Question

the internal measure of cuboidal room are 12m× 8m×4m×.find the total cost of whitewashing all four walls of a room, if the cost of white washing is rs5per m^2 .what will be the cost of white washing if the ceiling of the room is also whitewashed.
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Answer 1

Answer:

$\textsf{\orange{the total cost of whitewashing all}}\\\textsf{\orange{four walls of a room}}=\red{1280}$

$\rule{300}{2}$

$\sf\green{Given}\begin{cases}\sf\red{lenght(L)=12m} \\ \sf\orange{width(B)=8m} \\\sf\red{hieght(H)=4 m}\\\sf\blue{Rate=Rs.5per \:m^2}\end{cases}$

$\huge\star\underline\textsf{to find:- }$

$\textsf{the total cost of whitewashing all }\\\textsf{four walls of a room.}$

$\textsf{area \: of \: four \: wall \: of \: room \:}$=

$\boxed{\sf{\red {perimeter \: of \: the \: base \times \: hieght \: of \: room}}}$

$\large\star\underline\textsf{Formula used:- }$

$\boxed{\sf = 2(l + b) \times h}$

$\huge\underline\textsf{Explantion:- }$

$\leadsto\sf2(12 + 8) \times 4$

$\leadsto\sf2 \times 20 \times 4 = 160m {}^{2}$

$\small\underline\textsf{hence, }$

$\sf cost \: of \: white \: washing \: four \: walls \\\sf of \: room = (160 \times 5) = 800$

$\rule{300}{2}$

$\sf area \: of \: ceiling \: is \: 12 \times 8 = 96m {}^{2}$

$\rule{300}{2}$

$\sf cost \: of \: white \: washing \: the \: cieling \\\bf = (96 \times 5) = 480$

$\sf so \: the \: total \: cost \: of \: white \: washing \: \\\bf (800 + 480) = 1280$

$\rule{300}{2}$

$\textsf{\orange{the total cost of whitewashing all}}\\\textsf{\orange{four walls of a room}}=\red{1280}$

$\rule{300}{2}$

Answer 2

Answer :

$\bold{Cost\:of\:whitewashing\:4\:walls = Rs.800}$

$\bold{Total\: Cost = Rs. 1280 }$

Step-by-step explanation:

$\bf\red{Given:-}\begin{cases}\sf\red{Length = 12\:m}\\\sf\green{Breadth=8\:m}\\\sf\orange{Height = 4\:m}\\\sf\green{Rate = Rs.5 \: per\:{m}^{2} }\end{cases}$

$\bf\red{To\:find:-}\begin{cases}\sf\green{Cost\:of\:whitewashing\;4\:walls=?}\\\sf\purple{Cost\:of\:whitewashing\:including\:ceiling=?}\end{cases}$

$\rule{300}{1}$

$\bigstar\sf\red{ We\:know\:that:-}$

${\boxed{\sf\green{Area\:of\:4\:walls = 2(l + b)\times h}}}$

$\longrightarrow\sf\green{Area\:of\:4\:walls = 2(12 + 8)\times 4 }\\\\\longrightarrow\sf\red{Area\:of\:4\:walls = 2 \times 20 \times 4} \\\\\longrightarrow\sf\purple{Area\:of\:4\:walls = 160\:{m}^{2} }$

$\bigstar\sf\green{\ Cost\:of\:whitewashing = Area \times Rate } \\\\\longrightarrow\sf\red{Cost\:of\:whitewashing =Rs. (160 \times 5) } \\\\\longrightarrow\underline{\boxed{\bold{Cost\:of\:whitewashing = Rs.800}}}$

$\rule{100}{2}$

$\sf\red{We\:know\:that,}$

$\bigstar \ \underline{\boxed{\bold{Area\:of\:ceiling = l \times b}}}$

$\longrightarrow\sf\green{Area\:of\:ceiling = 12 \times 8 }\\\\\longrightarrow\sf\orange{Area\:of\:ceiling = 96\:{m}^{2}}$

$\bigstar\bold{\ Cost = Area \times Rate} \\\\\longrightarrow\sf\red{Cost = Rs.(96\times 5) } \\\\\longrightarrow\underline{\boxed{\bold{Cost = Rs.480 }}}$

$\rule{100}{2}$

$\bigstar\sf{\ Total\:cost = Rs.(800 + 480)} \\\\\longrightarrow\underline{\boxed{\bold{Total\:cost = Rs.1280}}}$