NCERT Solutions Class 6 Mathematics Chapter 10 Mensuration

Mensuration

Exercise: 10.1

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

Sol.1) Perimeter is the distance covered along the boundary forming a closed figure when you go round the figure once.
Perimeter is the sum of the sides of the polygon
a) Perimeter = 4 + 2 + 1 + 5 = 12 cm
b) Perimeter = 35 + 23 + 35 + 40 = 133 cm
c) Perimeter = 15 + 15 + 15 + 15 = 60 cm
d) Perimeter = 4 + 4+ 4 + 4 + 4 = 20 cm
e) Perimeter = 1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4 = 15 cm
f) Perimeter = 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 round with tape. What is the length of the tape required?
Sol.2) As the lid of the box is sealed around with a tape, we need to find the perimeter of the rectangular box.
Length of the rectangular box = 40 𝑐𝑚
Width of the box = 10 𝑐𝑚
Perimeter of a rectangle = 2 (length + width)
So, length of tape required = perimeter of the rectangular box = 2 (40 + 10) = 2(50) = 100cm

Q.3) A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Sol.3) Length of the table-top = 2 𝑚 25 𝑐𝑚 = 2 𝑚 + 0.25 𝑚 = 2.25 𝑚
Width of the table-top = 1 𝑚 50 𝑐𝑚 = 1 𝑚 + 0.50 𝑚 = 1.50 𝑚
We know that, Perimeter = 2(length + width)
= 2(2.25 + 1.50) = 2(3.75) = 7.50𝑚

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?
Sol.4) A frame is put around the photograph and hence to find the length of the wooden strip to frame we find the perimeter of the photograph.
Length of the photograph = 32 𝑐𝑚
Width of the photograph = 21 𝑐𝑚
Perimeter = 2(length + width)
= 2(32 + 21) = 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?
Sol.5) Length of the rectangular land = 0.7 𝑘𝑚
Width of the land = 0.5 𝑘𝑚
The rectangular land is to be fenced all around with wire. Hence, perimeter of the
rectangular land = length of the wire.
Perimeter = 2(length + width)
= 2(0.7 + 0.5) = 2(1.2) = 2.4 𝑘𝑚
Each side is to be fenced with 4 rows of wires. Therefore, total length of the wire needed is 4 times the perimeter.
Total length of wire needed = 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.
Sol.6) a) Perimeter = 3 𝑐𝑚 + 4 𝑐𝑚 + 5 𝑐𝑚 = 12 𝑐𝑚
b) An equilateral triangle is a triangle whose sides are equal.
Given, side of an equilateral triangle = 9 𝑐𝑚
Perimeter = 9 𝑐𝑚 + 9 𝑐𝑚 + 9 𝑐𝑚 OR 3 × 9 = 27 𝑐𝑚
c) An isosceles triangle is a triangle with two equal sides.
Given, equal sides = 8 cm and third side = 6 𝑐𝑚
Perimeter = 8 𝑐𝑚 + 8 𝑐𝑚 + 6 𝑐𝑚 OR (2 × 8) + 6 𝑐𝑚 = 22 𝑐𝑚

Q.7) Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Sol.7) Perimeter of a triangle = 10 𝑐𝑚 + 14 𝑐𝑚 + 15 𝑐𝑚
= 39 𝑐𝑚

Q.8) Find the perimeter of a regular hexagon with each side measuring 8 m.
Sol.8) A regular hexagon has six sides which are equal.
Given the measure of a side of a hexagon = 8 𝑚
Therefore, Perimeter of the hexagon = 6 × 8 = 48𝑚

Q.9) Find the side of the square whose perimeter is 20 m.
Sol.9) Perimeter of a square = 4 × side of a square
Given Perimeter = 20 𝑚
So, 20 𝑚 = 4 × 𝑠𝑖𝑑𝑒
Side of the square = 20/4 = 5𝑚

Q.10) The perimeter of a regular pentagon is 100 cm. How long is its each side?
Sol.10) A regular pentagon has five sides which are equal.
So, perimeter of a regular hexagon = 5 × side.
Given the perimeter of a regular hexagon = 100 𝑐𝑚
So, 100 𝑐𝑚 = 5 × side
Side = 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?
Sol.11) a) The string is used to form a square and hence the length of the string is the perimeter of the shape formed.
So perimeter of the square formed = 30 𝑐𝑚 = The length of the string.
Perimeter of a square is 4 × side
So, 4 × side = 30 𝑐𝑚
Side = 30/4 = 7.5 𝑐𝑚
b) The string is used to form an equilateral triangle and hence the length of the string is the perimeter of the shape formed.
So perimeter of the equilateral triangle formed = 30 𝑐𝑚 Perimeter of an equilateral triangle is 3 × side. [An equilateral triangle has three equal sides]
So, 3 × side = 30 𝑐𝑚
Side = 30/3 = 10 𝑐𝑚
c) The string is used to form a regular hexagon and hence the length of the string is the perimeter of the shape formed.
So perimeter of the regular hexagon formed = 30 𝑐𝑚
Perimeter of a regular hexagon is 6 × side. [A regular hexagon has six equal sides]
So, 6 × side = 30 𝑐𝑚
Side = 30/6 = 5𝑐𝑚

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?
Sol.12) Perimeter of the triangle = 36 𝑐𝑚
Sides of the triangle = 12 𝑐𝑚 and 14 𝑐𝑚
We know that the perimeter of a triangle = sum of all the sides.
That is, 36 𝑐𝑚 = 12 𝑐𝑚 + 14 𝑐𝑚 + third side
36 = 26 + third side
Third side = 36 – 26 = 10 𝑐𝑚
Hence, the measure of the third side of the triangle is 10 𝑐𝑚

Q.13) Find the cost of fencing a square park of side 250 𝑚 at the rate of 𝑅𝑠. 20 𝑝𝑒𝑟 𝑚𝑒𝑡𝑟𝑒
Sol.13) Side of the square = 250𝑚
The square park is to be fenced all around and hence we need to find the perimeter of the park, i.e. length of fencing required = perimeter of the square.
Perimeter of a square = 4 × side
= 4 × 250 = 1000 𝑚
So, Length of fencing required = 1000 𝑚
Now, cost of fencing per meter = 𝑅𝑠. 20
So, cost of fencing 1000 𝑚 = 1000 × 20 = 𝑅𝑠. 20,000

Q.14) Find the cost of fencing a rectangular park of length 175 𝑚 and breadth 125 𝑚 at the rate of 𝑅𝑠. 12 𝑝𝑒𝑟 𝑚𝑒𝑡𝑟𝑒.
Sol.14) Length of the rectangular park is 175 𝑚
Width of the rectangular park is 125 𝑚
Perimeter of the park = 2(length + width)
= 2(175 + 125)
= 2(300) = 600 𝑚
Cost of fencing per meter = 𝑅𝑠. 12
Cost of fencing the park is = 600 × 12 = 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?
Sol.15) Sweety and Bulbul run around the park and hence the distance covered by them is the perimeter of the park.
Sweety runs around a square park of side = 75𝑚
Perimeter of the park = 4 × side
= 4 × 75 = 300 𝑚
Bulbul runs around a rectangular park of length = 60 𝑚 and width 45 𝑚
Perimeter of the rectangular park = 2(length + width) = 2(60 + 45) = 2(105) = 210 𝑚
So, Distance covered by Sweety is 300 𝑚 and by Bulbul is 210 𝑚.
Therefore, Bulbul covers lesser distance.

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

Sol.16) a) Perimeter = 25 + 25 + 25 + 25 = 4 × 25 = 100 𝑐𝑚
b) Perimeter = 20 + 30 + 20 + 30 = 2(20 + 30) = 2(50) = 100 𝑐𝑚
c) Perimeter = 10 + 40 + 10 + 40 = 2(10 + 40) = 2(50) = 100 𝑐𝑚
d) Perimeter = 30 + 30 + 40 = 100 𝑐𝑚
Inference: The perimeters of the shapes are all equal.

Q.17) Avneet buys 9 square paving slabs, each with a side of ½ m. He lays them in the form of a square.

(a) What is the perimeter of his arrangement (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 [(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.)
Sol.17) a) The side of each square is ½ m and hence the length of the side of the square formed is 1/2 + 1/2 + 1/2 = 3/2 𝑚
So, Perimeter of the square formed is 4(3/2) = 1/2 𝑐𝑚
b) Shari arranges into cross form
So, the Perimeter of this shape = 1/2 + 1 + 1 + 1/2 + 1 + 1 + 1/2 + 1 + 1 + 1/2 + 1 + 1 = 10 𝑚
c) It is apparent that Shari’s arrangement has a greater perimeter.
d) No, it cannot be arranged such that the perimeter of the image is greater than 10 𝑐𝑚.'

Exercise: 10.2

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

Sol.1) The area of one full square is taken as 1 𝑠𝑞. 𝑢𝑛𝑖𝑡.
If it is a centimetre square sheet, then area of one full square will be 1 𝑠𝑞. 𝑐𝑚.
Ignore portions of the area that are less than half a square.
If more than half of a square is in a region, just count it as one square.
If exactly half the square is counted, take its area as 1/2 𝑠𝑞. 𝑚𝑒𝑡𝑟𝑒.
a) There are 9 full squares. So, Area of the shape = 9 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
b) There are 5 full squares and so the area of the shape is 5 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
c) There are 2 full squares and 4 exactly half squares. So, Area of the shape
= 2 + 4 × 1/2 = 2 + 2 = 4 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
d) There are 8 full squares and hence the area of the shape is 8 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
e) There are 10 full squares and hence the area of the shape is 10 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
f) There are 2 full squares and 4 exactly half squares. So, Area of the shape
= 2 + 4 × 1/2 = 2 + 2 = 4 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
g) There are 4 full squares and 4 exactly half squares. So, Area of the shape
= 4 + 4 × 1/2 = 4 + 2 = 6 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
h) There are 5 full squares and hence the area of the shape is 5 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
i) There are 9 full squares and hence the area of the shape is 9 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
j) There are 2 full squares and 4 exactly half squares. So, Area of the shape
= 2 + 4 × 1/2 = 2 + 2 = 4 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠. 
k) There are 4 full squares and 2 exactly half squares. So, Area of the shape
= 4 + 2 × 1/2
= 4 + 1 = 5 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
l) Less than half filled squares– 4; Area estimate = 0
More than half filled squares – 3; Area estimate = 3 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
Half-filled squares – 2; Area estimate = 2 × 1/2 = 1 𝑠𝑞. 𝑢𝑛𝑖𝑡
Full squares – 2; area estimate = 2 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
Area of the shape = 0 + 3 + 1 + 2 = 6 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
m) Less than half filled squares – 5; Area estimate = 0
More than half filled squares – 9; Area estimate = 9 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
Half-filled squares – 0; Area estimate = 0
Full squares – 5; area estimate = 5 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
Total area of the shape is 9 + 5 = 14 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
n) Less than half filled squares – 6; Area estimate = 0
More than half filled squares – 10; Area estimate = 10 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
Half-filled squares – 0; Area estimate = 0
Full squares – 8; area estimate = 8 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
Total area of the shape is 10 + 8 = 18 sq. units.

Exercise: 10.3

Q.1) 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) 2 m and 70 cm
Sol.1) Area of a rectangle = length × breadth
a) Length = 3 𝑐𝑚; breadth = 4 𝑐𝑚
Area = 3 × 4 = 12 𝑠𝑞. 𝑐𝑚
b) Length = 12 𝑚; breadth = 21 𝑚
Area = 12 × 21 = 252 𝑠𝑞. 𝑚
c) Length = 2 𝑘𝑚; breadth = 3 𝑘𝑚
Area = 2 × 3 = 6 𝑠𝑞. 𝑘𝑚
d) Length = 2 𝑚; breadth = 70 𝑐𝑚 = 0.7 𝑚
Area = 2 × 0.7 = 1.4 𝑠𝑞. 𝑚

Q.2) Find the areas of the squares whose sides are:
(a) 10 cm (b) 14 cm (c) 5 m
Sol.2) Area of a square = side × side
a) Side = 10 𝑐𝑚
Area = 10 × 10 = 100 𝑠𝑞. 𝑐𝑚
b) Side = 14 𝑐𝑚
Area = 14 × 14 = 196 𝑠𝑞. 𝑐𝑚
c) Side = 5 𝑚
Area = 5 × 5 = 25 𝑠𝑞. 𝑚

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?
Sol.3) a) Length = 9 𝑚; breadth = 6 𝑚
Area = 9 × 6 = 54 𝑠𝑞. 𝑚
b) Length = 17 𝑚; breadth = 3 𝑚
Area = 17 × 3 = 51 𝑠𝑞. 𝑚
c) Length = 4 𝑚; breadth = 14 𝑚
Area = 4 × 14 = 56 𝑠𝑞. 𝑚
c) has the largest area and b) has the smallest area

Q.4) The area of a rectangular garden 50 𝑚 long is 300 𝑠𝑞. 𝑚. find the width of the garden.
Sol.4) Length of the rectangular garden is 50 𝑚
Area = 300 𝑠𝑞. 𝑚
Area of a rectangle = length × breadth
i.e., 300 = 50 × 𝑏𝑟𝑒𝑎𝑑𝑡ℎ
Breadth (300/50)𝑚 = 6 𝑚
So, breadth (width) of the garden is 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 ?
Sol.5) To tile a rectangular plot, we need to find the area of the plot.
Given length of the plot = 500 𝑚
Width of the plot = 200 𝑚
So, area of the plot = 500 × 200 = 1,00,000 𝑠𝑞. 𝑚
The cost of tiling 100 𝑠𝑞. 𝑚 = 𝑅𝑠 8.
So, the cost of tiling 1,00,000 𝑠𝑞. 𝑚 is 8 × 1,00,000/100 = 𝑅𝑠. 8,000

Q.6) A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Sol.6) Length of the table-top = 2 𝑚
Width of the table-top = 1 𝑚 50 𝑐𝑚 = 1.50 𝑚
So, area of the table-top = length × breadth
= 2 × 1.50 = 3 𝑠𝑞. 𝑚

Q.7) A room is 4 𝑚 long and 3 𝑚 50 𝑐𝑚 wide. How many square metres of carpet are needed to cover the floor of the room?
Sol.7) Length of the room is 4 𝑚
Width of the room is 3 𝑚 50 𝑐𝑚 = 3.50 𝑚
To carpet the room, we need to find the area of the floor.
So, Area of the room = length × breadth = 4 × 3.50 = 14 𝑠𝑞. 𝑚
Therefore, 14 𝑠𝑞. 𝑚𝑒𝑡𝑟𝑒𝑠 of carpet is needed to cover the floor of the room.

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.
Sol.8) Given Length of the floor = 5 𝑚
Width of the floor = 4 𝑚
A square carpet is laid on the floor as shown below.

We need to find the area of the shaded portion.
Area of the square carpet = 3 × 3 = 9 𝑠𝑞. 𝑚
So, 9 𝑠𝑞. 𝑚 of the floor is covered with carpet.
Total area of the floor = 5 × 4 = 20 𝑠𝑞. 𝑚
So, area of the floor that is not carpeted = 20 – 9 = 11 𝑠𝑞. 𝑚

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?
Sol.9) Area of the piece of land = 5 × 4 = 20 𝑠𝑞. 𝑚
Area of each flower bed = 1 × 1 = 1 𝑠𝑞. 𝑚
Five square beds are dug on the land.

So, area of five such flower beds = 5 𝑠𝑞. 𝑚
So, area of the remaining part of the land is the shaded portion shown above.
Area of the remaining part = Area of the piece of land – area of the 5 flower beds.
= 20 – 5 = 15 𝑠𝑞. 𝑚

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

Length = 4 𝑐𝑚, Breadth = 3 𝑐𝑚;
Area = 4 × 3 = 12 𝑠𝑞. 𝑐𝑚
Area of B:
Length = 3 𝑐𝑚, Breadth2 + 1 = 3 𝑐𝑚
Area = 3 × 3 = 9 𝑠𝑞. 𝑐𝑚
Area of C:
Length = 2 + 1 + 1 = 4 𝑐𝑚, Breadth = (4 + 1) – 3 = 2 𝑐𝑚;
Area = 4 × 2 = 8 𝑠𝑞. 𝑐𝑚
Area of D:
Length = Breadth = 1 𝑐𝑚;
Area = 1 𝑠𝑞. 𝑐𝑚
Part D is a portion in both Part A and B.
So, Total area of the shape is = 12 + 9 + 8 – 1 = 28 𝑠𝑞. 𝑐𝑚
b) The splitting can be done as shown below:

Area of A:
Length = 3 𝑐𝑚, Breadth = 1 𝑐𝑚;
Area = 3 𝑠𝑞. 𝑐𝑚
Area of B:
Length = 3 𝑐𝑚, Breadth = 1 𝑐𝑚;
Area = 3 𝑠𝑞. 𝑐𝑚
Area of C:
Length = 3 𝑐𝑚; Breadth = 1 𝑐𝑚; Area = 3 𝑠𝑞. 𝑐𝑚
Total area of the shape = 3 + 3 + 3 = 9 𝑠𝑞. 𝑐𝑚

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

Sol.11) The splitting can be done as follows:

Area of A = 12 × 2 = 24 𝑠𝑞. 𝑐𝑚
Area of B = 8 × 2 = 16 𝑠𝑞. 𝑐𝑚
Total area = 24 + 16 = 40 𝑠𝑞. 𝑐𝑚
b) Area of A = 7 × 7 = 49 𝑠𝑞. 𝑐𝑚
Area of B = 7 × 21 = 147 𝑠𝑞. 𝑐𝑚
Area of C = 7 × 7 = 49 𝑠𝑞. 𝑐𝑚
Total area of the shape = 49 + 147 + 49 = 245 𝑠𝑞. 𝑐𝑚
c) Area of A = 5 × 1 = 5 𝑠𝑞. 𝑐𝑚
Area of B = 4 × 1 = 4 𝑠𝑞. 𝑐𝑚
Total area = 5 + 4 = 9 𝑠𝑞. 𝑐𝑚

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.
Sol.12) Length of the tile = 12 𝑐𝑚; Breadth of the tile = 5 𝑐𝑚
Area of one tile = 12 × 5 = 60 𝑠𝑞. 𝑐𝑚
a) Length of the rectangular region = 100 𝑐𝑚
Breadth of the rectangular region = 144 𝑐𝑚
Area of the rectangular region = 100 × 144 = 14400 𝑠𝑞. 𝑐𝑚
∴ number of tiles needed = 14400/60 = 240 𝑡𝑖𝑙𝑒𝑠
b) Length of the rectangular region = 70 𝑐𝑚
Breadth of the rectangular region = 36 𝑐𝑚
Area of the rectangular region = 70 × 36 = 2520 𝑠𝑞. 𝑐𝑚
Therefore, number of tiles needed = 2520/60 = 42 𝑡𝑖𝑙𝑒𝑠