Math, asked by soumyadeepmah5531, 10 months ago

Ravish wanted to make a temporary shelter for his 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?

Answers

Answered by nikitasingh79
5

Given:  Dimensions of the shelter :  

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

Required area of Tarpaulin to  make the shelter = Area of four sides + area of  the top of the car

Tarpaulin required to make the shelter = 2(l + b)× h + lb

= [2(4 + 3) × 2.5 + 4 × 3]

= (35 + 12)

= 47m²

Hence, 47 m² tarpaulin would be required to make the shelter.

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Answered by Anonymous
6

Given:  

  • L = 4 m
  • b = 3 m
  • h = 2.5 m

Required area of Tarpaulin to  make the shelter = Area of four sides + area of  the top of the car

Tarpaulin needed to make the shelter = 2(l + b)× h + lb

putting the value

= [2(4 + 3) × 2.5 + 4 × 3]

= (35 + 12)

= 47m²

Hence, 47 m² tarpaulin would be needed to make the shelter.

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