Mensuration Exercise 11.1

Class –VIII Mathematics (Ex. 11.1)

NCERT SOLUTION

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

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 field

= Perimeter of square field

 =  4  side

 m

Now Area of Square field

=

=  = 3600 m²

And Area of Rectangular field

= length  breadth = 80  40

= 3200

Hence, area of square field is larger.

NCERT Solutions for Class 8 Maths Exercise 11.1

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.

Ans.  Side of a square plot = 25 m

 Area of square plot =

=  = 625 m²

Length of the house = 20 m and

Breadth of the house = 15 m

 Area of the house = length  breadth

= 20  15 = 300 m²

Area of garden = Area of square plot

– Area of house

= 625 – 300 = 325 m²

 Cost of developing the garden per sq. m = ` 55

 Cost of developing the garden 325 sq. m = ` 55 325

= ` 17,875

Hence total cost of developing a garden around is ` 17,875.

NCERT Solutions for Class 8 Maths Exercise 11.1

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 = 20 m

Diameter of semi circle = 7 m

 Radius of semi circle =  = 3.5 m

Length of rectangular field

= 20 – (3.5 + 3.5) = 20 – 7 = 13 m

Breadth of the rectangular field = 7 m

 Area of rectangular field =

= 13  7 = 91

Area of two semi circles =

=  = 38.5 m²

Area of garden = 91 + 38.5 = 129.5 m²

Now Perimeter of two semi circles =  = 22 m

And Perimeter of garden

= 22 + 13 + 13

= 48 m

NCERT Solutions for Class 8 Maths Exercise 11.1

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. Given: Base of flooring tile = 24 cm

= 0.24 m

Corresponding height of a flooring tile

= 10 cm = 0.10 m

Now Area of flooring tile

= Base  Altitude

= 0.24  0.10

= 0.024 m²

 Number of tiles required to cover the floor

=

=

= 45000 tiles

Hence 45000 tiles are required to cover the floor.

NCERT Solutions for Class 8 Maths Exercise 11.1

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 = 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.