Math, asked by neelamyadav851006200, 1 day 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 manojkumar528435
0

Answer:

47m²

Step-by-step explanation:

length (l) of the shelter = 4m

breadth (b) of the shelter = 3m

height (h) of the shelter = 2.5m

area of tarpaulin = area of cuboid - area of

Answered by razeenrauf4
1

Answer:

47m²

Step-by-step explanation:

Length = 4m

Breadth = 3m

Height = 2.5m

Area of tarpaulin used = Area of cuboid - Area of bottom

The shelter covers the 4 sides and the top, the bottom part is left open.

Area of cuboid = 2 (lb + bh + lh)

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

= 2 ( 12 + 7.5 + 10)

= 2 x 29.5

= 59m²

Area of bottom = lxb

= 4 x 3

= 12m²

Area of tarpaulin used = 59 - 12 = 47m²

Parveen requires 47m² of tarpaulin to make a temporary shelter for his car

Hope it help

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