### Mensuration - Solutions 1

CBSE Class –VIII Mathematics
NCERT Solutions
CHAPTER - 11
Mensuration (Ex. 11.1)

1. A square and a rectangular field with measurements as given in the figure have the same perimeter.

Which field has a larger area?
AnsGiven: The side of a square = 60 m and the length of rectangular field = 80 m
According to question,
Perimeter of rectangular file = Perimeter of square field
2(l+b) = 4 X Side

(80 + b) = $\frac{240}{2}$
(80 + b) = 120
b = 120 - 80
b = 40 m
Hence, the breadth of the rectangular field is 40 m.
Now, Area of Square field= (Side)2
= (60)2 sq.m = 3600 sq.m
Area of Rectangular field = (length  breadth)
= 80  40 sq. m = 3200 sq. m
Hence, area of square field is larger.

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.

Ans. Side of a square plot = 25 m
Area of square plot = (Side)2 = (25)2 = 625 m2
Length and Breadth of the house is 20 m and 15 m respectively
Area of the house = (length x breadth )
= 20  15 = 300 m2
Area of garden = Area of square plot – Area of house
= (625 – 300) = 325 m2
Cost of developing the garden around the house is Rs.55
Total Cost of developing the garden of area 325 sq. m = Rs.(55 325)
= Rs.17,875

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 meters]
Ans. Given: Total length of the diagram = 20 m
Diameter of semi circle on both the ends = 7 m
Radius of semi circle = $\frac{Diameter}{2}$ = $\frac{7}{2}$= 3.5 m
Length of rectangular field = [Total length - (radius of semicircle on both side)]
={20 – (3.5 + 3.5)}
= 20 – 7 = 13 m
Breadth of the rectangular field = 7 m
Area of rectangular field = ( l x b)
= (13  7)  91
Area of two semi circles =
= 38.5 m2
Total Area of garden = (91 + 38.5)129.5 m2
Perimeter of two semi circles =
= 22 m
Hence, Perimeter of garden = (22 + 13 + 13)m = 48 m

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? [If required you can split the tiles in whatever way you want to fill up the corners]
Ans. Base of flooring tile = 24 cm 0.24 m
height of a flooring tile = 10 cm 0.10 m [1cm = 1/100 m]
Now, Area of flooring tile= Base  Altitude
= 0.24  0.10 sq. m
= 0.024 m2
Number of tiles required to cover the floor =
= 45000 tiles
Hence 45000 tiles are required to cover the floor.

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 where  is the radius of the circle.
= 1.4 cm
Circumference of semi circle =
4.4 cm

Total distance covered by the ant= (Circumference of semi circle + Diameter)
=( 4.4 + 2.8 )cm
= 7.2 cm
(b) Diameter of semi circle = 2.8 cm