Each will get \(\frac{1}{4}\) × 9 = \(\frac{9}{4}\) = 2\(\frac{1}{4}\) litres.

Question 2.
Six kilograms of rice was packed in five bags of the same size. How many kilograms of rice in each bag? What if it is packed in four bags?
Answer:
Each bag has one fifth of six kilograms. One fifth of 5 kilogram is 1 kilogram and then one fifth of remaining one kilogram. So, one and one-fifth of a kilogram.

As a part,
Each bag has \(\frac{1}{5}\) of 6, which is 1\(\frac{1}{5}\) kilograms.

As a product,
Each bag has \(\frac{1}{4}\) × 6 = \(\frac{6}{5}\) = 1\(\frac{1}{5}\) kilograms.
If it is packed in four bags,
Each bag has one-fourth of six kilograms. One-fourth of four kilograms is 1 kilogram and then one- fourth of remaining two kilograms. So, one and a half kilograms.

As a part,
Each bag has \(\frac{1}{4}\) of 6, which is 1\(\frac{1}{2}\) kilograms.
As a product,
Each bag has \(\frac{1}{4}\) × 6 = \(\frac{6}{4}\) = 1\(\frac{1}{2}\) kilograms.

Question 3.
A seven metre long string is divided into six equal pieces. What is the length of each piece? What if it is divided into three equal pieces?
Answer:
Each piece is one-sixth of seven metres. One-sixth of six metres is 1 metre and then one-sixth of the remaining one metre. So, one and one-sixth of a metre.

As a part,
Length of each piece is \(\frac{1}{6}\) of 7, which is 1\(\frac{1}{6}\) metres.

As a product,
Length of each piece = \(\frac{1}{6}\) × 7 = \(\frac{7}{6}\) = 1 \(\frac{1}{6}\) metres.
If it is divided into three equal pieces,
Each piece is one-third of seven metres. One-third of six metres is 2 metres and one-third of remaining one metre. So, two and one-third of a metre.

As a part,
Length of each piece is \(\frac{1}{3}\) of 7, which is 2\(\frac{1}{3}\) metres.
As a product,
Length of each piece \(\frac{1}{3}\) × 7 = \(\frac{7}{3}\) = 2\(\frac{1}{3}\) metres.
The calculations of the types of problems above can be done as follows:

Question 4.
We have to cut off \(\frac{3}{5}\) of a 7 metre long string. How long is this piece?
Answer:
Here we have to calculate \(\frac{3}{5}\) of 7.
\(\frac{3}{5}\) x 7 = \(\frac{3 \times 7}{5}=\frac{21}{5}=\frac{20+1}{5}\)
\(\frac{20}{5}+\frac{1}{5}\) = 4 + \(\frac{1}{5}\)
= 4\(\frac{1}{5}\)
So, we need to cut off 4\(\frac{1}{5}\) metres.

Page – 29

Question 1.
There are 35 children in a class. of them are girls. How many girls are there in the class?
Answer:
Number of girls = \(\frac{3}{5}\) × 35
= 3 × \(\frac{35}{5}\)
= 3 × 7
= 21
Or,
Number of girls = \(\frac{3}{5}\) × 35
= \(\frac{3 \times 35}{5}\)
= \(\frac{105}{5}\)
= 21

Question 2.
10 kilograms of rice is filled equally in 8 bags. If the rice in 3 such bags are taken together, how many kilograms would that be?
Answer:
Rice in 3 bags are taken together.
That is, we have to find- of 10 kilograms.
So, amount of rice = \(\frac{3}{8}\) × 10
= \(\frac{30}{8}=\frac{24+6}{8}\)
= \(\frac{24}{8}+\frac{6}{8}\)
= 3\(\frac{3}{4}\) kilograms.

Question 3.
The area of the rectangle in the figure is 27 square centimetres. It is divided into 9 equal parts.

What is the area of the darker part in square centimetres?
Answer:
5 of the 9 equal parts are darker.
So, we have to find \(\frac{5}{9}\) of 27 square centimetres.

Area of darker part = \(\frac{5}{9}\) × 27 = 5 × \(\frac{27}{9}\)
= 5 × 3 = 15 square centimeres.

Area of darker part = \(\frac{5}{9}\) × 27 = \(\frac{5 \times 27}{9}=\frac{135}{9}\)
= 15 square centimetres.

Page – 32

Question 1.
Draw rectangles and find these products.
(i) \(\frac{1}{2} \times \frac{1}{4}\)
(ii) \(\frac{1}{3} \times \frac{1}{6}\)
(iii) \(\frac{1}{5} \times \frac{1}{8}\)
Answer:
(i) \(\frac{1}{2} \times \frac{1}{4}\)
Draw a rectangle a divide it into 4 equal parts. Then divide one part into half.

Now extend the horizontal line.

So, \(\frac{1}{2} \times \frac{1}{4}\) = \(\frac{1}{8}\)

(ii) \(\frac{1}{3} \times \frac{1}{6}\)
Draw a rectangle and divide it into 6 equal parts. Then divide each part into 3 equal parts.

Now extend the horizontal line.

So, \(\frac{1}{3} \times \frac{1}{6}=\frac{1}{18}\)

(iii) \(\frac{1}{5} \times \frac{1}{8}\)
Draw a rectangle a divide it into 8 equal parts. Then divide one part into 5 equal parts.

Now extend the horizontal line.

So, \(\frac{1}{5} \times \frac{1}{8}=\frac{1}{40}\)

Question 2.
A one metre long string is divided into five equal parts. How long is half of each part in metres? In centimetres?
Answer:
When one metre long string divided into five equal parts each part is one- fifth of one metre.
Now half of each part is \(\frac{1}{2}\) of \(\frac{1}{5}\)of one metre.
∴ Length of each part = \(\frac{1}{2} \times \frac{1}{5}\) = \(\frac{1}{10}\) metre
= \(\frac{1}{10}\) × 100
= 10 centimetre.

Question 3.
One litre of milk is filled in two bottles of equal size. A quarter of the milk in one bottle was used to make tea. How many litres of milk were used for tea? In millilitres?
Answer:
When one litre of milk is filled in two bottles of equal size, each bottle contain half of a litre.
A quarter of the milk in one bottle was used to make tea, which is of of one litre.
Milk used for tea \(\frac{1}{4}\) of \(\frac{1}{2}\) one litre
= \(\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}\) litre
= \(\frac{1}{8}\) × 1000 = 125 millilitre.

Consider another type of problem:
Find of \(\frac{4}{5}\) of \(\frac{2}{3}\).
Answer:

Page – 34

Question 1.
A rope 2 metres long is cut into 5 equal pieces. What is the length of three quarters of one of the pieces in metres? In centimetres?
Answer:
When 2 metre long rope cut into 5 equal pieces, length of each piece is \(\frac{1}{5}\) of 2 metre, which is \(\frac{2}{5}\) metre.
Length of three quarters of one of the pieces is \(\frac{3}{4}\) of \(\frac{2}{5}\)
Required length of the piece = \(\frac{3}{4} \times \frac{2}{5}\)
= 3 × \(\frac{1}{4} \times \frac{1}{5}\) × 2
= 3 × \(\frac{1}{20}\) × 2
= \(\frac{6}{20}=\frac{3}{10}\)

Question 2.
4 bottles of the same size were filled with 3 litres of water. One of these was used to fill 5 cups of the same size. How much water is there in one such cup, in litres? And in millilitres?
Answer:
When 4 bottles of the same size were filled with 3 litres of water, each bottle has of 3 litre, which is \(\frac{3}{4}\) litre.
One of these was used to fill 5 cups of the same size. Then, amount of water in one cup is \(\frac{1}{5}\) of \(\frac{3}{4}\).
Amount of water in one cup = \(\frac{1}{5} \times \frac{3}{4}\)
= \(\frac{1}{5} \times \frac{1}{4}\) × 3
= \(\frac{1}{20}\) × 3
= \(\frac{3}{20}\)

Question 3.
A watermelon weighing four kilograms was cut into five equal pieces. One piece was again halved. What is the weight of each of these two pieces in kilograms? And in grams?
Answer:
When watermelon weighing 4 kg cut into five equal pieces, weight of each piece is \(\frac{1}{5}\) of 4 kg, which is \(\frac{4}{5}\)kg.
Each piece is again halved.
Then weight of each of these two pieces is \(\frac{1}{2}\) of \(\frac{4}{5}\)
Required weight = \(=\frac{1}{2} \times \frac{4}{5}\)
= \(=\frac{1}{2} \times \frac{1}{5}\) × 4
= \(\frac{1}{10}\) × 4
= \(\frac{4}{10}=\frac{2}{5}\) kilograms

Question 4.
A vessel full of milk is used to fill three bottles of the same size. Then the milk in each bottle was used to fill four cups of the same size. What fraction of the milk in the first vessel does each cup contain?
Answer:
When a vessel full of milk used to fill three bottles of same size, each bottle has one-third of milk.
When milk in each bottle was used to fill four cups of same size, each cup has \(\frac{1}{4}\) of \(\frac{1}{3}\) of the milk.
∴ fraction of milk in each cup = \(\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}\)

Question 5.
Draw a line AB of length 12 centimetres. Mark AC as \(\frac{2}{3}\) of AB. Mark AD as \(\frac{1}{4}\) of AC. What part of AB is AD?
Answer:

AD = \(\frac{1}{4}\) of AC
= \(\frac{1}{4}\) of \(\frac{2}{3}\) of AB
= \(\frac{1}{4} \times \frac{2}{3}\) of AB
= \(\frac{1}{4} \times \frac{1}{3}\) × 2 of AB
= \(\frac{1}{12}\) × 2 of AB
= \(\frac{2}{12}=\frac{1}{6}\) of AB

Question 6.
Calculate the following using multiplication:
(i) \(\frac{3}{7}\) of \(\frac{2}{5}\)
Answer:
\(\frac{3}{7}\) of \(\frac{2}{5}\) = \(\frac{3}{7} \times \frac{2}{5}\)
= 3 × \(\frac{1}{7} \times \frac{1}{5}\) × 2
= 3 × \(\frac{1}{35}\) × 2
= \(\frac{6}{35}\)

(ii) \(\frac{2}{3}\) of \(\frac{3}{4}\)
Answer:
\(\frac{2}{3}\) of \(\frac{3}{4}\) = \(\frac{2}{3} \times \frac{3}{4}\)
= 2 × \(\frac{1}{3} \times \frac{1}{4}\) × 3
= 2 × \(\frac{1}{12}\) × 3
= 6 × \(\frac{1}{12}=\frac{6}{12}=\frac{1}{2}\)

(iii) \(\frac{3}{5}\) of \(\frac{2}{7}\)
Answer:
\(\frac{3}{5}\) of \(\frac{2}{7}\) = \(\frac{3}{5} \times \frac{2}{7}\)
= 3 × \(\frac{1}{5} \times \frac{1}{7}\) × 2
= 3 × \(\frac{1}{35}\) × 2
= \(\frac{6}{35}\)

(iv) \(\frac{5}{6}\) of \(\frac{3}{10}\)
Answer:
\(\frac{5}{6}\) of \(\frac{3}{10}\) = \(\frac{5}{6} \times \frac{3}{10}\)
= 5 × \(\frac{1}{6} \times \frac{1}{10}\) × 3
= 5 × \(\frac{1}{60}\) × 3
= 15 × \(\frac{1}{60}=\frac{15}{60}=\frac{1}{4}\)

Page – 37

Question 1.
One and a half metres of cloth is needed for a shirt. How much cloth is required for five such shirts?
Answer:
Cloth needed for a shirt = 1 metres
Cloth needed for 5 shirts = 5 × 1\(\frac{1}{2}\)
= 5 × (1 + \(\frac{1}{2}\))
= (5 × 1) + (5 × \(\frac{1}{2}\))
= 5 + 2\(\frac{1}{2}\)
= 7\(\frac{1}{2}\) metres.
Or,
5 × 1\(\frac{1}{2}\) = 5 × \(\frac{3}{2}\)
= \(\frac{15}{2}\) = 7\(\frac{1}{2}\)

Question 2.
The price of one kilogram of okra is thirty rupees. What is the price of two and a half kilograms?
Answer:
Price of one kilogram of okra = 30 rupees
Price of two and a half kilograms of okra = 30 × 2\(\frac{1}{2}\)
= 30 × (2 + \(\frac{1}{2}\))
= (30 × 2) + (30 × \(\frac{1}{2}\))
= 60 + 15
= 75 rupees
Or,
30 × 2\(\frac{1}{2}\) = 30 × \(\frac{5}{2}\)
= \(\frac{30}{2}\) × 5
= 15 × 575 rupees

Question 3.
A person walks two and a half kilometres in an hour. At the same speed, how far will he walk in one and a half hours?
Answer:
Distance he walks in one hour = 2 \(\frac{1}{2}\) km
Distance he walks in one and a half hours = 1 \(\frac{1}{2}\) x 2\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{5}{2}\)
= \(\frac{3 \times 5}{2 \times 2}=\frac{15}{4}\)
= 3 \(\frac{3}{4}\) km

Question 4.
Roni has 36 stamps with her. Sahira says she has 2 times this. How many stamps does Sahira have?
Answer:
Number of stamps with Roni = 36
Number of stamps with Sahira = 2\(\frac{1}{2}\) × 36
= (2 + \(\frac{1}{2}\)) × 36
= (2 × 36) +(\(\frac{1}{2}\) × 36)
= 72 + 18
= 90
Or
2\(\frac{1}{2}\) × 36 = \(\frac{5}{2}\) × 36
= 5 × \(\frac{36}{2}\)
= 5 × 18
= 90

Question 5.
Joji works 4\(\frac{1}{2}\) hours each day. How many hours does he work in 6 days?
Answer:
Number of hours Joji works each day = 4 hours
Number of hours Joji work in 6 days = 6 × 4\(\frac{1}{2}\)
= 6 × (4 + \(\frac{1}{2}\))
= (6 × 4) + (6 × \(\frac{1}{2}\))
= 24 + 3
= 27 hours

Question 6.
Calculate the following:
(i) 4 times 5\(\frac{1}{3}\)
Answer:
4 times 5\(\frac{1}{3}\) = 4 × 5\(\frac{1}{3}\)
= 4 × (5 + \(\frac{1}{3}\))
= (4 × 5) + (4 × \(\frac{1}{3}\))
= 20 + 1\(\frac{1}{3}\)
= 21 \(\frac{1}{3}\)

(ii) 4\(\frac{1}{3}\) times 5
Answer:
4\(\frac{1}{3}\) times 5 = 4\(\frac{1}{3}\) × 5
= (4 + \(\frac{1}{3}\)) × 5
= (4 × 5) + (\(\frac{1}{3}\) × 5)
= 20 + 1\(\frac{2}{3}\)
= 21\(\frac{2}{3}\)

(iii) 1\(\frac{1}{2}\) times \(\frac{2}{3}\)
Answer:
1\(\frac{1}{2}\) times \(\frac{2}{3}\) = 1\(\frac{1}{2}\) × \(\frac{2}{3}\)
= \(\frac{3}{2} \times \frac{2}{3}\)
= \(\frac{3 \times 2}{2 \times 3}\)
= 1

(iv) \(\frac{2}{5}\) times 2\(\frac{1}{2}\)
Answer:
\(\frac{2}{5}\) times 2\(\frac{1}{2}\) = \(\frac{2}{5}\) × 2\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{2}{3}\)
= \(\frac{3 \times 2}{2 \times 3}\)
= 1

(v) 2\(\frac{1}{2}\) times 5\(\frac{1}{2}\)
Answer:
2\(\frac{1}{2}\) times 5\(\frac{1}{2}\) = 2\(\frac{1}{2}\) × 5\(\frac{1}{2}\)
= \(\frac{5}{2} \times \frac{11}{2}\)
= \(\frac{5 \times 11}{2 \times 2}\)
= \(\frac{55}{4}\)
= 13 \(\frac{3}{4}\)

Page – 42

Question 1.
The length and breadth of some rectangles are given below. Find the area of each:
(i) 3\(\frac{1}{4}\) centimetres, 4\(\frac{1}{2}\) centimetres
Answer:
3\(\frac{1}{4}\) centimetres, 4\(\frac{1}{2}\) centimetres
Area of the rectangle = 3\(\frac{1}{4}\) × 4\(\frac{1}{2}\)
= \(\frac{13}{4} \times \frac{9}{2}\)
= \(\frac{13 \times 9}{4 \times 2}\)
= \(\frac{117}{8}\)
= 14\(\frac{5}{8}\) square centimetres

(ii) 5\(\frac{1}{3}\) metres, 6\(\frac{3}{4}\) metres
Answer:
5\(\frac{1}{3}\) metres, 6\(\frac{3}{4}\) metres
Area of the rectangle = 5\(\frac{1}{3}\) × 6\(\frac{3}{4}\)
= \(\frac{16}{3} \times \frac{27}{4}\)
= \(\frac{16}{4} \times \frac{27}{3}\)
= 4 × 3 = 12

Question 2.
What is the area of a square of side 1 metres?
Answer:
Area of the square = 1\(\frac{1}{2}\) × 1\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{3}{2}\)
= \(\frac{3 \times 3}{2 \times 2}\)
= \(\frac{9}{4}\)
= 2\(\frac{1}{4}\) square metre

Question 3.
The perimeter of a square is 14 metres. What is its area?
Answer:
Perimeter of the square = 14 metres
4 × side = 14
∴ Side = \(\frac{14}{4}\)= 3\(\frac{1}{2}\) metre
Area = 3\(\frac{1}{2}\) × 3\(\frac{1}{2}\)
= \(\frac{7}{2} \times \frac{7}{2}\)
= \(\frac{7 \times 7}{2 \times 2}\)
= \(\frac{49}{4}\)
= 12\(\frac{1}{4}\) square metres

Intext Questions and Answers

Question 1.
A bottle holds a quarter litre of water. How much water is needed to fill three such bottles?
Answer:
Each bottle holds a quarter litre of water. So, water needed to fill three such bottles is three times a quarter litre, which is three-quarter litres.
This can be calculated as,
3 times \(\frac{1}{4}\) is \(\frac{3}{4}\) is \(\frac{3}{4}\)
As a product,
3 × \(\frac{1}{4}=\frac{3}{4}\)

Question 2.
The calculations in the type of problems above can be done easily as follows:
\(\frac{3}{4}\) litres of milk in a bottle; how many litres in 7 such bottles?
Answer:
Amount of milk in one bottle = \(\frac{3}{4}\) litres
Amount of milk in 7 such bottles = 7 times \(\frac{3}{4}\)
7 × \(\frac{3}{4}=\frac{7 \times 3}{4}=\frac{21}{4}\)

Split 21 as a multiple of 4.
\(\frac{21}{4}=\frac{20+1}{4}=\frac{20}{4}+\frac{1}{4}\)
= 5 + \(\frac{1}{4}\)
= 5\(\frac{1}{4}\)

Question 3.
A five metre long string is cut into three equal pieces. What is the length of each piece?
Answer:
3 metres cut into three equal pieces, each piece is 1 metre. The remaining 2 metre cut into three equal pieces, each piece must be two-third of a metre. So, the length of each piece is 1 metres.
In terms of number alone,
\(\frac{1}{3}\) of 5 is 1\(\frac{2}{3}\)
Writing it as a product,
\(\frac{1}{3}\) × 5 = \(\frac{5}{3}\) = 1\(\frac{2}{3}\)

Question 4.
Draw rectangle and find \(\frac{1}{3} \times \frac{1}{2}\)
Answer:
First, draw a rectangle and divide it into two equal part. Then each part is \(\frac{1}{2}\)

Then divide one part into three equal parts. Here, each part is \(\frac{1}{3} \times \frac{1}{2}\)

Now extend the horizontal line.

The rectangle is divided into 6 equal parts and so each part is
Hence, \(\frac{1}{3} \times \frac{1}{2}\) = \(\frac{1}{6}\)

Question 5.
Find 3 × 2\(\frac{1}{4}\)
Answer:
3 × 2\(\frac{1}{4}\) = 3 × (2 + \(\frac{1}{4}\))
= (3 × 2) + (3 × \(\frac{1}{4}\))
= 6 + \(\frac{3}{4}\)
= 6\(\frac{3}{4}\)

Another way,
3 × 2\(\frac{1}{4}\) = 3 × \(\frac{9}{4}\)
= \(\frac{27}{4}\)
= 6\(\frac{3}{4}\)
Consider another problem.

Question 6.
Find 3\(\frac{1}{2}\) × 2\(\frac{1}{4}\)
Answer:
3\(\frac{1}{2}\) × 2\(\frac{1}{4}\)
= \(\frac{7}{2} \times \frac{9}{4}\)
= \(\frac{7 \times 9}{2 \times 4}\)
= \(\frac{63}{8}=\frac{56+7}{8}=\frac{56}{8}+\frac{7}{8}\)
= 7\(\frac{7}{8}\)

Question 7.
Find the area of the rectangle having length 5\(\frac{1}{2}\) cm and breadth 3\(\frac{1}{3}\) cm.
Answer:

Now, divide the length 5\(\frac{1}{2}\) cm into 11 equal parts of length \(\frac{1}{2}\) cm.
Divide the breadth 3\(\frac{1}{3}\) cm into 10 equal parts of length \(\frac{1}{3}\) cm.

So, 11 × 10 = 110 rectangles in all, each of area \(\frac{1}{6}\) square centimetres.
∴ Area of rectangle = 110 × \(\frac{1}{6}\) = 18\(\frac{1}{3}\)
Now, 5\(\frac{1}{2}\) × 3\(\frac{1}{3}\)
= \(\frac{11}{2} \times \frac{10}{3}\)
= 11 × \(\frac{1}{2}\) × 10 × \(\frac{1}{3}\)
= 110 × \(\frac{1}{6}\)
= 18\(\frac{1}{3}\)
So, even if the lengths are in fractions, the area of a rectangle is still the product of the lengths of sides.

Class 7 Maths Chapter 2 Kerala Syllabus Fractions Questions and Answers

Question 1.
Calculate the following:
(i) \(\frac{2}{3}\) of 16
Answer:
\(\frac{2}{3}\) of 16
= \(\frac{2}{3}\) × 16
= \(\frac{2 \times 16}{3}=\frac{32}{3}\)
= 10\(\frac{2}{3}\)

(ii) \(\frac{4}{7}\) of 25
Answer:
\(\frac{4}{7}\) of 25
= \(\frac{4}{7}\) × 25
= \(\frac{4 \times 25}{7}=\frac{100}{7}\)
= 14\(\frac{2}{3}\)

(iii) \(\frac{2}{7}\) of \(\frac{1}{4}\)
Answer:
\(\frac{2}{7}\) of \(\frac{1}{4}\)
= \(\frac{2}{7}\) × \(\frac{1}{4}\)
= \(\frac{2}{7} \times \frac{1}{4}\)
= \(\frac{2 \times 1}{7 \times 4}=\frac{2}{28}\)
= \(\frac{1}{14}\)

(iv) 1\(\frac{1}{2}\) × 6\(\frac{2}{3}\)
Answer:
1\(\frac{1}{2}\) × 6\(\frac{2}{3}\)
= \(\frac{3}{2} \times \frac{20}{3}=\frac{3 \times 20}{2 \times 3}\)
= \(\frac{60}{6}\)
= 10

(v) 2\(\frac{3}{4}\) × \(\frac{5}{8}\)
Answer:
2\(\frac{3}{4}\) × \(\frac{5}{8}\)
= \(\frac{11}{4} \times \frac{5}{8}\)
= \(\frac{11 \times 5}{4 \times 8}=\frac{55}{32}\)
= 1\(\frac{23}{32}\)

(vi) \(\frac{4}{7}\) of \(\frac{3}{5}\)
Answer:
\(\frac{4}{7}\) of \(\frac{3}{5}\)
= \(\frac{4}{7}\) × \(\frac{3}{5}\)
= \(\frac{4 \times 3}{7 \times 5}\)
= \(\frac{12}{35}\)

Question 2.
12 metre long rope cut into 4 equal pieces.
(i) What is the length of each piece?
(ii) What if it is cut into 5 equal pieces?
Answer:
(i) Length of each piece = \(\frac{12}{4}\) = 3 metres

(ii) If it is cut into 5 equal pieces,
Length of each piece = \(\frac{12}{5}\) = 2\(\frac{2}{5}\) metres

Question 3.
Each piece of rope is \(\frac{7}{4}\) metres long. What is the total length of 8 such pieces?
Answer:
Lenth of each piece = \(\frac{7}{4}\) metres
Length of 8 pieces = \(\frac{7}{4}\) × 8 = 7 × \(\frac{8}{4}\)
= 7 × 2
= 14 metres

Question 4.
If 4 strings of length \(\frac{1}{3}\) metre were laid end to end, what would be the total length?
Answer:
Total length = \(\frac{1}{3}\) × 4 = \(\frac{4}{3}\)
= 1\(\frac{1}{3}\) metres

Question 5.
Suhara has 1 metre long silk ribbon. She gave half of it to Soumya. She in turn gave half of this to Reena. What is the length of the piece Reena got?
Answer:
Length of the piece Reena got = \(\frac{1}{2}\) of \(\frac{1}{2}\) of 1 metre
= \(\frac{1}{2}\) × \(\frac{1}{2}\)
= \(\frac{1}{4}\) metre

Question 6.
Half the children in a class are girls. A third of them are in the Math Club. What fraction of the total children are they?
Answer:
Number of girls in Math club = \(\frac{1}{3}\) of \(\frac{1}{2}\) of total children
= \(\frac{1}{3} \times \frac{1}{2}\) of total children
= \(\frac{1}{6}\) of total children

Question 7.
The length and breadth of some rectangles are given below. Calculate their areas.
(i) 4\(\frac{1}{2}\) cm, 3\(\frac{1}{4}\) cm
Answer:
Area = 4\(\frac{1}{2}\) × 3\(\frac{1}{4}\)
= \(\frac{9}{2} \times \frac{13}{4}=\frac{9 \times 13}{2 \times 4}\)
= \(\frac{117}{8}\)
= 14\(\frac{5}{8}\) cm²

(ii) 6\(\frac{3}{4}\) cm, 5\(\frac{1}{3}\) cm
Answer:
Area = 6\(\frac{3}{4}\) × 5\(\frac{1}{3}\)
= \(\frac{27}{4} \times \frac{16}{3}\)
= \(\frac{27}{3} \times \frac{16}{4}\)
= 9 × 4
= 36 cm²

Question 8.
What is the area of a square of side 1 metre?
Answer:
Area 1\(\frac{1}{2}\) x 1\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{3}{2}\)
= \(\frac{9}{4}\)
= 2\(\frac{1}{4}\) m²

Practice Questions

Question 1.
Each bottle has \(\frac{3}{4}\) litres of water. What is the quantity of water in 14 such bottles?
Answer:
\(\frac{21}{2}\) = 10\(\frac{1}{2}\) litres

Question 2.
A farmer plants the sapling of a plant at a uniform distance of cm. If he plants 27 such saplings in a row, find the total distance between the first and the last sapling.
Answer:
45 cm

Question 3.
In a class of 60 students, two-thirds are boys. How many girls are there in the class?
Answer:
20

Question 4.
The weight of an iron block is \(\frac{3}{4}\) kilogram. What is the total weight of such 17 blocks?
Answer:
\(\frac{51}{4}\) = 12\(\frac{3}{4}\) kilogram

Question 5.
There are some cans, each containing 3 litres of milk. The milk in each vessel is used to fill 5 identical bottles.
(i) How much milk is there in each bottle?
Answer:
\(\frac{3}{5}\) litres

(ii) How much milk in 3 such bottles?
Answer:
\(\frac{9}{5}\) = 1 \(\frac{4}{5}\) litres

(iii) In 10 bottles?
Answer:
6 litres

Question 6.
Calculate the following:
(i) \(\frac{1}{8} \times \frac{1}{5}\)
Answer:
\(\frac{1}{40}\)

(ii) \(\frac{1}{6} \times \frac{1}{7}\)
Answer:
\(\frac{1}{42}\)

(iii) \(\frac{2}{5} \times \frac{7}{9}\)
Answer:
\(\frac{14}{45}\)

(iv) \(\frac{4}{5}\) of \(\frac{2}{3}\)
Answer:
\(\frac{8}{15}\)

(v) \(\frac{2}{9}\) of \(\frac{3}{2}\)
Answer:
\(\frac{1}{3}\)

(vi) 1\(\frac{3}{4}\) of 4
Answer:
7

(vii) \(\frac{3}{8}\) of 2\(\frac{1}{2}\)
Answer:
\(\frac{15}{16}\)

(viii) 3\(\frac{1}{4}\) × 5\(\frac{2}{9}\)
Answer:
\(\frac{611}{36}\)

(ix) 4\(\frac{1}{7}\) × 3\(\frac{1}{8}\)
Answer:
\(\frac{725}{56}\)

Question 7.
If three litres of milk is equally divided among four persons, how much would each get?
Answer:
\(\frac{3}{4}\) litres

Question 8.
Six kilogram of rice is packed into four identical bags.
(i) How much rice is in each bag?
Answer:
\(\frac{3}{2}\) Kilograms

(ii) What if it is packed into two bags?
Answer:
3 Kilograms

Class 7 Maths Chapter 2 Notes Kerala Syllabus Fractions

Fractions are a way to represent parts of a whole. A fraction consists of a numerator and a denominator. The numerator is the number written above the fraction line, which tells how many equal parts you consider, and the denominator is the number written below the fraction line, which tells the total number of equal parts the whole thing has been cut into.

Multifold multiplication
If we have to find 4 times three-quarter, we calculate it in following way:
\(\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{3}{4}\) = 3
This can be easily calculated as:
4 × \(\frac{3}{4}=\frac{12}{4}\) = 3

Part and multiplication
If three litres of milk divided among four people, each will get \(\frac{1}{4}\) of 3 litres.
i.e, \(\frac{1}{4}\) of 3 = \(\frac{1}{4}\) × 3 = \(\frac{3}{4}\) litres.

Part of part
If we have to calculate \(\frac{1}{5}\) of \(\frac{1}{4}\), it can be done as as follows:
\(\frac{1}{5} \times \frac{1}{4}=\frac{1}{5 \times 4}=\frac{1}{20}\)
Now, if have to calculate \(\frac{3}{5}\) of \(\frac{2}{7}\),
\(\frac{3}{5} \times \frac{2}{7}=3 \times \frac{1}{5} \times \frac{1}{7} \times 2=\frac{3 \times 2}{5 \times 7}=\frac{6}{35}\)

To calculate 2\(\frac{1}{4}\) × 5,
2\(\frac{1}{4}\) × 5 = (2 + \(\frac{1}{4}\)) × 5
= (2 × 5) + (\(\frac{1}{4}\) × 5)
= 10 + 1\(\frac{1}{4}\)
= 11\(\frac{1}{4}\)

To calculate 1\(\frac{1}{2}\) × 3\(\frac{3}{4}\)
1\(\frac{1}{2}\) × 3\(\frac{3}{4}\) = \(\frac{3}{2} \times \frac{15}{4}\)
= \(\frac{3 \times 15}{2 \times 4}\)
= \(\frac{45}{8}=\frac{40+5}{8}=\frac{40}{8}+\frac{5}{8}\)
= 5 \(\frac{5}{8}\)
Even if the lengths are in fractions, the area of a rectangle is still the product of the lengths of sides.

Share And Fraction
If 4 litres of milk is divided equally among 3 persons, how much milk will each one get? First give 1 litre to each one. If the remaining 1 litre is divided among 3 persons, each will get \(\frac{1}{3}\) litre. So, in total each get 1\(\frac{1}{3}\) litres.
Numerically,
4 ÷ 3 = \(\frac{4}{3}\) = 1\(\frac{1}{3}\)

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 3 Triangles Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 3 Solutions Triangles

Class 7 Maths Chapter 3 Triangles Questions and Answers Kerala State Syllabus

Triangles Class 7 Questions and Answers Kerala Syllabus

Page 52

Question 1.
The sides of a triangle are natural numbers. If the lengths of two sides are 5 centimetres and 8 centimetres, what are the possible numbers which can be the length of the third side?
Answer:
We know that the sum of the lengths of any two sides is greater than the length of the third side.
So, 5 + 8 should be greater than the third side.
That is, 13 is greater than the third side.

So, the length of the third side should be less than 13.
Also, we know that the difference of the lengths of any two sides is smaller than the length of the third side.

So, 8 – 5 should be less than the third side.
3 should be lesser than the third side.

So, the length of the third side should be something greater than 3.
The natural numbers greater than 3 and lesser than 13 are; 4, 5, 6, 7, 8, 9, 10, 11 and 12.
These are the possible numbers that can be the length of the third side.

Question 2.
The lengths of the sides of a triangle are all natural numbers and two of the sides are 1 centimetre and 99 centimetres. What is the length of the third side?
Answer:
We know that the sum of the lengths of any two sides is greater than the length of the third side.
So, 1 + 99 should be greater than the third side.
That is, 100 is greater than the third side.
So, the length of the third side should be less than 100.
Also, we know that the difference of the lengths of any two sides is smaller than the length of the third side.
So, 99 – 1 should be less than the third side.
98 should be less than the third side.
So, the length of the third side should be something greater than 98.
The natural number greater than 98 and lesser than 100 is 99.
Thus the length of the third side is 99 cm.

Question 3.
Which of the following sets of three lengths can be used to draw a triangle?
(i) 4 centimetres, 6 centimetres, 10 centimetres
(ii) 3 centimetres, 4 centimetres, 5 centimetres
(iii) 10 centimetres, 5 centimetres, 4 centimetres
Answer:
(i) 4 + 6 = 10, which is not greater than the third side. So, it is not possible to draw a triangle with these lengths.
(ii) Here, the sum of any two sides is greater than the third side. So, we can draw a triangle with this set of lengths.
(iii) 5 + 4 = 9, which is not greater than the third side. So, it is not possible to draw a triangle with these lengths.

Question 4.
Draw these pictures:
(i)

Answer:
Steps to draw:

• Draw a square of side length 4 cm. (You can use the compass and the protractor)
• Draw the triangles of length 6 cm, 6 cm and 4 cm on all four sides of the square. (Measure 6 cm on your compass and draw an arc from each end of the sides of the square. join the meeting point of the arc’ to both ends of the sides)
• Join the third vertices of all the triangles.

(ii)

Answer:
Steps to draw:

• Draw a triangle of sides 12 cm, 10 cm, and 8 cm.
• Draw a triangle of sides 6 cm, 5 cm, and 4 cm. (Make sure that the midpoint of the side of length 6 cm is the upper vertex of the first triangle)
• Draw a triangle of sides 3 cm, 2.5 cm, and 2 cm. (Make sure that the midpoint of the side of length 3 cm is the upper vertex of the second triangle)
• Draw a line passing through the left vertices of all the triangles.
• Draw a line passing through the right vertices of all the triangles.

Intext Questions and Answers

Question 1.
Try to draw the pictures below:
(i)

Answer:
Step 1: Draw a triangle with equal side lengths as discussed above. (You are free to choose the side length)

Step 2: Extend the bottom side of the triangle to the right.

Step 3: Draw an equilateral triangle with this new line as the base.

Step 4: Draw the third triangle as shown in the figure.

Step 5: Delete the circles.

(ii)

Step 1: Draw a triangle with equal side lengths as discussed. (You are free to choose the side length)

Step 2: Draw a line of length the same as the side length of the triangle as shown in the figure.

Step 3: Draw an equilateral triangle with this new line as the base.

Step 4: Draw a line of length the same as the side length of the triangle as shown in the figure.

Step 5: Draw an equilateral triangle with this new line as the base.

Step 6: Delete the circles.

Question 2.
How do we draw a triangle with all its sides are equal?
Answer:
Let’s do it with the help of an example. Suppose we want to draw a triangle with all its sides equal to 3 cm. Following are the steps to draw it.
Step 1: Draw a line of length 3 cm

Step 2: Draw a circle with a radius of 3 cm (length of the given line) from the right end of the line.

Step 3: Draw a circle with a radius of 3 cm (length of the given line) from the left end of the. line.

Step 4: Both the circles intersect at two points. Choose any of them as the third vertex and complete the triangle.

Step 5: Delete the circles. Thus, we get the required triangle.

Question 3.
Draw the following triangle using circles.

Answer:
Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

In any triangle, the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle.

Question 4.
Can you draw these pictures by joining more and more small triangles?

Answer:
Draw two equilateral triangles of the same size as shown below.

Join all the outer vertices.

Draw the lines as shown.

Erase the unnecessary lines.

Consider the marked points.

Join the opposite points.

Shade it as shown in the text.

Question 5.
Draw the following figures using the methods we have used in this lesson.

Answer:
(i) Step 1:

Step 2:

Step 3:

Step 4:

(ii) Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

(iii) Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Step 6:

Step 7:

(iv) Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Step 6:

Class 7 Maths Chapter 3 Kerala Syllabus Triangles Questions and Answers

Question 1.
Which one of the following measures can be the sides of a triangle
(a) 3 cm, 4 cm, 7 cm
(b) 6 cm, 7 cm, 14 cm
(c) 4 cm, 5 cm, 10 cm
(d) 4 cm, 5 cm, 8 cm
Answer:
a) 3+4=7. It is not greater than the third side. So, it is not possible to draw such a triangle.
b) 6+7=13. It is not greater than the third side. So, it is not possible to draw such a triangle.
c) 5+49. It is not greater than the third side. So, it is not possible to draw such a triangle.
d) The sum of the lengths of any two sides is greater than the third side. So, it is possible to draw such a triangle.

Question 2.
Zeenath doing a project to find out the measures which can be used to draw a triangle. Which among the following measures can be used to draw a triangle?
(a) 8 cm 4 cm 3 cm
(b) 9 cm 5 cm 12 cm
(c) 5 cm 5 cm 10 cm
(d) 6 cm 7 cm 15 cm
Answer:
a) 4+37. It is not greater than the third side. So, it is not possible to draw such a triangle.
b) The sum of the lengths of any two sides is greater than the third side. So, it is possible to draw such a triangle.
c) 5+5 10. It is not greater than the third side. So, it is not possible to draw such a triangle.
d) 6+7=13. It is not greater than the third side. So, it is not possible to draw such a triangle.

Question 3.
Can a triangle have sides with lengths 6 cm, 5 cm and 9 cm?
Answer:
6 + 5 = 11
6 + 9 = 15
5 + 9 = 14
Here, sum of any two sides is greater than the third side.
Thus, a triangle can have sides with lengths 6 cm, 5 cm and 9 cm.

Question 4.
Is it possible to have a triangle with the following sides?
(i) 3 cm, 4 cm and 7 cm.
(ii) 7 cm, 7 cm and 7 cm.
(iii) 2 cm, 4 cm and 2 cm.
(iv) 3 cm, 5 cm and 7 cm.
Answer:
(i) 3 + 4 = 7, which is not greater than the third side. So, triangle is not possible.
(ii) All sides equal means equilateral triangle. An equilateral triangle can be of any length. So, triangle is possible.
(iii) 2 + 2 = 4, which is not greater than the third side. So, triangle is not possible.
(iv) 3 + 5 = 8
3 + 7 = 10
5 + 7 = 12
Here, sum of any two sides is greater than the third side. So, triangle is possible.

Question 4.
Draw the right triangle whose sides are 4 cm, 12 cm and 10 cm.
Answer:

Class 7 Maths Chapter 3 Notes Kerala Syllabus Triangles

We are already familiar with the concept of triangles. This chapter introduces a new idea called an equilateral triangle. We will also discuss how to draw different triangles using circles.
Following are the main topics discussed in this chapter.

Steps to draw an equilateral triangle using circles

• Draw a line of the given length
• Draw a circle with the same radius from the right end of the line.
• Draw a circle with the same radius from the left end of the line.
• Choose the intersecting point as the third vertex of the triangle

Steps to draw a non-equilateral triangle using circles

• Draw a line of length same as the first side of the triangle.
• Draw a semicircle with the length of the second side as the radius from one end of the line.
• Draw a semicircle with the length of the third side as the radius radius from other end of the line.
• Choose the intersecting point of these two circles as the third vertex of the triangle.

Relation between sides of the triangles
In any triangle;
the sum of the lengths of any two sides greater than the length of the third side.
the difference of the lengths of any two sides is smaller than the length of the third side.

Other ideas
A triangle having all three sides of the same length is called an equilateral triangle.
Sum of all the angles in a triangle is 180°
If only two angles are given we can draw more than one triangle with these.
If the length of two sides and the angle between them is specified, then there is only one triangle with this measure.

Lines And Math
Triangles with all sides equal are called equilateral triangles.
Eg:

In an equilateral triangles, all angles are equal to 60°

How can we draw triangles with unequal sides using circles?
Step 1: Draw a line of length same as the first side of the triangle.
Step 2: Draw a semicircle with the length of the second side as the radius from one end of the first line.
Step 3: Draw a semicircle with the length of the third side as the radius from the other end of the line.
Step 4: Choose the intersecting point of these two semi circles as the third vertex draw a triangle.

Relation between the length of sides of a triangle.
In any triangle;
the sum of the lengths of any two sides is greater than the length of the third side.
the difference of the lengths of any two sides is smaller than the length of the third side.

Angle Math
Sum of all the angles in a triangle is 180°
The total measure of the three angles of a triangle is 180°. This is called the angle sum property of a traingles

We can draw several triangles with two specified angles.
Eg:

To draw a specific triangle, we have to specify not only two angles, but the length of the side on which they stand also.
To draw more than one triangle with the same angles it is enough to draw lines parallel to the sides of the first triangle.
Eg:

Sides And Angles
If we specify the lengths of two sides of a triangle and the angle between them, we have the triangle. In a triangle with angles 30°, 60°, and 90°, the larger side is two times the smaller sides.
Eg:

How do we draw a triangle with all its sides are equal?

• Triangles with all sides equal are called equilateral triangles.
• How do we draw triangles with unequal sides using circles?
• In any triangle, the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle.

Relation between the length of sides of a triangle.

In any triangle;

• the sum of the lengths of any two sides is greater than the length of the third side.
• the difference of the lengths of any two sides is smaller than the length of the third side.
• Sum of all the angles in a triangle is 180°
• We can draw several triangles with two specified angles.
• To draw a specific triangle, we have to specify not only two angles, but the length of the side on which they stand also.
• If we specify the lengths of two sides of a triangle and the angle between them, we have the triangle.
  In a triangle with angles 30°, 60°, and 90o, the larger side is two times the smaller sides.