## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

- Class 8 Maths Mensuration Exercise 11.1
- Class 8 Maths Mensuration Exercise 11.2
- Class 8 Maths Mensuration Exercise 11.3
- Class 8 Maths Mensuration Exercise 11.4
- Mensuration Class 8 Extra Questions

**NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.1**

Ex 11.1 Class 8 Maths Question 1.

A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Solution:

Perimeter of figure (a) = 4 Ã— side = 4 Ã— 60 = 240 m

Perimeter of figure (b) = 2 [l + b]

Perimeter of figure (b) = Perimeter of figure (a)

2[l + b] = 240

â‡’ 2 [80 + b] = 240

â‡’ 80 + b = 120

â‡’ b = 120 – 80 = 40 m

Area of figure (a) = (side)^{2} = 60 Ã— 60 = 3600 m^{2}

Area of figure (b) = l Ã— b = 80 Ã— 40 = 3200 m^{2}

So, area of figure (a) is longer than the area of figure (b).

Ex 11.1 Class 8 MathsÂ Question 2.

Mrs Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of â‚¹ 55 per m^{2}.

Solution:

Area of the plot = side Ã— side = 25 m Ã— 25 m = 625 m^{2}

Area of the house = l Ã— b = 20 m Ã— 15 m = 300 m^{2}

Area of the garden to be developed = Area of the plot – Area of the house = 625 m^{2} – 300 m^{2} = 325 m^{2}

Cost of developing the garden = â‚¹ 325 Ã— 55 = â‚¹ 17875

Ex 11.1 Class 8 MathsÂ Question 3.

The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden. [Length of rectangle is 20 – (3.5 + 3.5) metres]

Solution:

Length of the rectangle = 20 – (3.5 + 3.5) = 20 – 7 = 13 m

Area of the rectangle = l Ã— b = 13 Ã— 7 = 91 m2

Area of two circular ends = 2(\(\frac { 1 }{ 2 }\) Ï€r^{2})

= Ï€r^{2}

= \(\frac { 22 }{ 7 }\) Ã— \(\frac { 7 }{ 2 }\) Ã— \(\frac { 7 }{ 2 }\)

= \(\frac { 77 }{ 2 }\) m^{2}

= 38.5 m^{2}

Total area = Area of the rectangle + Area of two ends = 91 m^{2} + 38.5 m^{2} = 129.5 m^{2}

Total perimeter = Perimeter of the rectangle + Perimeter of two ends

= 2 (l + b) + 2 Ã— (Ï€r) – 2(2r)

= 2 (13 + 7) + 2(\(\frac { 22 }{ 7 }\) Ã— \(\frac { 7 }{ 2 }\)) – 4 Ã— \(\frac { 7 }{ 2 }\)

= 2 Ã— 20 + 22 – 14

= 40 + 22 – 14

= 48 m

Ex 11.1 Class 8 MathsÂ Question 4.

A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m^{2}? (If required you can split the tiles in whatever way you want to fill up the corners).

Solution:

Area of the floor = 1080 m^{2} = 1080 Ã— 10000 cm^{2} = 10800000 cm^{2} [âˆµ 1 m^{2} = 10000 cm^{2}]

Area of 1 tile = 1 Ã— base Ã— height = 1 Ã— 24 Ã— 10 = 240 cm^{2}

Number of tiles required

= 45000 tiles

Ex 11.1 Class 8 MathsÂ Question 5.

An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, the circumference of a circle can be obtained by using the expression C = 2Ï€r, where r is the radius of the circle.

Solution:

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