Question

Praveen 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 2.5m, with base dimensions 4m×3m?​

Answer 1

Answer:

Dimension of the non-like structure = 4m×3m×2.5m

Tarpaulin only required for all the

for sides and top.

Thus Tarpaulin required = 2(l+b)×h+lb

=[2(4+3)×2.5+4×3]m

2

=[35+12]m

2

=47m

2

HOPE THIS WILL HELP YOU

ATTACHMENT AND WRITTEN BY ME

BOTH ARE SAME

Answer 2

Let l, b and h be the length, breadth and height of the shelter.

Given:

• l = 4m
• b = 3m
• h = 2.5m

Tarpaulin will be required for the top and four wall sides of the shelter.

$\sf$

★ Using formula,

$\boxed{ \pink{ \sf \: Area \: of \: tarpaulin \: required = 2(lh+bh)+lb}}$

• On putting the values of l, b and h, we get

$\sf \longrightarrow \: [2(4×2.5+3×2.5)+4×3] \: {m}^{2}$

$\sf \longrightarrow \: [2(10+7.5)+12] \: {m}^{2}$

$\sf \longrightarrow 14 \times 2.5 + 12 \: {m}^{2}$

$\sf \longrightarrow \: 35.0 + 12 \: {m}^{2}$

$\sf \longrightarrow \: 47 \: {m}^{2}$

$\sf$

Therefore, 47 m² tarpaulin will be required.