Question

A classroom is 6.8 m long, 3.5 m wide, and 270 cm high. Find the cost of plastering the 4 walls and ceiling at the rate of 712 per m2.​

Answer 1

Answer :-

$\:\\\large{\boxed{\sf{Firstly,\;let's\;understand\;the\;concept\;used\;:-}}}$

Here the concept of CSA of Cuboid and Area of Rectangle has been used. We see that we are given the dimensions of cuboid. Now we can find the 4 areas of walls as CSA of Cuboid. Then, we can find the Area of Ceiling as Area of Rectangular Plane. Now we can add these both to find the total area to be plastered. And the we can multiply it with the rate to find total cost of plastering.

Let's do it !!

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★ Formula Used :-

$\:\\\large{\boxed{\sf{CSA\;of\;Cuboid\;=\;\bf{2\:\times\:(Length\:+\:Breadth)\:\times\:Height}}}}$

$\:\\\large{\boxed{\sf{Area\;of\;Rectangle\;=\;\bf{Length\;\times\;Breadth}}}}$

$\:\\\large{\boxed{\sf{Area\;to\;be\;Plastered\;=\;\bf{CSA\;+\;Area\;of\;Rectangle}}}}$

$\:\\\large{\boxed{\sf{Total\;cost\;of\;Plastering\;=\;\bf{Area\;to\;be\;Plastered\;\times\;Rate_{(in\:per\:m^{2})}}}}}$

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★ Question :-

A classroom is 6.8 m long, 3.5 m wide, and 270 cm high. Find the cost of plastering the 4 walls and ceiling at the rate of 712 per m².

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★ Solution :-

Given,

» Length of Classroom = L = 6.8 m

» Breadth of Classroom = B = 3.5 m

» Height of Classroom = H = 270 cm = 2.7 m

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~ For the CSA of Classroom :-

This CSA is for four walls excluding ceiling and base.

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:CSA\;of\;Cuboid\;=\;\bf{2\:\times\:(Length\:+\:Breadth)\:\times\:Height}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:CSA\;of\;Cuboid\;=\;\bf{2\:\times\:(Length\:+\:Breadth)\:\times\:Height}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:CSA\;of\;Cuboid\;=\;\bf{2\:\times\:(6.8\:+\:3.5)\:\times\:2.7}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:CSA\;of\;Cuboid\;=\;\bf{2\:\times\:27.81\;\:=\;\:\underline{\underline{55.62\;\:cm^{2}}}}}}$

$\:\\\large{\boxed{\boxed{\tt{Area\;\;of\;\;Four\;\;Walls\;\;=\:\;\bf{55.62\;\;cm^{2}}}}}}$

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~ For the Area of Ceiling :-

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:Area\;of\;Rectangle\;=\;\bf{Length\;\times\;Breadth}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:Area\;of\;Rectangle\;=\;\bf{6.8\;\times\;3.5\:\;=\;\:\underline{\underline{23.8\;\:cm^{2}}}}}}$

$\:\\\large{\boxed{\boxed{\tt{Area\;\;of\;\;Ceiling\;\;=\:\;\bf{23.8\;\;cm^{2}}}}}}$

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~ For the Total Area to be Plastered :-

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:Area\;to\;be\;Plastered\;=\;\bf{CSA\;+\;Area\;of\;Rectangle}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:Area\;to\;be\;Plastered\;=\;\bf{55.62\;+\;23.8\;\:=\;\:\underline{\underline{79.42\;\:cm^{2}}}}}}$

$\:\\\large{\boxed{\boxed{\tt{Total\;\;Area\;\;for\;\;Plastering\;\;=\:\;\bf{79.42\;\;cm^{2}}}}}}$

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~ For the Total Cost of Plastering :-

$\:\\\qquad\large{\sf{:\Longrightarrow\;\;\:Total\;cost\;of\;Plastering\;=\;\bf{Area\;to\;be\;Plastered\;\times\;Rate_{(in\:per\:m^{2})}}}}$

$\:\\\qquad\large{\sf{:\Longrightarrow\;\;\:Total\;cost\;of\;Plastering\;=\;\bf{79.42\;\times\;712\:\;=\:\;\underline{\underline{Rs.\;\:56547.04}}}}}$

$\:\\\large{\underline{\underline{\rm{Thus,\;total\;cost\;of\;Plastering\;is\;\:\boxed{\bf{Rs.\;\; 56547.04}}}}}}$

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★ More to know :-

• Volume of Cuboid = Length × Breadth × Height

• Volume of Cube = (Side)³

• Perimeter of Rectangle = 2(L + B)

• Volume of Cylinder = πr²h

• Volume of Cone = ⅓ × πr²h

• Volume of Hemisphere = ⅔ × πr³

Answer 2

Answer in the attachment ↑↑