Question

The cost of papering the walls of a room 12 m long and 8 m broad at 80 paise
per sq. m. is Rs. 160. Find its height.
​

Answer 1

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{length \: of \: room \: = 12 \: m} \\ &\sf{breadth \: of \: room \: = 8 \: m} \\ &\sf{total \: cost \:of \: papering \: 4 \: walls = \: rs \: 160 }\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{height \: of \: the \: room}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  Let :-  \begin{cases} &\sf{height \: of \: room \: = \: h \: \: meter}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

☆ Given, cost of papering the wall is Rs 160 at 80 paisa per sq. m.

$\tt \:  \longrightarrow \: Area \: of \: wall \: to \: be \: papered \: \\ \: \: \: \: \: \: \: \: \: = \tt \:  \dfrac{cost \: of \: papering \: the \: wall}{cost \: of \: papering \: per \: meter} \\ \tt \:  \: \: \: \: \: = \dfrac{160}{0.80} \\ \: \: \: \: \: \: \: \: \: \: \: \: = 200 \: {m}^{2}$

$\tt \:  \longrightarrow \: Length \: of \: room \: = \: 12 \: m$

$\tt \:  \longrightarrow \: Breadth \: of \: room \: = \: 8 \: m$

$\bf \:  Area \: of \: four \: walls \: = 2(l \: + \: b )\: \times h$

$\tt \:  \longrightarrow \: 200 = 2 \times (12 + 8) \times h$

$\tt \:  \longrightarrow \: 200 = 40h$

$\tt \:  \longrightarrow \: h \: = \: \dfrac{200}{40}$

$\tt \:  \longrightarrow \: h \: = 5 \: m$

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More info:

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length ²+breadth ²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²

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