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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). - CBSE Class 9 - Mathematics

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Question

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?

Solution

Length (l) of shelter = 4 m

Breadth (b) of shelter = 3 m

Height (h) of shelter = 2.5 m

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

Area of Tarpaulin required = 2(lh + bh) + l b

= [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.

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Mathematics Class 9 (2018 to Current)
Chapter 13: Surface Area and Volumes
Ex. 13.10 | Q: 8 | Page no. 213

Video TutorialsVIEW ALL [1]

Solution 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). Concept: Surface Area of a Cuboid and a Cube.
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