## NCERT Solutions For Class 6 Maths Chapter 10 Mensuration

Here at MentorAtHome, We provide NCERT Solutions For Class 6 Maths Chapter 10 Mensuration. NCERT solutions help students to improve their performance, and attain better results. In NCERT Solutions Class 6 Maths Chapter 10 there is 3 exercise.

NCERT Solutions For Class 6 Maths Chapter 10 Mensuration, will discuss perameter, area and volume. NCERT Solutions For Class 6 Maths Chapter is important for basic understanding.

### NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercises 10.1

Class 6th Maths Chapter 10 Solutions ex 10.1

Q.1. Find the perimeter of each of the following figures:

Ans:-
(a) Perimeter = Sum of all the sides
= 4 cm + 2 cm + 1 cm + 5 cm = 12 cm
(b) Perimeter = Sum of all the sides
= 40 cm + 35 cm + 23 cm + 35 cm
= 133 cm or 1.33 m
(c) Perimeter = Sum of all the sides
= 15 cm + 15 cm + 15 cm + 15 cm = 15 cm x 4 = 60 cm
(d) Perimeter = Sum of all the sides
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm – 4 cm x 5 = 20 cm
(e) Perimeter = Sum of all the sides
= 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm + 1 cm
= 15 cm
(f) Perimeter = Sum of all the sides
1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm
= 52 cm

Q.2. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all around with tape. What is the length of the tape required?
Ans:-
Total Length of required tape = Perimeter of rectangle lid
= 2 (40 + 10) = 2 (50)
= 2 x 50 = 100 cm

Q.3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Ans:-
Length of table-top = 2 m 25 cm
Breadth of table-top = 1 m 50 cm
Perimeter of the table top = 2 [length + breadth]
= 2 (2.25 + 1.50) = 2 (3.75)
= 2 × 3.75 = 7.5 m

Q.4. What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Ans:-
Length of the strip = 32 cm
Breadth of the strip = 21 cm
Perimeter = 2 [length + breadth]
= 2 (32 + 21) = 2 (53)
= 2 × 53 = 106 cm

Q.5. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Ans:-
Length of the rectangular piece of land = 0.7 km
Breadth of the rectangular piece of land = 0.5 km
Perimeter = 2 [length + breadth]
= 2 (0.7 + 0.5) = 2 (1.2)
= 2 × 1.2 = 2.4 km
Each side is to be fenced with 4 rows
= 4 × 2.4 = 9.6 km

Q.6. Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm, and 5 cm

(b) An equilateral triangle of side 9 cm
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Ans:-
(a) the perimeter of the given triangle
= The sum of all sides of the triangle
= 3 + 4 + 5 = 12 cm
(b) the perimeter of the equilateral triangle
= 3 × side
= 3 x 9 = 27 cm
(c) Perimeter of the isosceles triangle
= Sum of all the sides of the triangle
= (8 + 8 + 6) = 22 cm

Q.7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Ans:-
The perimeter of a triangle
= Sum of all the sides of the triangle
= 10 + 14 + 15 = 39 cm

Q.8. Find the perimeter of a regular hexagon with each side measuring 8 m
Ans:-
The perimeter of a regular hexagon
= 6 x side
= 6 x 8 = 48 m.

Q.9. Find the side of the square whose perimeter is 20 m.
Ans:-
The perimeter of a square = 4 x side
20 = 4 x side
side = 20 ÷ 4 = 5 m
The side of the square is 5 m

Q.10. The perimeter of a regular pentagon is 100 cm. How long is each side?
Ans:-
The perimeter of the regular pentagon = 100 cm
Number of sides in regular pentagon = 5
Length of each side = Perimeter ÷ Number of sides
= 100 ÷ 5 = 20 cm.

Q.11. A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Ans:-
(a) Length of string = 30 cm
Number of equal sides in a square = 4
Length of each side
= 30 ÷ 4 = 7.50 cm.
(b) Length of string = 30 cm
Number of equal sides in equilateral triangle = 3
Length of each side
= 30 ÷ 3 = 10 cm
(c) Length of string = 30 cm
Number of equal sides in regular hexagon = 6
Length of each side
= 30 cm ÷ 6 = 5 cm

Q.12. The two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Ans:-
Let the third side of the triangle = x
Given two sides of triangle 12cm and 14cm
The perimeter of triangle = 36 cm
The perimeter of triangle = Sum of all the sides of the triangle
12 + 14 + x = 36
26 + x = 36
x = 36 – 26
x = 10 cm

Q.13. Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per meter.
Ans:-
Side of square = 250 m
The perimeter of square = 4 × side
= 4 × 250 = 1000 m
Cost of fencing = ₹ 20 per. m
Cost of fencing for 1000 m
= ₹ 20 × 1000 = ₹ 20,000

Q.14. Find the cost of fencing a rectangular park of length 175 cm and breadth 125 m at the rate of ₹ 12 per meter.
Ans:-
Length of the rectangular park = 175 m
Breadth of the rectangular park = 125 m
Perimeter of the park = 2 [length + breadth]
= 2 (175 + 125) = 2 (300)
= 2 × 300 = 600 m
Cost of fencing = ₹ 12 per m
Cost of fencing for 600 m
12 x 600 = ₹ 7200

Q.15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Ans:-
Side of the square park = 75 m
its perimeter = 4 x 75 = 300 m
Perimeter of the rectangular park = 2 [length + breadth]
= 2 [60 + 45 ]
= 2 x 105 = 210 m.
Since 210 m < 300 m.
So, Bulbul covers less distance.

Q.16. What is the perimeter of each of the following figures? What do you infer from the answers?

Ans:-
(a) Perimeter of the square = = 4 × side
= 4 x 25 cm = 100 cm
(c) Perimeter of the rectangle
= 2 [40 cm + 10 cm] = 2 x 50 cm = 100 cm
(b) Perimeter of the rectangle
= 2 [30 cm + 20 cm] = 2 x 50 cm = 100 cm
(d) Perimeter of the triangle = Sum of all sides
= 30 cm + 30 cm + 40 cm = 100 cm
All the figures have same perimeter.

17. Avneet buys 9 square paving slabs, each with a side of 1 / 2 m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [fig 10.7(i)]?

(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig 10.7 (ii)]?
(c) Which has a greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e they cannot be broken.)
Ans:-
(a) Side of square = 3 × side
= 3 × 1 / 2 = 3 / 2 m
Perimeter of Square = 4 × 3 / 2
= 2 × 3 = 6 m
(b) Perimeter = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 = 10 m
(c) The arrangement in the form of the cross has a greater perimeter
(d) Perimeters greater than 10 m cannot be determined.

### NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercises 10.2

NCERT Class 6 Maths Chapter 10 Solutions ex 10.2

Q.1. Find the areas of the following figures by counting squares:

Ans:-
(a) Number of full squares = 9
Area of 1 square = 1
∴ Area of 9 squares = 9 x 1 = 9 sq. units.
(b) Number of full squares = 5
Area of 1 square = 1
∴ Area of the figure = 5 x 1 = 5 sq. units
(c) Number of full squares = 2
Number of half squares = 4
∴ Area of the covered figure = 2 x 1 + ( 4 x 1 / 2)
= 2 + 2 = 4 sq. units
(d) Number of full squares = 8
Area of 1 square = 1
∴ Area of the covered figure = 8 x 1 = 8 sq. units.
(e) Number of full squares = 10
Area of 1 square = 1
Area covered by the figure = 10 x 1 = 10 sq. units.
(f) Number of full squares = 2
Number of half squares = 4
∴ Area of the covered figure = 2 x 1 + (4 x 1 / 2)
= 2 + 2 = 4 sq. units.
(g) Number of full squares = 4
Number of half squares = 4
∴ Area of the covered figure = 4 x 1 + (4 x 1 / 2)
= 4 + 2 = 6 sq. units.
(h) Number of full squares = 5
Area of 1 square = 1
∴ Area of the covered figure = 5 x 1 = 5 sq. units.
(i) Number of full squares = 9
Area of 1 square = 1
∴ Area of the covered figure = 9 x 1 = 9 sq. units.
(j) Number of full squares = 2
Number of half squares = 4
∴ Area of the covered figure = 2 x 1 + (4 x 1 / 2)
= 2 + 2 = 4 sq. units.
(k) Number of full squares = 4
Number of half squares = 2
∴ Area of the covered figure = 4 x 1 + (2 x 1 / 2)
= 4 + 1 = 5 sq. units
(l) Number of full squares = 4
Number of squares more than half = 3
Number of half squares = 2
∴ Area of the covered figure = 4 x 1 + 3 x 1 + (2 x 1 / 2)
= (4 + 3 + 1) = 8 sq. units.
(m) Number of full squares = 6
Number of more than half squares = 8
Area of the covered figure = (6 x 1 + 8 x 1)
= 6 + 8 = 14 sq. units.
(n) Number of full squares = 9
Number of more than half squares = 9
∴ Area of the covered figure= (9 x 1 + 9 x 1)
= 9 + 9 = 18 sq. units.

### NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercises 10.3

Ch 10 Class 6 Maths Solutions ex 10.3

Q.1. Find the area of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm
Ans:-
We know that Area of rectangle = Length × Breadth
(a) Length of the rectangle = 3 cm
Breadth of the rectangle = 4 cm
= 3 cm x 4 cm = 12 cm2
(b) Length of the rectangle = 12 m
breadth of the rectangle = 21 m
= 12 m x 21 m = 252 m2
(c) Length of the rectangle = 2 km
breadth of the rectangle 3 km
= 2 km x 3 km = 6 km2
(d) Length of the rectangle = 2 m
breadth of the rectangle = 70 cm or 0.70 m
= 2 m x 0.70 m = 1.40 m2

Q.2. Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
Ans:-
We know that Area of the square = Side x Side
(a) 10 cm
= 10 cm x 10 cm = 100 cm2
(b) 14 cm
= 14 cm x 14 cm = 196 cm2
(c) 5 m
= 5 m x 5 m = 25 m2

Q.3. The length and breadth of the three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?
Ans:-
We know that Area of rectangle = Length × Breadth
(a) Length of the rectangle = 9 m
breadth of the rectangle = 6 m
= 9 m x 6 m = 54 m2
(b) Length of the rectangle = 17 m
breadth of the rectangle = 3m
= 17 m x 3 m 51 m2
(c) Length of the rectangle = 4 m
breadth of the rectangle = 14 m
= 4 m x 14 m = 56 m2
Rectangle (c) has the largest area, i.e., 56 m2  and (b) has the smallest area, i.e., 51m2

Q.4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Ans:-
Length of the rectangular garden = 50 m
Area of the rectangular garden = 300 m2
Area of rectangle = length × width
300 = 50 × width
width = 300 / 50
width = 6 m

Q.5. What are the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m.?
Ans:-
Length of the rectangular plot = 500 m
and the breadth = 200 m
Area of the plot = length x breadth
= 500 m x 200 m = 100000 sq m
Now rate of tiling the plot = ₹8 per 100 sq m
= (8 × 1,00,000) / 100 = ₹ 8000
Hence the required cost = ₹8000

Q.6. A table-top measures 2 m by 1 m 50 cm. What is its area in square meters?
Ans:-
Length of the table-top = 2 m
breadth of the table-top = 1 m 50 cm or 1.50 m
Area of the table-top = length x breadth
= 2 m x 1.50 m = 3 m2

Q.7. A room is 4 m long and 3 m 50 cm wide. Howe, many square meters of carpet is needed to cover the floor of the room?
Ans:-
Length of the room = 4 m
breadth of the room = 3 m 50 cm = 3.5 m
Area of the room = length x breadth
= 4 m x 3.5 m = 14 m2

Q.8. A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Ans:-
Length of the floor = 5 m
breadth of the floor = 4 m
Area of the floor = length x breadth
= 5m x 4m = 20 m2
Side of the carpet = 3 m
Area of the square carpet = side x side
= 3 m x 3 m = 9 m2
Area of the floor which is not carpeted
= 20 m2  – 9 m2 = 11 m2

Q.9. Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Ans:-
Side of the square flower bed = 1 m.
Area of the flower bed = side x side
Area of 1 square flower bed = 1 m x 1 m = 1 m2
Area of 5 square flower beds = 1 m x 5 m = 5 m2
Now length of the land = 5 m
breadth of the land = 4 m
Area of the land = length x breadth
=5m x 4m = 20 m2
Area of the remaining part of the land
= 20 m2 – 5 m2 = 15 m2

Q.10. By splitting the following figures into rectangles, find their areas (The measures are given in centimeters).

Ans:-
(A)

Area of the rectangle I = length x breadth
= 4 cm x 3 cm = 12 cm2
Area of the rectangle II = length x breadth
= 3 cm x 2 cm = 6 cm2
Area of the rectangle III = length x breadth
= 4 cm x 1 cm = 4 cm2
Area of the rectangle IV = length x breadth
= 3 cm x 2 cm = 6 cm2
Total area of the given figure
= 12 cm2 + 6 cm2 + 4 cm2 + 6 cm2 = 28 cm2
(b)

Area of the rectangle I length x breadth
= 12 cm x 2 cm = 24 cm2
Area of the rectangle II length x breadth
= 8 cm x 2 cm = 16 cm2
Area of rectangle III length x breadth
= 3 cm x 1 cm = 3 cm2
∴ Total area of the given figure
= 3 cm2 + 3 cm2 + 3 cm2 = 9 cm2

Q.11. Split the following shapes into rectangles and find their areas. (The measures are given in centimeters)

Ans:-
(a) Splitting the given figure into the rectangles I and II, we get

Area of the rectangle I
= 12 cm x 2 cm = 24 cm2
Area of the rectangle II
= 8 cm x 2 cm = 16 cm2
∴ Total area of the figure
= 24 cm2 + 16 cm2 = 40 cm2
(b) Splitting the given figure into the rectangles I, II and III, we get

Area of the rectangle I
= 7 cm x 7 cm = 49 cm2
Area of the rectangle II
= 21 cm x 7 cm = 147 cm2
Area of the rectangle III
= 7 cm x 7 cm = 49 cm2
∴ Total area of the figure
= 49 cm2 + 147 cm2 + 49 cm2 = 245 cm2
(c) Splitting the given figure into the rectangles I, and II, we get

Area of the rectangle I
= 5 cm x 1 cm = 5 cm2
Area of the rectangle II
= 4 cm x 1 cm = 4 cm2
∴ Total area of the figure
= 5 cm2 + 4 cm2 = 9 cm2

Q.12. How many tiles whose length and breadth are 12 cm and 5 cm, respectively will be needed to fit in a rectangular region whose length and breadth are respectively?
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm
Ans:-
Length of one tile = 12 cm
Breadth of one tile = 5 cm
Area of 1 tile = length x breadth
= 12 cm x 5 cm = 60 cm2

(a) Length of the region = 144 cm
Breadth of the region = 100 cm
Area of the region = length x breadth
= 144 cm x 100 cm = 14400 cm2
∴ Number of tiles = (Area of rectangle) / (Area of one tile)
= 14400 cm2 ÷ 60 cm2 = 240 tiles

(b) Length of the region = 70 cm
Breadth of the region = 36 cm
Area of the rectangular region = length x breadth
= 70 cm x 36 cm = 2520 cm2
∴ Number of tiles = (Area of rectangle) / (Area of one tile)
= 2520 cm2 ÷ 60 cm2 = 42 tiles.

NCERT Solutions For Class 6 Maths Chapter 10 mensuration are based on NCERT Books. You can download ncert book for class 6 maths chapter 10. Ch 10 class 6 maths solutions are designed by our subject expert team.

## What We Learn?

1. Perimeter is the distance covered along the boundary forming a closed figure
when you go round the figure once.
2. (a) Perimeter of a rectangle = 2 × (length + breadth)
(b) Perimeter of a square = 4 × length of its side
(c) Perimeter of an equilateral triangle = 3 × length of a side
3. Figures in which all sides and angles are equal are called regular closed figures.
4. The amount of surface enclosed by a closed figure is called its area.
5. To calculate the area of a figure using a squared paper, the following conventions