Math, asked by sivaprakash1120, 11 months ago

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 x 3 m?

Answers

Answered by amitnrw
55

Answer:

tarpaulin would be required to make the shelter = 47 m²

Step-by-step explanation:

Total Area required

= 4 * 2.5  + 4 * 2.5  + 3 * 2.5  + 3 * 2 .5  + 4 * 3

= 10 + 10 + 7.5 + 7.5 + 12

= 47

stitching margins are very small, and therefore negligible

tarpaulin would be required to make the shelter = 47 m²

Answered by 23saurabhkumar
45

Answer:

Area of the tarpaulin required, A = 47 square metre

Step-by-step explanation:

In the question,

There are 5 faces in total with the front face capable of rolling up.

Height of the shelter, h = 2.5 m

Length of the shelter, l = 4 m

Width of the shelter, w = 3 m

So,

Area of the cover required, A = 2(h x w) + 2(l x h) + (l x w)

A = 2(2.5 x 3) + 2(4 x 2.5) + (4 x 3)

A = 15 + 20 + 12

A = 47 m square.

Therefore, the area of the tarpaulin required to make the entire cover is equal to 47 metre square.

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