Question

8.
N
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 3m?​

Answer 1

Answer:

YES

Step-by-step explanation:

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

$\huge \underline \mathbb {SOLUTION:-}$

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.

Using formula, Area of tarpaulin required = 2(lh + bh)+lb

Put the values of l, b and h, we get

= [2(4 × 2.5 + 3 × 2.5) + 4 × 3] m²

= [2(10 + 7.5) + 12]m²

= 47 m²

• Therefore, 47 m² tarpaulin will be required