Class 8 Maths

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?

mensuration 1

Answer: Perimeter of Square `= 4 xx \text(side) = 4 xx 60 = 240` sq metres
Area of Square = Side² `= 60^2 = 3600` square metre
Perimeter of Rectangle = 2(length + breadth)
Or, `240 = 2(80 + text(breadth))`
Or, `80 + text(breadth) = 120`
Or, breadth `= 120-80 = 40` metres
Area of rectangle = length x breadth `= 80 xx 40 = 3200` square metres
Now it is clear that the area of the square field is greater than the area of the rectangular field.


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 Rs 55 per m².

mensuration 2

Answer: Area of the square plot = Side² `= 25^2 = 625` square metre
Area of the house construction part = length x breadth
`= 20 xx 15 = 300` sq m
So, area of the garden `= 625-300=325` square metre
Cost of developing the garden = Area x Rate
`= 300 xx 55 = 16500` rupees


Question 3: The shape of a garden is rectangular in the middle and semi circular 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].

mensuration 3

Answer: Area of the rectangular part = length x breadth
`= 20 xx 7 = 140` sq m
Area of Semicircular portions: `= п xx \r^2`

Perimeter of the shape `= 22 + 20 + 20 = 62` metres


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).

Answer: Area of the Parallelogram = base x height
`= 24 xx 10 = 240` sq cm

Required number of tiles `=text(Area of floor)/text(Area of tile)`

`=(1080xx100xx100)/(240)=45000`

(area of floor is converted into square cms)

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, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.

mensuration 6

Answer: (a) Perimeter `=(πd)/(2)+d`

`=(22)/(7)xx(2.8)/(2)+2.8=4.4+2.8=7.2` cm

(b) Perimeter `=(πd)/(2)+2xx\text(breadth)+text(length)`

`=4.4+3+2.8=10.2` cm

(Perimeter of semicircular part is same as in (a))

(c) Perimeter `=(πd)/(2)+2xx\text(slant height)`

`=4.4+4=8.4` cm

So, the food shape in (a) requires the ant to cover the least distance.

Ex 11.1

Ex 11.2

Ex 11.3

Ex 11.4