The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22m and 120 m. The advertisements yield an earning of Rs. 5000 per m2 per year. A company hired both walls for 3 months. How much rent did it pay?

Foundation

Mathematics Foundation

Area of Triangle

Question

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22m and 120 m. The advertisements yield an earning of Rs. 5000 per m² per year. A company hired both walls for 3 months. How much rent did it pay?

Rs 2200000

Rs 3300000

Rs 1100000

Rs 6600000

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

The lengths of the sides of the walls are 122 m, 22 m and 120 m.
We have,
1222 = 1202 + 202 = 14884
So, walls are in the form of a right triangle.
$∴$    Area of two walls = 2 x $(\frac{1}{2} \times Base \times Height)$
⇒   Area of two walls = 2 x $(\frac{1}{2} \times 120 \times 22)$ = 2640 m²
We have,
       Yearly rent = Rs 5000 per m²
$∴$    Monthly rent = Rs $(\frac{5000}{12})$ per m²
Hence, rent paid by the company for 3 months = Rs $(\frac{5000}{12} \times 3 \times 2640)$ = Rs 3300000.

Please subscribe our Youtube channel to unlock this solution.