Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 11 Letter Math Questions and Answers Notes Pdf to clear their doubts.
SCERT Class 6 Maths Chapter 11 Solutions Letter Math
Class 6 Kerala Syllabus Maths Solutions Chapter 11 Letter Math Questions and Answers
Letter Math Class 6 Questions and Answers Kerala Syllabus
Addition and Subtraction
Intext Questions (Page No. 162-163)
Question 1.
Now consider the polygon given in the textbook.
Do you see any relation between the number of sides of a polygon, the number of corners, the number of lines, and the number of triangles?
Answer:
First, we check the relation between the number of sides and the number of lines.
The number of lines is 3 less than the number of corners.
That is, Number of lines = Number of corners – 3
In any polygon, the number of corners is the same as the number of sides.
That means, Number of lines = Number of sides – 3
Let s be the number of sides of the polygon and l be the number of lines of the polygon.
Then we can write the relation as l = s – 3
We can say this in another two forms, that is, s = l + 3
s – l = 3
(If we add 3 to the number of lines, we get the number of sides.
The difference between the number of sides and the number of lines will results into 3.)
Now, let’s look at the relation between the number of sides of a polygon and the triangles formed within it.
The number of triangles is 2 less than the number of sides.
Let t be the number of triangles and s be the number of sides.
Then the relation between them can be written as follows.
s – t = 2
t = s – 2
s = t + 3
What is the relation between the number of triangles and the number of lines within them?
If we draw 1 lines we get 2 triangles.
If we draw 2 lines, we get 3 triangles.
That means the number of triangles is 1 more than the number of lines.
This relation can be expressed using letters in different forms, such as
l + 1 = t,
t – 1 = l,
t – l = 1
The last two relations can be explained as follows.
The number of lines is 1 less than the number of triangles.
The difference between the number of triangles and the number of lines is 1.
Addition and Subtraction (Page No. 165)
Question 1.
The pictures below show a line drawn from a point on another line:
(i) In what ways can we say the relation between the angles marked on the left and right of the top line?
(ii) If we denote the measure of the angle on the left by l and the measure of the angle on the right by r, then in what ways can we write the relation between the numbers l and r?
(iii) In all cases, l and r are between what numbers?
Answer:
(i) The sum of the angles on the left and right side of the slant line is 180°.
The angle on the right side can be obtained by subtracting the angle on the left side from 180°.
The angle on the left side can be obtained by subtracting the angle on the right side from 180°.
(ii) Angle on the left = l
Angle on the right = r
Here, l + r = 180
180 – l = r
180 – r = l
(iii) In all cases, the values of l & r are between 0° and 180°.
Question 2.
(i) Each side of a square of perimeter 40 centimetres is extended by 1 centimetre to make a larger square. By how much is the perimeter increased?
(ii) If the sides of a square of perimeter 60 centimetres are extended like this, what would be the perimeter of the large square?
(iii) Is the difference in perimeters of any square, and the larger square made by extending each side by 1 centimetre, the same number for all squares? Why?
(iv) Is it also true for rectangles that are not squares?
(v) If the perimeter of the larger rectangle is denoted by p and the perimeter of the original small rectangle is denoted by s, in what ways can we write the relation between these two numbers?
Answer:
(i) If the side is extended by 1 cm, then its perimeter also increases by 4 cm.
Perimeter = 40 cm
Perimeter of the larger square = 44 cm
Increase in the perimeter = 44 – 40 = 4
(ii) Perimeter = 60 cm
Each side is extended by 1 cm, then its perimeter is increased by 4 cm.
Perimeter of the larger square = 64 cm
(iii) Extending each side by 1 cm increases each of the 4 sides by 1 cm, so the perimeter always increases by 4 cm.
(iv) Yes. The perimeter increase is also 4cm for any rectangle when each side is increased by 1 cm.
(v) p = s + 4
s = p – 4
p – s = 4
Letter Multiplication
Intext Questions (Page No. 167)
Question 1.
Now, suppose we make triangles like this?
Write the number of triangles and the number of matchsticks in a table like this:
Answer:
In this problem, the relation between the number of triangles and the number of matchsticks is different.
First, we can complete the table:
From this, we can see that the relation between the triangles and the number of matchsticks is like this.
If one is added to twice the number of triangles, we get the number of matchsticks required to make the triangle.
This relation can be written in this form.
m = 2t + 1
t = \(\frac{m-1}{2}\)
\(\frac{m-1}{t}\) = 2
Difference in the number of matchsticks:
The number of matchsticks used to make 3 triangles in this way is 3 × 3 = 9.
If the 3 triangles are made in this form, the number of matchsticks used is 3 × 2 + 1 = 7.
The difference in the number of triangles in two different ways is 1.
The difference remains the same for any number of triangles.
4 triangles:
Method-1: 4 × 3 = 12
Method-1: 4 × 2 + 1 = 9
Difference = 3
5 triangles:
Method-1: 5 × 3 = 15
Method-1: 5 × 2 + 1 = 11
Difference = 4
Therefore, the relation is that if the triangles are made in two different ways, then the number of matchsticks needed is one less than the number of triangles.
Let’s denote the number of triangles by t and the difference in the number of matchsticks needed in the two ways by d.
That means we get d = t – 1
If 10 triangles are made in these two different ways, then
Difference in the number of matchsticks = 10 – 1 = 9
If it is 100, then the difference in the number of matchsticks = 100 – 1 = 99
Letter Multiplication (Page No. 172-174)
Question 1.
We can also make squares with matchsticks:
(i) As we go on making such squares, what is the relation between the number of squares and the number of matchsticks used?
(ii) Write this relation, denoting the number of squares by 5 and the number of matchsticks by m.
(iii) What kind of numbers are s and m in this?
(iv) For a multiple of 4, how do we calculate the number of squares that can be made with that many matchsticks?
(v) Write this calculation also using letters.
Answer:
(i)
The number of matchsticks is 4 times the number of squares.
(ii) The number of squares is denoted by s
Number of matchsticks is m = 4s
(iii) Both s and m are whole numbers.
m is always an even number, since each square requires 4 matchsticks.
m is always a multiple of 4.
(iv) For getting the number of squares, divide the number of matchsticks used by 4.
(v) If the number of squares is denoted by s
The number of matchsticks is m
s = \(\frac {m}{4}\)
Question 2.
We can make matchstick squares like this also:
(i) Write as a table the number of squares and the number of matchsticks needed to make squares like this, and find the relation between these numbers.
(ii) Explain this relation by drawing pictures showing how the squares are made.
(iii) Write this relation denoting the number of squares by s and the number of matchsticks used by m.
(iv) What kind of numbers are s and m in this?
(v) If we take a number that leaves a remainder of 1 on division by 3, how do we calculate the number of squares that can be made like this using that many matchsticks?
(vi) Write this calculation also using letters.
Ans:
(i)
(ii)
For the first square, 1 matchstick is joined with 3 matchsticks, we get 4 matchsticks
For two squares, join 3 and 1 with 3 matchsticks, we get 7 matchsticks.
For three squares, join two times 3 and 1 with 3 matchsticks, we get 10 matchsticks (That means 3 times 3 and 1).
Continuing like this.
For ten squares, join nine times 3 and 1 with 3 matchsticks, we get 30 matchsticks (that means, 10 times 3 and 1).
In general, we can say that 1 is added to 3 times the number of squares.
(iii) Number of squares be s
The number of matchsticks is m
Therefore, the relation is m = 3s + 1
(iv) Both s and m are natural numbers.
The value of m is 3, ’s are added with 4.
That means the value is the number that is divisible by 3, and the remainder is 1.
(v) If some matchsticks m leaves a remainder of 1 when divided by 3, then we can make squares like this.
That means (Number of matchsticks – 1)
(vi) s = \(\frac{m-1}{3}\)
Question 3.
(i) When an equal number of matchstick squares are made in the two different ways as in the first and second questions, what is the difference in the number of matchsticks used?
(ii) Denoting the number of squares as s and the difference in the number of matchsticks used in the two ways as d, how do we write the relation between s and d?
(iii) Explain the reason for this relation.
(iv) One kid uses 25 matchsticks to make squares in the second way. How many matchsticks would be needed to make the same number of squares first?
Answer:
(i) For making the squares in the second method, the value obtained by subtracting 1 from the number of squares in the first method can be removed from the number of matchsticks used in the first method.
To make 5 squares, subtract 4 from 20, we get 16 (20 – 4 = 16).
Matchsticks are enough for making it in the second method.
(ii) s – d = 1
(iii) In the first method, number of matchsticks is 4 times the number of squares.
In the second method, 1 is added to 3 times the number of squares.
That means, in the first method, it is in the form of: 4 + 4 + 4 + 4 + ….
In the second method, it is in the form of: 4 + 3 + 3 + 3 + …
From the second square onwards, 1 is decreased from it.
So the difference in the matchsticks is 1 less than the total number of squares.
(iv) The number of matchsticks used for making 25 squares in the second method is 25 × 3 + 1 = 76
The number of matchsticks used for making 25 squares in the first method is 25 × 4 = 100
Question 4.
See these pictures:
The first picture is a square of sides 1 centimetre. The second picture is a rectangle made by joining together two such squares. The third picture is a rectangle made by joining together three such squares.
(i) Calculate the perimeter of each.
(ii) What is the perimeter of the rectangle made by joining together four such squares like this? What about the rectangle made with five squares?
(iii) Make a table of the number of squares used to make the rectangles and the perimeter of the rectangles, starting from the first. What is the relation between the number of squares and the perimeter? Explain the reason for this relation?
(iv) Denoting the number of squares by s and the perimeter by p, write the relation between them.
(v) To get a rectangle of perimeter 1 metre, how many squares must be joined like this?
Ans:
(i) (a) 4 cm
(b) 6 cm
(c) 8 cm
(ii) 10 cm, 12 cm
(iii)
The perimeter of the square is obtained by adding 2 to twice the number of squares.
When we join squares side by side to make a rectangle, the two lengths of the rectangle increase by 2 (1 + 1 = 2)
But the width remains the same at 1 cm.
The length of the square is 1 cm.
By considering the 2 sides, we get twice the squares, and also it increases by 2 when considering the two widths of 2 cm.
Thus, we get the perimeter of the rectangle.
(iv) p = 2s + 2
(v) From this relation,
Number of squares = (perimeter – 2) ÷ 2
That is s = \(\frac{p-2}{2}\)
⇒ s = \(\frac{100-2}{2}=\frac{98}{2}\)
⇒ s = 49
49 Squares must be joined.
Class 6 Maths Chapter 11 Kerala Syllabus Letter Math Questions and Answers
Class 6 Maths Letter Math Questions and Answers
Question 1.
(i) Find the number of matchsticks used to make this hexagon.
(ii) If the hexagons are joined like this, then the number of matchsticks used to make this is.
(iii) If 10 such patterns are joined together. How many matchsticks are used to make this?
(iv) How can we say this relation? Explain this with letters.
(v) If there is 48 sticks joined together, then how many patterns are there?
(vi) How can we say this relation? Explain this with letters.
(vii) How can we explain the relation between the hexagon and the number of matchsticks used? Express it in letters also.
Answer:
(i) 6
(ii) 12
(iii) 10 × 6 = 60
(iv) The number of matchsticks used is 6 times the number of hexagons.
Let h be the number of hexagons.
m be the number of matchsticks
m = 6h
(v) 48 ÷ 6 = 8
(vi) If the number of matchsticks used is divided by 6, we get the number of hexagons.
h = \(\frac {m}{6}\)
(vii) If the number of matchsticks is divided bythe number of hexagons, we get 6.
\(\frac {m}{h}\) = 6
Question 2.
If the hexagon is arranged in this way.
(i) To form a pattern of 3 hexagons joined together, how many matchsticks are used?
(ii) If it is 10, how many matchsticks are there?
(iii) How can we say these relations? Express it in letters.
Answer:
(i) 16
(ii) 51
(iii) Let the number of hexagons be h
Number of matchsticks is m
m = 5h + 1
h = \(\frac{n-1}{5}\)
\(\frac{n-1}{h}\) = 5
Question 3.
Some portions of a circle is shaded.
(i) The angle measure of the shaded portion is 40°. What is the angle measure of the remaining portion?
(ii) If the shaded portion is 120°, what is the remaining measure?
(iii) How can we express the relation between the angles of the shaded portion and the unshaded portion?
(iv) Explain it in letters.
Ans:
(i) 360° – 40° = 320°
(ii) 360° – 120° = 240°
(iii) If the angle measure of the shaded portion is subtracted from 360°, we get the angle measure of the unshaded portion.
If the angle measure of the unshaded portion is subtracted from 360°, we get the angle measure of the shaded portion.
If we take the sum of the angle measures of both the shaded and unshaded portions, we get 360°.
(iv) If the shaded portion is s
Unshaded portion is u, we get,
360 – s = u
360 – u = s
s + u = 360
Question 4.
The image of a pentagon is given below.
(i) How many matchsticks are needed to make this pattern?
(ii) If 3 pentagons are joined together like this.
How many matchsticks are needed to make this pattern?
(iii) If the number of pentagons is 8, how many matchsticks are used?
(iv) How can we explain the relation between the number of pentagons and the number of sticks used?
(v) Express this relation using letters.
Answer:
(i) 5
(ii) 4 + 4 + 4 + 1 = 13
(iii) 8 × 4 + 1 = 33
(iv) If the number of pentagons is multiplied by 4 and 1 is added, we get the number of matchsticks used.
(v) Let the number of pentagons be p, and the number of matchsticks be n.
n = 4p + 1
p = \(\frac{n-1}{4}\)
\(\frac{n-1}{p}\) = 4
Class 6 Maths Chapter 11 Notes Kerala Syllabus Letter Math
→ If three points are on the same straight line, then the distance between the points at the two ends is equal to the sum of the distances from the end points to the point in between.
→ The triangles are formed by arranging the matchsticks in a triangular shape.
→ The number of matchsticks is equal to the number of triangles multiplied by 3.
m = 3t
→ To get the number of triangles, divide the number of matchsticks by 3.
t = \(\frac {m}{3}\)
→ If we divide the number of matchsticks by the number of triangles, we get 3.
\(\frac {m}{t}\) = 3
In the previous chapter, we discussed some of the unchanging relationships. In this chapter, we are going to discuss the unchanging relationships that can be expressed algebraically. This unit introduces the basic ideas needed to represent such relationships using letters.
Addition and Subtraction
Three points are marked on a straight line. Two points are at the ends of the line, and one point lies between them. If the total length of this line is 8 cm and the distance from the left end to the point that is between the end points are 3 cm. Then the line is divided into two parts, that is, 3 cm and 5 cm. The length of the line at the left end is 3 cm, and at the right end is 5 cm.
By adding the left and right ends, we get 8 cm. Or subtracting 5 cm (that means the right end) from 8 cm, we get 3 cm (that is the length of the left end). This can be described like this:
If the length of the left end is subtracted from 8 cm, we get the length of the right end.
Similarly, if we subtract the length of the right end from 8 cm, we get the length of the left end.
This relationship is always true, even when the position of the point that is between the endpoints is changed.
These relations can be concluded like this:
Distance from the left end = 8 – Distance from the right end
Distance from the right end = 8 – Distance from the left end.
Distance from the left end + Distance from the right end = 8 cm
We can rewrite the above mentioned relations with letters.
If we consider the distance from the left end as l and the distance from the right end as r.
Then we can write the above-mentioned relations like this.
l = 8 – r
r = 8 – l
l + r = 8
In this relation, the value of l and r can be any value, but the sum of the two values must be 8.
Now consider the line problem from the textbook.
In a 5 cm long line, mark a point 1 cm from the right. At what distance from the left end should we mark it?
5 – 1 = 4 cm
Subtracting the distance from the right end from 5 centimetres gives the distance from the left end.
5 – Distance from the right end = Distance from the left end
Let’s see how can we represent the above relation using letters,
Consider the distance from the right end = r
Distance from the left end = l
Then we get, 5 – r = l
Depends on the position of the point, the value of l and r changes, but the relationship between them remains the same.
We can say it in another way:
Distance from the right end + Distance from the left end = 5
r + l = 5
5 – Distance from the left end = Distance from the right end.
5 – l = r
In general, we can say that,
If three points are on the same straight line, then the distance between the points at the two ends is equal to the sum of the distances from the endpoints to the point in between.
For example, consider the line and the points given in it.
Total length of the line is 10 centimetres. A point is marked 4 centimetres from its right end. Then the point is 6 centimetres from its left end.
4 + 6 = 10 cm
10 – 4 = 6 cm
10 – 6 = 4 cm
If we write this using the letters we get,
Let the total length be d, we get the following relations.
d = r + l
r = d – l
l = d – r
In this, we can take l, r, and d as any three numbers, but the sum of l and r should be d.
Letter Multiplication
Let’s consider the triangular problem given in the textbook.
Here, the triangles are formed by arranging the matchsticks in a triangular shape.
Therefore, the number of matchsticks is equal to the number of triangles multiplied by 3.
Shortening this statement, using letters:
Let the number of triangles be t, and the number of matchsticks be m.
Then the relation can be written as, m = t × 3
It can also be written as m = 3t
Similarly, we can say the other relations also.
To get the number of triangles, divide the number of matchsticks by 3.
That is, t = \(\frac {m}{3}\)
If we divide the number of matchsticks by the number of triangles, we get 3.
That is \(\frac {m}{t}\)= 3