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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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