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 \mathrm{~m} $, with base dimensions $ 4 \mathrm{~m} \times 3 \mathrm{~m} $ ?

AcademicMathematicsNCERTClass 9

Given:

Parveen wanted to make a temporary shelter for her car by making a box-like structure with a tarpaulin that covers all four sides and the top of the tire car (with the front face as a flap which can be rolled up). 

The shelter of height $2.5\ m$ with base dimensions $4\ m \times 3\ m$.

To do:

We have to find the area of the tarpaulin required.

Solution:

Length of the base $(l) = 4\ m$

Breadth of the base $(b) = 3\ m$

Height $(h) = 2.5\ m$

Therefore,

Area of tarpaulin used $=$ Area of the walls $+$ Area of the roof

$=2 h(l+b)+lb$

$=2 \times 2.5(4+3)+4 \times 3$

$=5 \times 7+12$

$=35+12$

$=47 \mathrm{~m}^{2}$

The area of the tarpaulin required is $47\ m^2$.

raja
Updated on 10-Oct-2022 13:42:09

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