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The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $13 m, 14 m$ and $15 m$ . The advertisements yield an earning of $Rs 2000$ per $m^{2}$ per year. A company hired one of its walls for 6 months. How much rent did it pay?

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Rs . 82000

Rs . 43000

Rs . 84000

Rs . 85000

answer is C.

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Detailed Solution

Given sides of triangle are

13 m, 14 m

and

15 m

.
Semi perimeter is,

s = \frac{a + b + c}{2}

where a, b, c are the sides of triangle.

\Rightarrow s = \frac{13 + 14 + 15}{2}

\Rightarrow s = \frac{42}{2}

\Rightarrow s = 21

Now, area of triangle by Heron’s Formula is,

A = \sqrt{s (s - a) (s - b) (s - c)}

\Rightarrow A = \sqrt{21 (21 - 13) (21 - 14) (21 - 15)}

\Rightarrow A = \sqrt{21 (8) (7) (6)}

\Rightarrow A = \sqrt{7056}

\Rightarrow A = 84 m^{2}

Analyse further,

1 m^{2}

costs

Rs . 2000

per year.
Therefore, cost for 6 months is,

\Rightarrow \frac{1}{2} \times 84 m^{2} \times Rs . 2000

\Rightarrow Rs . 84000

Thus, the company will have to pay

= Rs . 84000

as rent.
Therefore, option (3) is correct.

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