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Ncert Solutions for Mensurations Class 8 CBSE Part 1


In this page we have NCERT book Solutions for Class 8th Maths:Mensurations 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

So

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.

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].


Answer: Area of the rectangular part = length x breadth

= 20 x 7 = 140 sq m

Area of Semicircular portions: = ? x r2

Here ?=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 ? 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.

 

Answer

 

  1.  In the first case, Perimeter of the shape

    = Perimeter of Semicircular part + Diameter of the part

    = ?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

    = ?r +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.

 


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