NCERT Solutions for Mensuration Class 8 Maths Exercise 11.1

In this page we have NCERT Solutions for Mensuration Class 8 Maths Chapter 11 for EXERCISE 1 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Answer: Perimeter of Square   is given by
= 4 x side = 4 x 60 = 240 m
Area of Square = Side² = 60² = 3600 m2
Now Perimeter of Rectangle is given by
= 2(length + breadth)
Or, = 2(80 + breadth)
Now both the square and rectangle has the same perimeter
240=2(80 + breadth)
80 + breadth = 120
Breadth = 120-80 = 40 m
Area of rectangle = length x breadth = 80 x 40 = 3200 m2
Now it is clear that the area of the square field is greater than the area of the rectangular field.
Question 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.
Ncert Solutions for Mensurations Class 8
Answer: Area of the square plot = Side² = 25² = 625 m2
Area of the house construction part = length x breadth
= 20 x 15 = 300 m2
So, area of the garden = Area of square plot – area of house in construction=625-300=325 m2
Cost of developing the garden = Area  of Garden x Rate
= 300 x 55 = 16500 rupees
Question 3
 The shape of a garden is rectangular in the middle and semicircular 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) m].

Ncert Solutions for Mensurations Class 8
Answer: Area of the rectangular part = length x breadth
= 20 x 7 = 140 sq m
Area of Semicircular portions: = π x r2
Here π=22/7   and radius =7/2=3.5 m
So area of two Semicircular portions: =Area of circle= ? x r2
= (22/7) × 3.5 ×3.5= 37.5
So Total area of the garden=
Area of rectangular portion + Area of circle
=140+37.5=177.5 m2
Now the perimeter of shape= Perimeter of semicircular portion + Length+ Perimeter of semicircular portion + length
=Perimeter of Circle + 2 Length
= 2 π r + 40
=22+40=62 m

Question 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 m2? (If required you can split the tiles in whatever way you want to fill up the corners).
Answer: Area of the Parallelogram = base x height
= 24 x 10 = 240 cm2
So required Number of tiles = Area of the Floor/Area of the Tiles
= 1080 X100 X100/240  =45000
(area of floor is converted into square cm)
Question 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 c = 2πr, where r is the radius of the circle.
NCERT Solutions for Mensuration Class 8 Maths Chapter 11
  1.  In the first case, Perimeter of the shape = Perimeter of Semicircular part + Diameter of the part
    = π r + 2r
    =(22/7) × (1.4)  + 2.8
    =7.2 cm
  2. In the second case, Perimeter of the shape = Perimeter of Semicircular part + 2Length + Breath
    = πr +2 L + B
    =4.4 + 3 +2.8
    =10.2 cm
  3. In the third case, Perimeter of the shape = Perimeter of Semicircular part + 2(slant height)
    =4.4 + 4=8.4 cm

So, the food shape in (a) requires the ant to cover the least distance.

Download this assignment as pdf
link to this page by copying the following text
Also Read

Class 8 Maths Class 8 Science

Latest Articles
Synthetic Fibres and Plastics Class 8 Practice questions

Class 8 science chapter 5 extra questions and Answers

Mass Calculator

3 Fraction calculator

Garbage in Garbage out Extra Questions7