**Download NCERT Solutions for Class 6 Maths Chapter 10 – Mensuration**

**NCERT Solutions for Class 6 Maths Chapter 10 – Mensuration comprises of the 3 Exercises**

**NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.1**

**Ques. 1. Find the perimeter of each of the following figures:**

**Sol.1.** Perimeter of a figure = sum of measurements of all sides

**a)**

Perimeter = (4+2+1+5) cm

= 12 cm

**(b)**

Perimeter = (35+23+35+40) cm

= 133 cm

**(c)**

Perimeter = (15+15+15+15) cm

= 60 cm

**(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) × 4 cm

= 52 cm

**Ques. 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. **Length of the tape = perimeter of the rectangular lid

= 2 × (l + b)

= 2 × (40 + 10) cm

= 100 cm

**Ques. 3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?**

**Sol.3. **Perimeter of the table top = 2 × (l + b)

l = 2 m 25 cm

= 2.25 m

b = 1 m 50 cm

= 1.50 m

Perimeter = 2 × (2.25 + 1.50)

= 7.50 m

= 7 m 50 cm

**Ques. 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. **Length of wooden strip required to frame the photograph = Perimeter of photograph = 2 × (l + b)

l = 32 cm

b = 21 cm

Perimeter = 2 × (32 + 21) cm

= 106 cm

**Ques. 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 wire needed = 4 × Perimeter of the land = 4 × 2 × (l + b)

l = 0.7 km

b = 0.5 km

Length of wire needed = 4 × 2 × (0.7 + 0.5) km

= 8 × 1.2 km

= 9.6 km

**Ques. 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 + 4 + 5) cm

= 12 cm

(b) Perimeter = (3 × 9) cm

= 27 cm

(c) Perimeter = (8 × 2 + 6) cm

= 22 cm

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

**Sol.7. **Perimeter of the triangle = (10 + 14 + 15) cm

= 39 cm

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

**Sol.8.** No. of sides of a hexagon = 6

Length of each side = 8 m

Perimeter = (6 × 8) m

= 48 m

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

**Sol.9.** No. of sides of a square = 4

Let the length of each side be ‘a’.

Perimeter = 4a

4a = 20 m

a = m

= 5 m

Hence, the square has a side length of 5 m.

**Ques. 10. The perimeter of a regular pentagon is 100 cm. How long is its each side?
Sol.10.** No. of sides of a pentagon = 5

Let the length of each side be ‘a’.

Perimeter = 5a

5a = 100 cm

a = cm

= 20 cm

Hence, the regular pentagon has a side length of 20 cm.

**Ques. 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)** No. of sides of a square = 4

Let the length of each side be ‘a’.

Perimeter = 4a

4a = 30 cm

a = cm

a = 7.5 cm

Hence, the square has a side length of 7.5 cm.

**(b)** No. of sides of a triangle = 3

Let the length of each side be ‘a’.

Perimeter = 3a

3a = 30 cm

a = cm

= 10 cm

Hence, the equilateral triangle has a side length of 10 cm.

**(c)** No. of sides of a hexagon = 6

Let the length of each side be ‘a’.

Perimeter = 6a

6a = 30 cm

a = cm

= 5 cm

Hence, the square has a side length of 5 cm.

**Ques. 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 a triangle = sum of all 3 sides

Let the third side be ‘a’.

Perimeter = (12 + 14 + a) cm

(12 + 14 + a) cm = 36 cm

26 + a = 36

a = 10 cm

Hence, the third side of the triangle measures 10 cm.

**Ques. 13. Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.**

**Sol.13. **Perimeter of square park = 4a

= 4 × 250 m

= 1000 m

Cost of fencing per metre = Rs 20/m

Cost of fencing the park = Rs 20 × 1000

= Rs 20,000

**Ques. 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.** Perimeter of rectangular park = 2 × (l + b)

= 2 × (175 + 125) m

= 600 m

Cost of fencing per metre = Rs 12/m

Cost of fencing the park = Rs 12 × 600

= Rs 7200

**Ques. 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.** Distance covered by Sweety = Perimeter of the square park

= 4a

= 4 × 75 m

= 300 m

Distance covered by Bulbul = Perimeter of the rectangular park

= 2 × (l + b)

= 2 × (60 + 45) m

= 210 m

Hence, Bulbul covers less distance than Sweety.

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

**Sol.16. **

(a) Perimeter = 25 × 4 cm

= 100 cm

(b) Perimeter = 2 × (40 + 10) cm

= 100 cm

(c) Perimeter = 2 × (30 + 20) cm

= 100 cm

(d) Perimeter = 2 × (30 + 40) cm

= 100 cm

All figures have same perimeter.

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

**(a) What is the perimeter of his arrangement [Figure (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 [(Figure (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) Perimeter of arrangement (i) = 4 × (3 × ) m

= 6 m

(b) Perimeter of arrangement (ii) = 20 × m

= 10 m

(c) Arrangement (ii), i.e., cross has a greater perimeter.

(d) No, this is the maximum perimeter of 10 m. This can also be achieved by straight arrangement of the slabs in a line.

**NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.2**

**Ques. 1. Find the areas of the following figures by counting square:**

**Sol.1. **

**(a)** Area = No. of full filled squares

= 9 sq units

**(b)** Area = No. of full filled squares

= 5 sq units

**(c)** Area = No. of full filled squares + No. of half filled squares

= 2 + 4 × sq units

= 4 sq units

**(d)** Area = No. of full filled squares

= 8 sq units

**(e)** Area = No. of full filled squares

= 10 sq units

**(f)** Area = No. of full filled squares + No. of half filled squares

= 2 + 4 × sq units

= 4 sq units

**(g)** Area = No. of full filled squares + No. of half filled squares

= 4 + 4 × sq units

= 6 sq units

**(h)** Area = No. of full filled squares

= 5 sq units

**(i)** Area = No. of full filled squares

= 9 sq units

**(j)** Area = No. of full filled squares + No. of half filled squares

= 2 + 4 × sq units

= 4 sq units

**(k)** Area = No. of full filled squares + No. of half filled squares

= 4 + 2 × sq units

= 5 sq units

**(l)** Area = No. of full filled squares + No. of half filled squares

= 5 + 6 × sq units

= 8 sq units

**(m)** Area = No. of full filled squares + No. of half filled squares

= 7 + 14 × sq units

= 14 sq units

**(n)** Area = No. of full filled squares + No. of half filled squares

= 10 + 16 × sq units

= 18 sq units

**NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.3**

**Ques. 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.**

**(a)** Area of the rectangle = l × b

= 3 × 4 sq cm

= 12 sq cm

**(b)** Area of the rectangle = l × b

= 12 × 21 sq m

= 252 sq m

**(c)** Area of the rectangle = l × b

= 2 × 3 sq km

= 6 sq km

**(d)** Area of the rectangle = l × b

= 2 × 0.70 sq m

= 1.40 sq m

**Ques. 2. Find the areas of the squares whose sides are:
**

**(a) 10 cm**

(b) 14 cm

(c) 5 m

(b) 14 cm

(c) 5 m

**Sol.2.**

**(a)** Area of the square = a²

= 10 × 10 sq cm

= 100 sq cm

**(b)** Area of the square = a²

= 14 × 14 sq cm

= 196 sq cm

**(c)** Area of the square = a²

= 5 × 5 sq m

= 25 sq m

**Ques. 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

(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)** Area of the rectangle = l × b

= 9 × 6 sq m

= 54 sq m

**(b)** Area of the rectangle = l × b

= 17 × 3 sq m

= 51 sq m

**(c)** Area of the rectangle = l × b

= 4 × 14 sq m

= 56 sq m

(c) has the largest area and (b) has the smallest area.

**Ques. 4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the ****garden.**

**Sol.4. **Area of the rectangular garden = l × b

300 sq m = 50 × b sq m

b = m

= 6 m

Hence, the width of the garden = 6 m.

**Ques. 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.** Area of the rectangular plot = l × b

= 500 × 200 sq m

= 1,00,000 sq m

Cost of tiling per hundred sq m = Rs 8

Cost of tiling the plot = Rs 8 ×

= Rs 8000

**Ques. 6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?**

**Sol.6.** Area of the rectangular table top = l × b

= 2 × 1.50 sq m

= 3 sq m

**Ques. 7. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is ****needed to cover the floor of the room?**

**Sol.7. **Area of the rectangular floor = l × b

= 4 × 3.50 sq m

= 14 sq m

Hence, 14 sq metres of carpet are needed to cover the floor of the room.

**Ques. 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.** Area of floor = 5 × 4 sq m

= 20 sq m

Area of square carpet = 3 × 3 sq m

= 9 sq m

Area of floor that is not carpeted = (20 – 9) sq m

= 11 sq m

**Ques. 9. Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and ****4m wide. What is the area of the remaining part of the land?**

**Sol.9.** Area of land = 5 × 4 sq m

= 20 sq m

Total area covered by square flower beds = 5 × 1 × 1 sq m

= 5 sq m

Area of remaining part of land = (20 – 5) sq m

= 15 sq m

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

**Sol.10. **

**(a)** Area = [(4 × 3) + (4 × 1) + (3 × 2) + (2 × 2) + (2 × 1)] sq cm

= 28 sq cm

**(b)** Area = [(3 × 1) + (3 × 1) + (3 × 1)] sq cm

= 9 sq cm

**Ques. 11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)**

**Sol.11.**

**(a)** Area = [(10 × 2) + (10 × 2)] sq cm

= 40 sq cm

**(b)** Area = [(7 × 7) × 5] sq cm

= 245 sq cm

**(c)** Area = [(5 × 1) + (4 × 1)] sq cm

= 9 sq cm

**Ques. 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.** Area of a tile = 12 × 5 sq cm

= 60 sq cm

**(a)** Area of rectangular region = 100 × 144 sq cm

= 14,400 sq cm

No. of tiles required =

= 240 tiles

**(b)** Area of rectangular region = 70 × 36 sq cm

= 2520 sq cm

No. of tiles required =

= 42 tiles

**Maths – NCERT Solutions Class 6**

**NCERT Solutions Class 6**

**This Chapter contains the Exercises relating to the following topics, which are discussed in Chapter 10 – Mensuration Class 6 NCERT book : –**

- 10.1 Introduction
- 10.2 Perimeter
- 10.2.1 Perimeter of a rectangle
- 10.2.2 Perimeter of regular shapes
- 10.3 Area
- 10.3.1 Area of a rectangle
- 10.3.2 Area of a square

**Download NCERT Solutions for Class 6 Maths Chapter 10 – Mensuration**

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