Question

"Question 8 Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?

Class 9 - Math - Surface Areas and Volumes Page 213"

Answer 1

Dimensions of the box- like structure = 4m × 3m × 2.5

Tarpaulin is only required for all the four sides and top.

Thus, Tarpaulin required = 2(l+b)×h + lb
  = {2(4+3)×2.5 + 4×3} m2
  = 47m2

Answer 2

Given:

Dimensions of the shelter

l=4m , b= 3m ,h= 2.5m

 

Required area of Tarpaulin to  make the shelter=

Area of four sides + area of  the top of the car

Tarpaulin required = 2(l+b)×h + lb

= [2(4+3)×2.5 + 4×3]

= (35+12)
= 47m²

Hence,Tarpaulin required to make the shelter=47m²

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Hope this will help you...