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 4m x 3 m?

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Answer 1

Answer:

?4?

Step-by-step explanation:

this is your answer hope it will help you.

Answer 2

$\large{\red{\underline{\underline{\textsf{\maltese\:{\red{Given:}}}}}}}$

$\sf{Length\:(l)\:of \:the \:shelter = 4m}$

$\sf{Breadth\:(b) \:of \:the \:shelter = 3m}$

$\sf{Height\:(h) \:of \:the \:shelter = 2.5m}$

$\large{\red{\underline{\underline{\textsf{\maltese\:{\red{To\:Find:}}}}}}}$

$\sf{Tarpaulin\: would \:be\: required\: to\: make\: the \:shelter}$

$\large{\red{\underline{\underline{\textsf{\maltese\:{\red{Concept:}}}}}}}$

$\sf{Tarpaulin \:will\: be\: required\: for\: the\: top\: and\: four}$ $\sf{wall\: sides\: of\: the \:shelter.}$

$\sf{Using\: formula, }$

$\sf{Area \:of \:tarpaulin \:required = 2(lh+bh)+lb}$

$\large{\red{\underline{\underline{\textsf{\maltese\:{\red{Solution:}}}}}}}$

$\sf{On \:substituting \:the\: values\: in \:the \:formula,\: we\: get}$

$\sf{= [2(4×2.5+3×2.5)+4×3] m^2}$

$\sf{= [2(10+7.5)+12]m^2}$

$\sf{= 47\:m^2}$

$\sf{Therefore, \:47 \:m^2 \:tarpaulin \:will \:be\: required.}$