Question

praveen 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 . 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×5m ????​

Answer 1

Answer:

Height-2.5m

base- 5m×4m

formula-1/2×base×height

put it on formula- 1/2×5m×4m×2.5m=25m

Answer is 25m

Answer 2

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

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

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

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

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

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

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

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

$\sf{Using\: formula, }$

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

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

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

$\sf{= [2(4×2.5+3×2.5)+4×3] m^2}$

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

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

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