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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} a year. A company hired one of its walls for 6 months. How much did it pay?

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} a year. A company hired one of its walls for 6 months. How much rent did it pay?

Answers (1)

[Rs. 84000]

Here \; sides\; of \; triangular\; walls\; are \; a = 14 m, b = 15 m \; and \; c = 13 m

We have to find the area of this triangle

Using Heron’s formula, semi-perimeter:

S=\frac{a+b+c}{2}=\frac{14+15+13}{2}=\frac{42}{2}=21m

The\; area\; of\; triangular\; wall =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}

=\sqrt{21\left ( 21-13\right )\left ( 21-14 \right )\left ( 21-15 \right )}

={\sqrt{21\times 8\times 7\times 6}}

= \sqrt{3\times 7\times 2\times 2\times 2\times 7\times 2\times 3}

= \sqrt{\left ( 2\times 2\times 2\times 2 \right )\times \left ( 3\times 3\times 7\times 7 \right )}

= 2\times 2\times 3\times 7=84m^{2}

Given that the advertisements yield earning per m2 for 1 year=RS 2000

Earnings\; per\; m^{2} \; per\; month =Rs.\left ( \frac{2000}{12} \right )

Earnings\; per\; m^{2}\; for\; 6\; months =Rs.\; \left ( \frac{2000}{12}\times 6 \right )

Earnings\; for\; 84 m^{2}\; for\; 6 \; months =Rs\left ( \frac{2000}{2}\times 84 \right )

Rent \; the\; company \; has\; to\; pay = Rs. 2000 \times 42 = Rs. \; 84000

Hence\; company\; has\; to\; pay\; Rs.\; 84000.

 

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