Chapter 11 Mensuration Exercise 11.1
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)2 = 60 × 60 = 3600 m2
Area of figure (b) = l × b = 80 × 40 = 3200 m2
So, area of figure (a) is longer than the area of figure (b).
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
Cost of developing the garden = ₹ 325 × 55 = ₹ 17875
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:
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 = 1 × base × height = 1 × 24 × 10 = 240 cm2
Number of tiles required
= 45000 tiles
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:
