**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 × side = 4 × 60 = 240 m

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

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

2[l + b] = 240

**⇒**** 2 [80 + b] = 240
**

**⇒**

**80 + b = 120**

**⇒**

**b = 120 – 80 = 40 m**

Area of figure (a) = (side)

Area of figure (b) = l × b = 80 × 40 = 3200 m

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

Area of figure (a) = (side)

^{2}= 60 × 60 = 3600 m^{2}Area of figure (b) = l × b = 80 × 40 = 3200 m

^{2}So, area of figure (a) is longer than the area of figure (b).

**Ex 11.1 Class 8
Maths **

**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 ₹ 55 per m ^{2}.**

Solution:

Area of the plot = side × side = 25 m × 25 m = 625 m

Area of the house = l × b = 20 m × 15 m = 300 m

Area of the garden to be developed = Area of the plot – Area of the house = 625 m

Cost of developing the garden = ₹ 325 × 55 = ₹ 17875

Solution:

Area of the plot = side × side = 25 m × 25 m = 625 m

^{2}Area of the house = l × b = 20 m × 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 = ₹ 325 × 55 = ₹ 17875

** **

**Ex 11.1 Class 8 Maths**

** 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 × b = 13 × 7 = 91 m2

Area of two circular ends = 2(
πr

= πr

=
×
×

=
m

= 38.5 m

Total area = Area of the rectangle + Area of two ends = 91 m

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

= 2 (l + b) + 2 × (πr) – 2(2r)

= 2 (13 + 7) + 2(
×
) – 4 ×

= 2 × 20 + 22 – 14

= 40 + 22 – 14

= 48 m

Solution:

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

Area of the rectangle = l × b = 13 × 7 = 91 m2

Area of two circular ends = 2(

^{2})= πr

^{2}=

=

^{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 × (πr) – 2(2r)

= 2 (13 + 7) + 2(

= 2 × 20 + 22 – 14

= 40 + 22 – 14

= 48 m

**Ex 11.1 Class 8
Maths **

**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 × 10000 cm^{2} = 10800000
cm^{2} [**∵**** 1 m ^{2}
= 10000 cm^{2}]**

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

Number of tiles required

= 45000 tiles

**Ex 11.1 Class 8
Maths **

**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 = 2πr, where r is the radius of the circle.

Solution:

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 = 2πr, where r is the radius of the circle.

Solution:

0comments