# NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Written by Team Trustudies
Updated at 2021-02-11

## NCERT solutions for class 6 Maths Chapter 10 Mensuration Exercise 10.1

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

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

(a) Perimeter = Sum of all the sides
= 1 + 2 + 4 + 5
= 12 cm
(b) Perimeter = Sum of all the sides
= 23 + 35 + 35 + 40
= 133 cm
(c) Perimeter = Sum of all the sides
= 15 + 15 + 15 + 15
= 60 cm
(d) Perimeter = Sum of all the sides
= 4 + 4 + 4 + 4 + 4
=20 cm
(e) Perimeter = Sum of all the sides
= 1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4
= 15 cm
(f) Perimeter = Sum of all the sides
= 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3
= 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?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Length of required tape = Perimeter of rectangle
= 2 (40 + 10)
= 2 × (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?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Length of table top = 2 m 25 cm = 2.25 m
Breadth of table top = 1 m 50 cm = 1.50 m
Perimeter of 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?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Required length of wooden strip = Perimeter of photograph
= 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?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Perimeter of the field = 2 (Length + Breadth)
= 2 (0.7 + 0.5)
= 2 (1.2)
= 2 × 1.2
= 2.4 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.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

(a) Perimeter of triangle = 3 + 4 + 5
= 12 cm
(b) Perimeter of an equilateral triangle = 3 × side
= 3 × 9
= 27 cm
(c) Perimeter of isosceles triangle = 8 + 8 + 6
= 22 cm

Q.7 Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Perimeter of triangle = 10 + 14 + 15
= 39 cm

Q.8 Find the perimeter of a regular hexagon with each side measuring 8 m.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Perimeter of hexagon = 6 × 8 = 48 m

Q.9 Find the side of the square whose perimeter is 20 m.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Perimeter of square = 4 × side
20 = 4 × side
Side =$\frac{20}{4}$
Side = 5 m

Q.10 The perimeter of a regular pentagon is 100 cm. How long is its each side?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Perimeter of regular pentagon = 100 cm
5 × side = 100 cm
Side = $\frac{100}{5}$
Side = 20 cm

Q.11 A piece of strings 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?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

(a) Perimeter of square = 30 cm
4 × side = 30
Side = $\frac{30}{4}$
Side = 7.5 cm
(b) Perimeter of an equilateral triangle = 30 cm
3 × side = 30
Side = $\frac{30}{10}$
Side = 10 cm
(c) Perimeter of a regular hexagon = 30 cm
6 × side = 30
Side = $\frac{30}{6}$
Side = 5 cm

Q.12 Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Let a cm be the third side
Perimeter of triangle = 36 cm
12 + 14 + a = 36
26 + a = 36
a = 36 – 26
a = 10 cm

Q.13 Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Side of square = 250 m
Perimeter of square = 4 × side
= 4 × 250
= 1000 m
Cost of fencing = Rs 20 per m
Cost of fencing for 1000 m = Rs 20 × 1000
= Rs 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 metre.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Length = 175 cm
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (175 + 125)
= 2 × (300)
= 2 × 300
= 600 m
Cost of fencing = 12 × 600 = Rs 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?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Perimeter of square = 4 × side
= 4 × 75
= 300 m
$?$Distance covered by Sweety is 300 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (60 + 45)
= 2 × 105
= 210 m
Hence, Bulbul covers less distance than Sweety.

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

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

(a) Perimeter of square = 4 × side
= 4 × 25
= 100 cm
(b) Perimeter of rectangle = 2 (40 + 10)
= 2 × 50
= 100 cm
(c) Perimeter of rectangle = 2 (Length + Breadth)
= 2 (30 + 20)
= 2 (50)
= 2 × 50
= 100 cm
(d) Perimeter of triangle = 30 + 30 + 40
= 100 cm
All the figures have same perimeter.

Q.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 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.)

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

(a) Side of square = 3 × side
= 3 × $\frac{1}{2}$
= 3 / 2 m
Perimeter of Square = 4 × $\frac{3}{2}$
= 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 cross has greater perimeter
(d) Perimeters greater than 10 m cannot be determined.

## NCERT solutions for class 6 Maths Chapter 10 Mensuration Exercise 10.2

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

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

(a) Number of full squares = 9
Area of 1 square = 1 sq unit
$?$Area of 9 squares = 9 x 1 sq unit
= 9 sq units.
So, the area of the portion covered by 9 squares = 9 sq units
(b) Number of full squares = 5
$?$ Area of the figure = 5 x 1 sq unit = 5 sq units
(c) Number of full squares = 2
Number of half squares = 4
$?$ Area of the covered figure
= 2 x 1 + 4 x $\frac{1}{2}$ = 2 + 2 = 4 sq units
(d) Number of full squares = 8
$?$ Area of the covered portion of the figure
= 8 x 1 sq unit = 8 sq units.

(e) Number of full squares = 10
Area covered by the figure = 10 x 1 sq unit = 10 sq units.
(f) Number of full squares = 2
Number of half squares = 4
$?$ Area of the covered figure
= (2 x 1 + 4 x $\frac{1}{2}$)= (2 + 2) sq units = 4 sq units.
(g) Number of full squares = 4
Number of half squares = 4
$?$ Area of the covered figure = (4 x 1 + 4 x 12)
= (4 + 2) sq units = 6 sq units.

(h) Number of full squares = 5
$?$ Area of the covered figure
= 5 x 1 sq unit = 5 sq units.

(i) Number of full squares = 9
$?$ Area of the covered figure
= 9 x 1 sq units = 9 sq units.

(j) Number of full squares = 2
Number of half squares = 4
$?$ Area of the covered figure
=(2 x 1 + 4 x $\frac{1}{2}$) sq units
= (2 + 2) sq units = 4 sq units.
(k) Number of full squares = 4
Number of half squares = 2
$?$ Area of the covered figure
= (4 x 1 + 2 x $\frac{1}{2}$)sq units
= (4 + 1)sq units = 5sq 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 $\frac{1}{2}$ sq units
= (4 + 3 + 1) sq units = 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) sq units
= (6 + 8) sq units = 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) sq units = (9 + 9) sq units = 18 sq units.

## NCERT solutions for class 6 Maths Chapter 10 Mensuration Exercise 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

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Area of rectangle = Length × Breadth
(a) l = 3 cm and b = 4 cm
Area = l × b = 3 × 4 = 12 $c{m}^{2}$
(b) l = 12 m and b = 21 m
Area = l × b = 12 × 21
= 252 ${m}^{2}$
(c) l = 2 km and b = 3 km
Area = l × b = 2 × 3
= 6 $k{m}^{2}$
(d) l = 2 m and b = 70 cm = 0.70 m
Area = l × b = 2 × 0.70
= 1.40 ${m}^{2}$

Q.2 Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

(a) Area of square = $sid{e}^{2}$
= ${10}^{2}=100c{m}^{2}$
(b) Area of square =$sid{e}^{2}$ $={14}^{2}=196c{m}^{2}$
(c) Area of square = $sid{e}^{2}$
$={5}^{2}=25c{m}^{2}$

Q.3 The length and breadth of 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?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

(a) Area of rectangle = l × b
$=9×6=54{m}^{2}$
(b) Area of rectangle = l × b
$=17×3=51{m}^{2}$
(c) Area of rectangle = l × b
$=4×14=56{m}^{2}$
Area of rectangle 56 ${m}^{2}$ i.e (c) is the largest area and area of rectangle 51 ${m}^{2}$ i.e (b) is the smallest area

Q.4 The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Area of rectangle = length × width
300 = 50 × width
width = 300 / 50
width = 6 m

Q.5 What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs 8 per hundred sq m.?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Area of land = length × breadth
$=500×200=1,00,000{m}^{2}$
Cost of tiling 1,00,000 sq m of land = (8 × 1,00,000) / 100
= Rs 8000

Q.6 A table top measures 2 m by 1 m 50 cm. What is its area in square metres?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

l = 2m
b = 1m 50 cm = 1.50 m
Area = l × b = 2 × 1.50
$=3{m}^{2}$

Q.7 A room is 4 m long and 3 m 50 cm wide. Howe many square metres of carpet is needed to cover the floor of the room?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

l = 4m
b = 3 m 50 cm = 3.50 m
Area = l × b = 4 × 3.50
=14 ${m}^{2}$

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.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Area of floor = l × b = 5 × 4
$=20{m}^{2}$
Area of square carpet = 3 × 3
$=9{m}^{2}$
Area of floor that is not carpeted = 20 – 9
$=11{m}^{2}$

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?

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Area of flower square bed = 1 × 1
$=1{m}^{2}$
Area of 5 square bed = 1 × 5
$=5{m}^{2}$
Area of land = 5 × 4
$=20{m}^{2}$
Remaining part of the land = Area of land – Area of 5 square bed
$=20–5=15{m}^{2}$

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

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

(a) Area of HKLM = 3 x 3 =$9c{m}^{2}$
Area of IJGH = 1 x 2 = 2 $c{m}^{2}$
Area of FEDG = 3 x 3 = 9 $c{m}^{2}$
Area of ABCD = 2 x 4 = 8 $c{m}^{2}$
Total area of the figure = 9 + 2 + 9 + 8 = 28 $c{m}^{2}$
(b) Area of ABCD = 3 x 1 = 3 $c{m}^{2}$
Area of BDEF = 3 x 1 = 3 $c{m}^{2}$
Area of FGHI = 3 x 1 = 3 $c{m}^{2}$
Total area of the figure = 3 + 3 + 3 = 9 $c{m}^{2}$

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

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

(a) Area of rectangle ABCD = 2 x 10 = 20 $c{m}^{2}$
Area of rectangle DEFG = 10 x 2 = 20 $c{m}^{2}$
Total area of the figure = 20 + 20 = 40 $c{m}^{2}$

(b) There are 5 squares each of side 7 cm.
Area of one square = 7 x 7 = 49$c{m}^{2}$
Area of 5 squares = 49 x 5 = 245 $c{m}^{2}$
(c) Area of rectangle ABCD = 5 x 1 = 5 $c{m}^{2}$
Area of rectangle EFGH = 4 x 1 = 4 $c{m}^{2}$
Total area of the figure = 5 + 4 $c{m}^{2}$

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

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

(a) Area of rectangle = 100 × 144
= 14400 cm
Area of one tile = 5 × 12
$=60c{m}^{2}$
Number of tiles = $\frac{\left(Areaofrectangle\right)}{\left(Areaofonetile\right)}$
$=\frac{14400}{60}$ = 240
Hence, 240 tiles are needed
(b) Area of rectangle = 70 × 36
$=2520c{m}^{2}$
Area of one tile = 5 × 12
$=60c{m}^{2}$
Number of tiles = $\frac{\left(Areaofrectangle\right)}{\left(Areaofonetile\right)}$
=$\frac{2520}{60}=42$

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