NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 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 the figure (a) = 4 × side = 4 × 60 = 240 m
Perimeter of the figure (b) = 2[l + b] = 2[80 + b] m
Perimeter of the figure (a) = Perimeter of the figure (b)
240 = 2[80 + b]
80 + b = 120
b = 120 – 80 = 40 m
Area of the figure (a) = (side)2 = 60 × 60 = 3600 m2
Area of the figure (b) = l × b = 80 × 40 = 3200 m2
So, the area of the figure (a) is larger than the area of the 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 m2.

Solution:
Area of the plot = side × side = 25 m × 25 m = 625 m2
Area of the house = l × b = 20 m × 15 m = 300 m2
Area of the garden to be developed = Area of the plot – Area of the house

                                                           = 625 m2 – 300 m2 = 325 m2
The total cost of developing the garden = ₹ 325 × 55 = ₹ 17,875

 

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(1/2 πr2)
= πr2
22/7 × 7/2 × 7/2
77/2 m2
= 38.5 m2
Total area of the garden = Area of the rectangle + Area of two ends

                                        = 91 m2 + 38.5 m2 = 129.5 m2
Total perimeter of the garden = 13 + 13 + 2 × (πr)
= 26 + 2(22/7 × 7/2)
= 26 + 22
= 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 m2? (If required you can split the tiles in whatever way you want to fill up the corners).

Solution:
Area of the floor = 1080 m2 = 1080 × 10000 cm2 = 10800000 cm2 

[ 1 m2 = 10000 cm2]
Area of 1 tile = base × height = 24 × 10 = 240 cm2
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:



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