Parveen wanted to make a temporary shelter for her car, by making box-like structure with tarpoulin 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 $4 m \times 3 m$ ?

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Solution

Given that $l = 4 m; b = 3 m; h = 2.5 m$

$\Rightarrow$ Required Area $= l b + 2 b h + 2 l h$

$\Rightarrow$ Required Area $= (4 \times 3) + 2 (3 \times 2.5) + 2 (4 \times 2.5)$

$\Rightarrow$ Required Area $= (12 + 15 + 20) m^{2}$

$\Rightarrow$ Required Area $= 47 m^{2}$

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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?

Q. Question 8
Parveen wanted to make a temporary shelter for her car, by making a box-like structure with a 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 many tarpaulins would be required to make the shelter of height 2.5m, with base dimensions

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Q. Parween 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

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, with base dimensions

4 m \times 3 m

Given that l=4m;b=3m;h=2.5m⇒ Required Area =lb+2bh+2lh⇒ Required Area =(4×3)+2(3×2.5)+2(4×2.5)⇒ Required Area =(12+15+20)m2⇒ Required Area =47m2

Given that $l = 4 m; b = 3 m; h = 2.5 m$

$\Rightarrow$ Required Area $= l b + 2 b h + 2 l h$

$\Rightarrow$ Required Area $= (4 \times 3) + 2 (3 \times 2.5) + 2 (4 \times 2.5)$

$\Rightarrow$ Required Area $= (12 + 15 + 20) m^{2}$

$\Rightarrow$ Required Area $= 47 m^{2}$