By practising the NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.1 students get an overall idea about the perimeter of rectangles and various regular shapes like squares and equilateral triangles. The solutions are solved in an elaborate manner to help students ace the exam. Students who wish to perform well in Mathematics exams can use NCERT Solutions Class 6 Maths Chapter 10 Mensuration Exercise 10.1 provided here, and prepare exceptionally well for their exams. These solutions can help them understand the right method of solving the questions.
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.1
Access NCERT Solutions for Class 6 Chapter 10: Mensuration Exercise 10.1
1. Find the perimeter of each of the following figures:
Solutions:
(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
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?
Solutions:
Length of required tape = Perimeter of rectangle
= 2 (Length + Breadth)
= 2 (40 + 10)
= 2 (50)
= 100 cm
∴ The required length of tape is 100 cm
3. A table top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table top?
Solutions:
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
∴ The perimeter of the table top is 7.5 m
4. What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm, respectively?
Solutions:
The required length of wooden strip = Perimeter of photograph
= 2 (Length + Breadth)
= 2 (32 + 21)
= 2 (53)
= 2 × 53
= 106 cm
∴ The required length of the wooden strip is 106 cm
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?
Solutions:
Perimeter of the field = 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
∴ The total length of the required wire is 9.6 km
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.
Solutions:
(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
7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solutions:
Perimeter of triangle = 10 + 14 + 15
= 39 cm
∴ The perimeter of the triangle is 39 cm
8. Find the perimeter of a regular hexagon with each side measuring 8 m.
Solutions:
Perimeter of hexagon = 6 × 8
= 48 m
∴ The perimeter of the regular hexagon is 48 m
9. Find the side of the square whose perimeter is 20 m.
Solutions:
Perimeter of square = 4 × side
20 = 4 × side
Side = 20 / 4
Side = 5 m
∴ The side of the square is 5 m
10. The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solutions:
The perimeter of the regular pentagon = 100 cm
5 × side = 100 cm
Side = 100 / 5
Side = 20 cm
∴ The side of the pentagon is 20 cm
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?
Solutions:
(a) Perimeter of square = 30 cm
4 × side = 30
Side = 30 / 4
Side = 7.5 cm
(b) Perimeter of an equilateral triangle = 30 cm
3 × side = 30
Side = 30 / 3
Side = 10 cm
(c) Perimeter of a regular hexagon = 30 cm
6 × side = 30
Side = 30 / 6
Side = 5 cm
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?
Solutions:
Let x cm be the third side
Perimeter of triangle = 36 cm
12 + 14 + x = 36
26 + x = 36
x = 36 – 26
x = 10 cm
∴ The third side is 10 cm
13. Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Solutions:
Side of square = 250 m
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
14. Find the cost of fencing a rectangular park of length 175 cm and breadth 125 m at the rate of ₹ 12 per metre.
Solutions:
Length = 175 cm
Breadth = 125 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (175 + 125)
= 2 (300)
= 2 × 300
= 600 m
Cost of fencing = 12 × 600
= 7200
∴ The cost of fencing is ₹ 7,200
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?
Solutions:
Perimeter of square = 4 × side
= 4 × 75
= 300 m
∴ The distance covered by Sweety is 300 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (60 + 45)
= 2 (105)
= 2 × 105
= 210 m
∴ The distance covered by Bulbul is 210 m
Hence, Bulbul covers less distance than Sweety.
16. What is the perimeter of each of the each of the following figures? What do you infer from the the answers?
Solutions:
(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 the 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 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.)
Solutions:
(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 a cross has greater perimeter
(d) Perimeters greater than 10 m cannot be determined.
Also, explore –
NCERT Solutions for Class 6 Maths Chapter 10
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