### 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

2(l+b) = 4 X Side

(80 + b) =

(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

= (60)2 sq.m = 3600 sq.m

Area of Rectangular field = (length breadth)

= 80 40 sq. m = 3200 sq. m

= 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

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]

[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 = = = 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.

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*
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.