# NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1

These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.1

Question 1.
Find the perimeter of each of the following figures: (a) Perimeter = Sum of all the sides = 4 cm + 2 cm + 1 cm + 5 cm = 12 cm
(b) Perimeter = Sum of all the sides = 23 cm + 35 cm + 40 cm + 35 cm = 133 cm
(c) Perimeter = Sum of all the sides =15 cm + 15 cm + 15 cm + 15 cm = 60 cm
(d) Perimeter = Sum of all the sides = 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm
(e) Perimeter = Sum of all the sides 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm
(f) Perimeter = Sum of all the sides = 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm = 52 cm Question 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?
Total length of tape required
= Perimeter of rectangle
= 2 (length + breadth) = 2 (40 + 10)
= 2 × 50 = 100 cm = 1 m
Thus, the total length of tape required is 100 cm or 1 m.

Question 3.
A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
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 = 7.5 m
Thus, the perimeter of tabletop is 7.5 m.

Question 4.
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Length of wooden strip = Perimeter of photograph
Perimeter of photograph
= 2 x (length + breadth)
= 2 (32 + 21) = 2 × 53cm =106 cm
Thus, the length of the wooden strip required is equal to 106 cm. Question 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?
Since the 4 rows of wires are needed.
Therefore the total length of wires is equal to 4 times the perimeter of rectangle.
Perimeter of field = 2 × (length + breadth)
= 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 km
= 2.4 × 1000 m = 2400 m
Thus, the length of wire = 4 × 2400 = 9600 m = 9.6 km

Question 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.
(a) Perimeter of ΔABC = AB + BC + CA = 3 cm + 5 cm + 4 cm = 12 cm (b) Perimeter of equilateral ABC
= 3 × side = 3 × 9 cm = 27 cm (c) Perimeter of AABC = AB + BC + CA = 8 cm + 6 cm + 8 cm = 22 cm Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Perimeter of triangle
= Sum of all three sides
= 10 cm + 14 cm + 15 cm = 39 cm
Thus, the perimeter of triangle is 39 cm Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Perimeter of Hexagon
= 6 × length of one side
= 6 × 8 m = 48 m
Thus, the perimeter of hexagon is 48 m.

Question 9.
Find the side of the square whose perimeter is 20 m.
Perimeter of square = 4 × side
⇒ 20 = 4 × side
⇒ Side = $$\frac{20}{4}$$ = 5 cm
Thus, the side of square is 5 cm.

Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Perimeter of regular pentagon = 100 cm
⇒ 5 × side = 100 cm
⇒ Side = $$\frac{100}{5}$$ = 20 cm
Thus, the side of regular pentagon is 20 cm. Question 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?
Length of string = Perimeter of each figure
(a) Perimeter of square = 30 cm
⇒ 4 × side = 30 cm
⇒ Side = $$\frac{30}{4}$$ = 7.5 cm 4
Thus, the length of each side of square is 7.5 cm.

(b) Perimeter of equilateral triangle = 30 cm
⇒ 3 × side = 30 cm
⇒ Side = $$\frac{30}{3}$$ = 10 cm
Thus, the length of each side of equilateral triangle is 10 cm.

(c) Perimeter of hexagon = 30 cm
⇒ 6 × side = 30 cm
⇒ Side = $$\frac{30}{6}$$ = 5 cm
Thus, the side of each side of hexagon is 5 cm.

Question 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?
Let the length of third side be X cm.
Length of other two side are 12 cm and 14 cm.
Now, Perimeter of triangle = 36 cm
⇒ 12 +14 + X = 36
⇒ 26 + X = 36
⇒ X = 36 – 26 = 10 cm
Thus, the length of third side is 10 cm Question 13.
Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Side of square = 250 m
Perimeter of square
= 4 × side = 4 × 250 = 1000 m
Since, cost of fencing of per meter = ₹20
Therefore, the cost of fencing of 1000 meters = 20 × 1000 = ₹ 20,000

Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of × 12 per metre.
Length of rectangular park = 175 m
Breadth of rectangular park = 125 m
Perimeter of park
= 2 × (length + breadth)
= 2 × (175 + 125) = 2 × 300 = 600 m
Since, the cost of fencing park per meter = ₹ 12
Therefore, the cost of fencing park of 600 m = 12 × 600 = ₹ 7,200

Question 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?
Distance covered by Sweety
= Perimeter of square park
Perimeter of square = 4 × side
= 4 × 75 = 300 m
Thus, distance covered by Sweety is 300 m. Now, distance covered by Bulbul =
Perimeter of rectangular park
Perimeter of rectangular park
= 2 × (length + breadth)
= 2 × (60 + 45) = 2 × 105 = 210 m
Thus, Bulbul covers the distance of 210 m and Bulbul covers less distance. Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers? (a) Perimeter of square
= 4 × side = 4 × 25 = 100 cm

(b) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (40 + 10) = 2 × 50 = 100 cm

(c) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (30 + 20) = 2 × 50 = 100 cm

(d) Perimeter of triangle = Sum of all sides
= 30 cm + 30 cm + 40 cm = 100 cm
Thus, all the figures have same perimeter. Question 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 (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 (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.) (a) The arrangement so formed in Fig (i) is a square of side equal to 3 × side of the square slab
∴ Perimeter of his arrangement in Fig (i)
= 4 × ( 3 × side of the square slab)
= 4 × ( 3 × 0.5) m = 4 × 1.5 m = 6 m

(b) The cross arrangement so formed in Fig (ii) by 9 square slabs, has perimeter, which consists of 20 length segments each equal to side of I a square slab i.e. 0.5 m
∴ Perimeter of her arrangement in Fig (ii)
= 20 × ( side of the square slab)
= 20 × 0.5 m = 10 m

(c) ∵ 10m > 6m
∴ Cross arrangement has greater perimeter

(d) ∵ Total number of tiles = 9
∴ The following arrangement will also have greater perimeter. Since, this arrangement is in the form of rectangle with length and breadth as $$\frac { 9 }{ 2 }$$ m and $$\frac { 1 }{ 2 }$$ respectively.
∴ Perimeter = 2 $$\left(\frac{9}{2}+\frac{1}{2}\right)$$
= 2 × = 10m $$\left[\frac{10}{2}\right]$$ = 10m

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