# NCERT Solutions for Class 8 Maths Exercise 11.1 ## myCBSEguide App

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NCERT solutions for Class 8 Maths Mensuration ## NCERT Solutions for Class 8 Maths Mensuration

###### 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.

## NCERT Solutions for Class 8 Maths Exercise 11.1

NCERT Solutions Class 8 Mathematics PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 8 Mathematics includes text book solutions from Class 8 Maths Book . NCERT Solutions for CBSE Class 8 Maths have total 16 chapters. 8 Maths NCERT Solutions in PDF for free Download on our website. Ncert class 8 solutions PDF and Maths ncert class 8 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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