Mensuration - Solutions 1
CBSE Class –VIII Mathematics
NCERT Solutions
CHAPTER - 11
Mensuration (Ex. 11.1)
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?

Which field has a larger area?
Ans. Given: 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







Hence, the breadth of the rectangular field is 40 m.
Now, Area of Square field= (Side)2
= (60)2 sq.m = 3600 sq.m
= (60)2 sq.m = 3600 sq.m
Area of Rectangular field = (length
breadth)
= 80
40 sq. m = 3200 sq. m

= 80

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

Ans. Side of a square plot = 25 m

Length and Breadth of the house is 20 m and 15 m respectively

= 20
15 = 300 m2

Area of garden = Area of square plot – Area of house
= (625 – 300) = 325 m2



= 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]
[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

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

= (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


= 

= 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.



Ans. (a) Radius = 

= 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

Radius =
= 1.4 cm

Radius =

Circumference of semi circle = 

=
4.4 cm


Total distance covered by the ant= (1.5 + 2.8 + 1.5 + 4.4)
10.2 cm

(c) Diameter of semi circle = 2.8 cm

Radius = 

= 1.4 cm
Circumference of semi circle = 

= 
4.4 cm


Total distance covered by the ant= (2 + 2 + 4.4) = 8.4 cm
Hence for figure (b) food piece, the ant would take a longer round.