NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

NCERT
SOLUTIONS
CLASS - 6TH

aglase .co

Class : 6th
Subject : Maths
Chapter : 10
Chapter Name : Mensuration

Exercise 10.1

Q1 Measure and write the length of the four sides of the top of your study table.
AB = ____ cm
BC = ____ cm
CD = ____ cm
DA = ____ cm
Now, the sum of the lengths of the four sides
=AB + BC + CD + DA
= ___ cm +___ cm +___ cm +___ cm
= _____ cm
What is the perimeter?

Answer. Missing

Page : 206 , Block Name : Try These

Q2 Measure and write the lengths of the four sides of a page of your notebook. The sum of the
lengths of the four sides
= AB + BC + CD + DA
= ___ cm +___ cm +___ cm +___ cm
= _____ cm

Answer. Missing

Page : 206 , Block Name : Try These

Q3 Meera went to a park 150 m long and 80 m wide. She took one complete round on its boundary.

What is the distance covered by her?

Answer. Missing

Page : 206 , Block Name : Try These

Q4 Find the perimeter of the following gures:
(a)

Perimeter = AB + BC + CD + DA
= __+__+__+__
= _____

(b)

Perimeter = AB + BC + CD + DA
= __+__+__+__
= ______

(c)

Perimeter = AB + BC + CD + DE + EF + FG + GH +HI + IJ + JK + KL + LA
= __ + __ +__ + __ + __ + __ + __ + __ +__+ __ + __ + __
= ______

(d)

Perimeter = AB + BC + CD + DE + EF + FA
= __ + __ + __ + __ + __ + __
= ____________

Answer. Missing

Page : 207 , Block Name : Try These

Q1 Find the perimeter of the following rectangles:

Answer. Missing

Page : 208 , Block Name : Try These

Q1 Find various objects from your surroundings which have regular shapes and nd their
perimeters

Answer. Missing

Page : 212 , Block Name : Try These

Q1 Find the perimeter of each of the following gures:

Answer. Perimeter of a polygon is equal to the sum of the lengths of all sides of that polygon.
(a) perimeter =(4 + 2 +1 + S) Cm = 12 Cm
(b) Perimeter = (23 + 35 + 40 + 35) cm = 133 cm
(c) perimeter =(15 + 15 + 15 + 15) cm= 60cm
(d) Perimeter = (4 + 4 + 4 + 4 + 4) cm = 20 cm
(e) Perimeter =(1 +4 + 0.5 + 2.5 + 2.5 + 0.5 + 4) cm = 15 cm
(f) Perimeter (1+3+2+3+4+1+3+2+3+4+1+3+2+3+4+1+3+2+3+ 4) =52 cm

Page : 212 , Block Name : Exercise 10.1

Q2 The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the
length Of the tape required?

Answer. Length (I) of rectangular box = 40 cm
Breadth (b) of rectangular box = 10 cm
Length of tape required = Perimeter of rectangular box
= 2 (l + b) = 2(40 + 10) = 100 cm

Page : 212 , Block Name : Exercise 10.1

Q3 A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

Answer. Length (l) Of table-top = 2 m 25 cm 2 + 0.25 = 2.25 m
Breadth (b) of table-top = 1 m 50 cm = 1 + 0.50 = 1 .50 m
Perimeter of table-top = 2 (l + b)
= 2 x (2.25 + 1.50)
= 2 x 3.75 = 7.5 m

Page : 212 , Block Name : Exercise 10.1

Q4 What is the length of the wooden strip required to frame a photograph of length and
breadth 32 cm and 21 cm respectively?

Answer. Length (I) Of photograph = 32 cm
Breadth (b) of photograph = 21 cm
Length of wooden strip required = Perimeter of Photograph
= 2 x (32 + 21) -2 x 53
= 2 x (l + b)
= 2 x (32 + 21) = 2 x 53 = 106 cm

Page : 212 , Block Name : Exercise 10.1

Q5 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?

Answer. Length (l) Of land = 0.7 km
Breadth (b) of land = 0.5 km
Perimeter = 2 x (l + b)
= 2 x (0.7 + 0.5) = 2 x 1.2 = 2.4 km
Length of wire required = 4 x 2.4 = 9.6 km

Page : 212 , Block Name : Exercise 10.1

Q6 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.

Page : 213 , Block Name : Exercise 10.1

Answer. (a) Perimeter = (3 + 4 + 5) cm = 12 cm
(b) Perimeter of an equilateral triangle = 3 x Side Of triangle
= (3 x 9) cm = 27 cm
(c) Perimeter = (2 x 8) + 6 = 22 cm

Page : 213 , Block Name : Exercise 10.1

Q7 Find the perimeter Of a triangle with sides measuring 10 cm. 14 cm and 15 cm.

Answer. Perimeter of triangle = Sum of the lengths of all sides of the triangle
perimeter = 10 + 14 + 15 = 39 cm

Page : 213 , Block Name : Exercise 10.1

Q8 Find the perimeter of a regular hexagon with each side measuring 8 m.

Answer. Perimeter Of regular hexagon = 6 x Side of regular hexagon
Perimeter Of regular hexagon = 6 x 8 = 48 m

Page : 213 , Block Name : Exercise 10.1

Q9 Find the side of the square whose perimeter is 20 m.

Answer. Perimeter of square = 4 x Side
20 = 4 x Side
Side= 20

4
= 5m

Page : 213 , Block Name : Exercise 10.1

Q10 The perimeter Of a regular pentagon is 100 cm. How long is its each side?

Answer. Perimeter of regular pentagon = 5 x Length of side
100 = 5 x Side
Side= 100

5
= 20cm

Page : 213 , Block Name : Exercise 10.1

Q11 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?

Answer. (a) Perimeter = 4 x Side
30 = 4 x Side
Side= 30

4
= 7.5cm

(b) Perimeter = 3 x Side
30 = 3 x Side
Side= 30

3
= 10cm

(c) Perimeter = 6 x Side
30 = 6 x Side
Side= 30

6
= 5cm

Page : 213 , Block Name : Exercise 10.1

Q12 Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is
its third side?

Answer. Perimeter of triangle = Sum of all sides of the triangle
36 12 + 14 + Side
36 = 26 + Side
Side = 36 - 26 = 10 cm
Hence, the third side of the triangle is 10 cm.

Page : 213 , Block Name : Exercise 10.1

Q13 Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.

Answer. Length of fence required = Perimeter of the square park
= 4 x Side
= 4 x 250 = 1000 m
Cost for fencing 1m of square park = Rs 20
Cost for fencing 1000 m of square park = 1000 x 20
= Rs 20000

Page : 213 , Block Name : Exercise 10.1

Q14 Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of
Rs 12 per metre.

Answer. Length (l) of rectangular park = 175 m
Breadth (b) of rectangular park 125 m
Length of wire required for fencing the park = Perimeter of the park
= 2 x (l + b)
= 2 x (175 + 125)
= 2 x 300
= 600 m
Cost for fencing 1 m of the park = Rs 12
Cost for fencing 600 m of the square park =600 x 12
= Rs 7200

Page : 213 , Block Name : Exercise 10.1

Q15 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?

Answer. Distance covered by Sweety = 4 x Side of square park
= 4 x 75 = 300 m
Distance covered by Bulbul = 2 x (60 + 45)
= 2 x 105 = 210 m

Therefore, Bulbul covers less distance.

Page : 213 , Block Name : Exercise 10.1

Q16 What is the perimeter of each of the following gures? What do you infer from the
answers?

Answer. (a) Perimeter of square 4 x 25 100 cm
(b) Perimeter of rectangle = 2 x (10 + 40) = 100 cm
(c) Perimeter of rectangle 2 x (20 + 30) 100 cm
(d) Perimeter of triangle — 30 + 30 + 40 = 100 cm
It can be inferred that all the gures have the same perimeter.

Page : 213 , Block Name : Exercise 10.1

Q17 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 [ gure (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 [ gure (ii)]?
(c) Which has greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you nd
a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be
broken.)

Answer. (a) Side of square = (3 ×
1 3
)m = m
2 2

Perimeter of Square = 4 × 3

2
= 6m

(b) Perimeter of cross = 0.5+1+1+0.5+1+1+0.5+1+1+0.5+1+1=10m

(c) The arrangement in the form of a cross has a greater perimeter.
(d) Arrangements with perimeters greater than 10 m cannot be determined.

Page : 213 , Block Name : Exercise 10.1

Q1 Draw any circle on a graph sheet. Count the squares and use them to estimate the area of the
circular region.

Answer. Missing

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Q2 Trace shapes of leaves, ower petals and other such objects on the graph paper and nd their
areas.

Answer. Missing

Page : 215 , Block Name : Try These

Exercise 10.2

Q1 Find the areas of the following gures by counting square:

Answer. (a) The gure contains 9 fully lled squares only. Therefore, the area of
this gure will be 9 square units.
(b) The gure Contains S fully lled Squares only. Therefore, the area of this gure will
be 5 square units.
(c) The gure contains 2 fully lled squares and 4 half- lled squares. Therefore, the area of this
gure will be 4 square units.
(d) The gure contains 8 fully lled squares only. Therefore, the area of this gure will
be 8 square units.
(e) The gure contains 10 fully lled squares only. Therefore, the area of this gure will
be 10 square units.

(f) The gure contains 2 fully lled squares and 4 half- lled squares. Therefore, the area
of this gure will be 4 square units.
(g) The gure contains 4 fully lled squares and 4 half- lled squares. Therefore, the area of this
gure will be 6 square units.
(h) The gure contains 5 fully lled squares only. Therefore, the area Of this gure will
be 5 square units.
(i) The gure contains 9 fully lled squares only. Therefore, the area of this gure will be
9 square units.
(j) The gure contains 2 fully lled squares and 4 half- lled squares. Therefore, the area
of this gure will be 4 square units.
(k) The gure contains 4 fully lled squares and 2 half- lled squares. Therefore, the area of this
gure will be 5 square units.
(l) From the given gure, it can be observed that,

Total area = 2 + 6 = 8 square units
(m) From the given gure, it can be observed that,

Total area = 5 + 9 = 14 square units
(n) From the given gure, it can be observed that,

Total area = 8 + 10 = 18 square units.

Page : 216 , Block Name : Exercise 10.2

Exercise 10.3

Q1 Find the areas 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) 2m and 70cm

Answer. It is known that, Area of rectangle = Length x Breadth
(a) l = 3 cm
b = 4 cm
Area I x b = 3 x 4 = 12 cm 2

(b) l = 12 m
b = 21 m
Area = / x b = 12 x 21 = 252 m 2

(c) I = 2 km
b= 3 km
Area = I x b = 2 x 3 = 6 km
2

(d) l = 2m
b = 70 cm = 0.70 m
Area = 1 x b = 2 x 0.70 = 1.40 m 2

Page : 219 , Block Name : Exercise 10.3

Q2 Find the areas of the squares whose sides are:
(a) 10 cm (b) 14 cm (c) 5 m

Answer. It is known that,
Area Of square = (Side) 2

(a) Side = 10 cm
Area = (10) -100cm
2 2

(b) Side = 14 cm
Area =(14) = 196cm
2 2

(c) Side = 5 m

Area = (5) = 25 m 2

Page : 219 , Block Name : Exercise 10.3

Q3 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?

Answer. It is known that, Area of rectangle= Length x Breadth
(a) l = 9 m , b = 6 m
Area = l x b = 9 x 6 = 54m 2

(b) l = 17 m , b = 3 m
Area = l x b = 17 x 3 = 51m 2

(c) l = 4 m , b = 14 m
Area = l x b = 4 x 14 = 56m 2

It can be seen that rectangle (c) has the largest area and rectangle (b) has the smallest
area.

Page : 219 , Block Name : Exercise 10.3

Q4 The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden

Answer. Given, area of the rectangular garden = 300 sq m and length of the rectangular garden =
50 m We know that,
∴ Area of the rectangular garden = Length × Breadth
300 sq m = 500 m x Breadth
Breadth = 300 sq m / 50
=6m
Hence, the breadth or width of the garden is 6 m.

Page : 219 , Block Name : Exercise 10.3

Q5 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?

Answer. Area of rectangular plot 500 x 200 100000m 2

Cost of tiling per 100m = Rs 8
2

Cost of tiling per 100000 m = 2 8
× 100000
100

= Rs 8000

Page : 219 , Block Name : Exercise 10.3

Q6 A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?

Answer. Length (l) = 2 m
Breadth (b) = 1 m 50 cm = (1 + 50

100
) m = 1.5m

Area = l x b = 2 x 1.5 = 3m 2

Page : 219 , Block Name : Exercise 10.3

Q7 A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover
the oor of the room?

Answer. Length (l) = 4 m
Breadth (b) = 3 m 50 cm = 3.5 m
Area = l x b = 4 x 3.5 = 14m 2

Page : 219 , Block Name : Exercise 10.3

Q8 A oor is 5 m long and 4 m wide. A square carpet Of sides 3 m is laid on the oor. Find the area
of the oor that is not carpeted.

Answer. Length (l) = 5 m
Breadth (b) = 4 m
Area of oor = l x b = 5 x 4 = 20 m 2

Area covered by the carpet = (Side ) 2

=3m =9m
2 2

Area not covered by the carpet = 20 - 9 = 11 m 2

Page : 219 , Block Name : Exercise 10.3

Q9 Five square ower 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?

Answer. Area of the land = 5 x 4 = 20 Side 2

Area occupied by 5 ower beds = S x ( Side ) x ( 1 ) = 5 m
2 2 2

Area of the remaining part = 20 - 5 = 15 m 2

Page : 219 , Block Name : Exercise 10.3

Q10 By splitting the following gures into rectangles, nd their areas (The measures are
given in centimetres).

Answer. (a) The given gure can be broken into rectangles as follows.

Area of 1st rectangle = 4 x 2 = 6 cm
2

Area of 2nd rectangle =6 x 1 =6 cm2

Area of 3rd rectangle = 3 x 2 = 6 cm
2

Area of 4th rectangle = 4 x 2 = 8 cm
2

Total area of the complete gure = 8 + 6 + 6 + 8 = 28 cm 2

(b) The given gure can be broken into rectangles as follows.

Area of 1st rectangle = 3 x 1 = 3 cm
2

Area of 2nd rectangle = 3 x 1 = 3 cm 2

Area of 3rd rectangle = 3 x 1 = 3 cm
2

Total area of the complete gure = 3+ 3 + 3 = 9 cm 2

Page : 219 , Block Name : Exercise 10.3

Q11 Split the following shapes into rectangles and nd their areas. (The measures are given in
centimetres).

Answer. (a) The given gure can be broken into rectangles as follows.

Area of 1st rectangle = 12 x 2 = 24cm 2

Area of 2nd rectangle = 8 x 2 = 16 cm 2

Total area of the complete gure = 24 + 16 = 40 cm 2

(b) The given gure can be broken into rectangles as follows.

Area of 1st rectangle = 21 x 7 = 147cm 2

Area of 2nd rectangle = 7 x 7 = 49 cm 2

Area of 3rd rectangle = 7 x 7 = 49 cm 2

Total area of the complete gure = 147 + 49 + 49 = 250 cm 2

(c) The given gure can be broken into rectangles as follows.

Area of 1st rectangle = 5 x 1 = 5 cm 2

Area of 2nd rectangle = 4 x 1 = 4 cm 2

Total area of the complete gure = 5 + 4 = 9 cm 2

Page : 220 , Block Name : Exercise 10.3

Q12 How many tiles whose length and breadth are 12 cm and 5 cm respectively will be
needed to t in a rectangular region whose length and breadth are respectively:
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm

Answer. (a) Total area of the region = 100 x 144 = 14400 cm 2

Area of one tile = 12 x 5 = 60 cm 2

Number of tiles required= 14400

60
= 240

Therefore, 240 tiles are required.

(b) Total area of the region = 70 x 36 = 2520 cm 2

Area of one tile = 60 cm2

Number of tiles required= 2520

60
= 42

Therefore, 42 tiles are required.

Page : 220 , Block Name : Exercise 10.3