  # NCERT Solutions for Class 8 Maths Chapter 11 (Ex 11.1) Mensuration

## NCERT Solutions for Class 8 Chapter 11 Mensuration -Free PDF Download

Free PDF download of NCERT Solutions Maths Class 8 Solutions Chapter 11 – Mensuration solved by Expert Maths Teachers on CoolGyan.Org. All Chapter 11 – Mensuration Questions with Solutions for NCERT to help you to revise complete Syllabus and Score More marks.
Maths Revision Notes for Class 8

 Chapter Name Mensuration Chapter Chapter 11 Exercise Exercise 11.1 Class Class 8 Subject Maths NCERT Solutions Board CBSE TEXTBOOK CBSE NCERT Category NCERT Solutions

# NCERT SOLVED

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) = 24022402 (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 = Diameter2Diameter2 = 7272= 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.