NCERT Solutions Class 6 Maths Chapter 10 Mensuration Ex 10.1

NCERT Solutions Class 6 Maths
Chapter – 10 (Mensuration)

The NCERT Solutions in English Language for Class 6 Mathematics Chapter – 10 Mensuration Exercise 10.1 has been provided here to help the students in solving the questions from this exercise.

Chapter 10: Mensuration

Exercise – 10.1

1. Find the perimeter of each of the following figures:
$NCERT Class 6 Math Solution$

Solutions:

(a) Perimeter = Sum of all the sides
= 4 cm + 2 cm + 1 cm + 5 cm
= 12 cm

(b) Perimeter = Sum of all the sides
= 40 cm + 35 cm + 23 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
= 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm + 1 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

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) cm
= 2 (50) cm
= 2 x 50 cm
= 100 cm
∴ 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) m
= 2 (3.75) m
= 2 × 3.75 m
= 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:

Required length of wooden strip = Perimeter of photograph
= 2 (Length + Breadth)
= 2 (32 + 21) cm
= 2 (53) cm
= 2 × 53 cm
= 106 cm
∴ 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) km
= 2 (1.2) km
= 2 × 1.2 km
= 2.4 km
Each side is to be fenced with 4 rows = 4 × 2.4
= 9.6 km
∴ 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) We know that the perimeter of the given triangle = The sum of all sides of the triangle
∴ Perimeter of the triangle = 3 cm + 4 cm + 5 cm = 12 cm

(b) We know that the perimeter of the given triangle
= Sum of all the sides of the triangle
= (9 + 9 + 9) = 27 cm

(c) Perimeter of the given isosceles triangle
= Sum of all the sides of the triangle
= (8 + 8 + 6) cm = 22 cm

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

Solutions:

Perimeter of a triangle = Sum of all the sides of the triangle
= 10 cm + 14 cm + 15 cm
= 39 cm
∴ The perimeter of triangle is 39 cm

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

Solutions:

Perimeter of a regular hexagon = 6 x side = 6 x 8 m = 48 m.
∴ Perimeter of regular hexagon is 48 m

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

Solutions:

Perimeter of a square = 4 x side
20 = 4 x side
∴ side = 20 m ÷ 4 = 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:

Perimeter of the regular pentagon = 100 cm
Number of sides in regular pentagon = 5
∴ Length of each side = Perimeter ÷ Number of sides
= 100 cm ÷ 5 = 20 cm.
∴ Side of the pentagon is 20 cm.

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?

Solutions:

(a) Length of string = 30 cm
Number of equal sides in a square = 4
∴ Length of each side of the square = 30 cm ÷ 4 = 7.50 cm.

(b) Length of string = 30 cm
Number of equal sides in equilateral triangle = 3
∴ Length of each side of the equilateral triangle = 30 cm ÷ 3 = 10 cm

(c) Length of string = 30 cm
Number of equal sides in regular hexagon = 6
∴ Length of each side of the regular hexagon = 30 cm ÷ 6 = 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 m and breadth 125 m at the rate of ₹ 12 per metre.

Solutions:

Length = 175 m
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
∴ 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 m
= 300 m
∴ Distance covered by Sweety is 300 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (60 + 45) m
= 2 (105) m
= 2 × 105 m
= 210 m
∴ 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?
$NCERT Class 6 Math Solution$

Solutions:

(a) Perimeter of the square = 25 cm + 25 cm + 25 cm + 25 cm
= 4 x 25 cm
= 100 cm

(b) Perimeter of the rectangle = 30 cm + 20 cm + 30 cm + 20 cm
= 2 [30 cm + 20 cm]
= 2 x 50 cm
= 100 cm

(c) Perimeter of the rectangle = 40 cm + 10 cm + 40 cm + 10 cm
= 2 [40 cm + 10 cm]
= 2 x 50 cm
= 100 cm

(d) Perimeter of the triangle = Sum of all sides
= 30 cm + 30 cm + 40 cm
= 100 cm

All the figures have same perimeter.

17. Avneet buys 9 square paving slabs, each with a side of $\mathbf{\frac{1}{2}}$ m. He lays them in the form of a square.
$NCERT Class 6 Math Solution$
(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) Perimeter of his arrangement = 4 × Length of one side
= $4 \times \left ( \frac{1}{2} + \frac{1}{2} + \frac{1}{2}\right )$ m

= $4 \times \left ( \frac{3}{2} \right )$ m

= 2 x 3 m
= 6 m

(b) Perimeter of her arrangement = Sum of the lengths of all the sides.

$= \frac{1}{2} m + 1m +1m+\frac{1}{2}m+1m+1m+\frac{1}{2}m+1m+1m+\frac{1}{2}m+1m+1m$

= 8m + $\frac{4}{2}m$

= 8m + 2m
= 10 m

(d) Yes ! there is a way of getting an even greater perimeter. It is shown below:
$NCERT Class 6 Math Solution$
Because side of a square = $\frac{1}{2}$ m

Therefore Length = $\frac{9}{2}$ m and Breadth = $\frac{1}{2}$ m

Perimeter = 2 (Length + Breadth)

= $2 \left ( \frac{9}{2} + \frac{1}{2}\right )$ m

= $2 \left ( \frac{10}{2} \right )$ m

= 10 m
The above arrangement will also have the greater perimeter.

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NCERT Solutions Class 6 Maths Chapter 10 Mensuration Ex 10.1