Question

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Answer 1

Answer :-

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of TSA and Area of Rectangle has been used. We know that the base area of the boxed type shelter for car will be equal to the area of rectangle. And we know that base is left empty since their is no tarpaulin used. First we can find the TSA of Cuboid (shelter) and then we can subtract area of base from it. This is will give the area of Tarpaulin used.

Let's do it !!

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★ Equations Used :-

$\\\;\boxed{\sf{TSA\;of\;Cuboid\;=\;\bf{2(LB\;+\;BH\;+\;LH)}}}$

$\\\;\boxed{\sf{Area\;of\;Rectangle\;=\;\bf{L\;\times\;B}}}$

$\\\;\boxed{\sf{Area\;of\;Tarpaulin\;Used\;=\;\bf{TSA\;of\;Cuboid\;-\;Area\;of\;Base}}}$

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★ Solution :-

Given,

» Length of the Shelter = L = 4 m

» Breadth of the Shelter = B = 3 m

» Height of the Shelter = H = 2.5 m

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~ For the TSA of the Shelter ::

$\\\;\;\;\sf{:\rightarrow\;\;TSA\;of\;Cuboid\;=\;\bf{2(LB\;+\;BH\;+\;LH)}}$

$\\\;\;\;\sf{:\rightarrow\;\;TSA\;of\;Cuboid_{(Shelter)}\;=\;\bf{2[(4\;\times\;3)\;+\;(3\;\times\;2.5)\;+\;(4\;\times\;2.5)]}}$

$\\\;\;\;\sf{:\rightarrow\;\;TSA\;of\;Cuboid_{(Shelter)}\;=\;\bf{2(29.5)}}$

$\\\;\;\;\bf{:\rightarrow\;\;TSA\;of\;Cuboid_{(Shelter)}\;=\;\bf{59\;\;m^{2}}}$

$\\\;\underline{\boxed{\tt{TSA\;of\;Shelter\;=\;\bf{59\;\;m^{2}}}}}$

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~ For the Area of Tarpaulin used ::

$\\\;\;\;\sf{:\mapsto\;\;Area\;of\;Tarpaulin\;Used\;=\;\bf{TSA\;of\;Cuboid_{(Shelter)}\;-\;Area\;of\;Rectangle_{(Base)}}}$

Also,

• Area of Base = Area of Rectangle

And,

$\\\;\;\;\sf{:\rightarrow\;\;Area\;of\;Rectangle_{(Base)}\;=\;\bf{L\;\times\;B}}$

By applying the values, we get,

$\\\;\;\;\sf{:\mapsto\;\;Area\;of\;Tarpaulin\;Used\;=\;\bf{TSA\;of\;Cuboid\;-\;(Length\;\times\;Breadth)}}$

$\\\;\;\;\sf{:\mapsto\;\;Area\;of\;Tarpaulin\;Used\;=\;\bf{59\;-\;(4\;\times\;3)}}$

$\\\;\;\;\sf{:\mapsto\;\;Area\;of\;Tarpaulin\;Used\;=\;\bf{59\;-\;12}}$

$\\\;\;\;\bf{:\mapsto\;\;Area\;of\;Tarpaulin\;Used\;=\;\bf{47\;\;m^{2}}}$

$\\\;\large{\underline{\underline{\rm{Thus,\;area\;of\;tarpaulin\;used\;is\;\;\boxed{\bf{47\;\;m^{2}}}}}}}$

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★ More Formulas to know :-

$\\\;\sf{\leadsto\;\;CSA\;of\;Cuboid\;=\;2(L\;+\;B)\;\times\;H}$

$\\\;\sf{\leadsto\;\;Volume\;of\;Cuboid\;=\;Length\;\times\;Breadth\;\times\;Height}$

$\\\;\sf{\leadsto\;\;Volume\;of\;Square\;=\;(Side)^{3}}$

$\\\;\sf{\leadsto\;\;TSA\;of\;Square\;=\;6\;\times\;(Side)^{2}}$

$\\\;\sf{\leadsto\;\;CSA\;of\;Square\;=\;4\;\times\;(Side)^{2}}$

$\\\;\sf{\leadsto\;\;Area\;of\;Square\;=\;(Side)^{2}}$