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The floor of a rectangular hall has a perimeter $ 250 \mathrm{~m} $. If the cost of painting the four walls at the rate of $ Rs. 10 \mathrm{per} \mathrm{m}^{2} $ is $ Rs. 15000 $, find the height of the hall.
[Hint: Area of the four walls = Lateral surface area.]
Given:
The floor of a rectangular hall has a perimeter \( 250 \mathrm{~m} \) and the cost of painting the four walls at the rate of \( Rs. 10 \mathrm {per}\mathrm {m}^{2} \) is \( Rs. 15000 \).
To do:
We have to find the height of the hall.
Solution:
Let the length, breadth and height of the rectangular hall be $l, b$ and $h$ respectively.
We know that,
The perimeter of the rectangle $=2(l+b)$
We have,
Area of the four walls = Lateral surface area and
The perimeter of the rectangle is $250\ m$.
Therefore,
The area of the wall $=2lh+bh$
This implies,
The area of the four walls $=2(l+b)h$
$=2(l+b)h$
$=250h\ m^2$
We also have,
The cost of painting the wall per square meter $=Rs.\ 10$
The cost of painting the four walls per square meter $=$ area of the four walls $\times$ the cost of painting the wall per square meter
This implies,
$=250h\ m^2\times Rs.\ 10$
$=Rs.\ 2500h\ m^2$
We have,
The cost of painting the four walls at the rate of \( Rs. 10 \mathrm{per} \mathrm{m}^{2} \) $=Rs.\ 15000$
This implies,
$Rs.\ 15000=Rs.\ 2500h$
$h=\frac{Rs.\ 2500}{Rs.\ 15000}$
$h=6$
Therefore,
The height of the wall is $6\ m$.
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