Question

The triangular side walls of a flyover have been used for advertisements. The sides of
the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield a earning of 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?​

Answer 1

Solution :-

☆ Area of ∆ ABC = 1/2 b × h

→ 1/2 120 × 22

→ 1320m²

○ 1m² = Rs 5000 per year.

○ 1320m² = Rs 5000 × 1320.

❃ Now In 12 month rent 5000 × 1320

○ 1 month = 5000 × 1320/12

○ For 3 month = 5000 × 1320 × 3/12

→ 55000/3

→ Rs 1650000 Ans.

Hope it helps you !!!

Answer 2

Given:-

• The sides of the walls are 122m, 22 m and 120 m.
• The advertisements yield a earning of 5000 per m^2 per year.
• A company hired one of its walls for 3 months.

To Find:-

• How much rent did it pay?

Solution:-

The perimeter of a triangle is equal to the sum of its three sides is denoted by 2S.

• 2s = (a+b+c)

•s = (a+b+c)/2

Using Formula:

$\sf \: semiperimeter \: of \: triangle = \frac{(a + b + c)}{2} \\ \\ \sf \: area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)}$

where,

• a, b and c are three sides of triangle

$\sf \: s = \frac{(a + b + c)}{2} \\ \\ \sf \: s = \frac{(122 + 22 + 120)}{2} \\ \\ \sf s = \frac{264}{2} = 132 \: m$

Now,

$\sf \: area \: of \: the \: wall = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \sf \: = \sqrt{132(132 - 122)(132 - 22)(132 - 120)} \\ \\ \sf \: = \sqrt{132 \times 10 \times 110 \times 12} \\ \\ \sf \: = \sqrt{11 \times 12 \times 10 \times 11 \times 10 \times 12} \\ \\ \sf \: = \sqrt{11 \times \times 12 \times 12 \times 10 \times 10} \\ \\ \sf \: = 11 \times 12 \times 10 \\ \\ \sf \: = 1320 {m}^{2}$

Now,

Earning on 1m^2 per year = 5000 rupees

Earning on 1320m^2 per year = 1320 * 5000 = 6600000 rupees

Now, earning in 1320m^2 in 12 months = 6600000 rupees

Earning in 3 months = 660000 *3/12 = 1650000 rupees

Hence,the rent paid by the company for 3 months is 1650000 rupees.