2.

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 an

earning of 5000 per mº per year. A company hired one of its walls for 3 months. How

much rent did it pay?

# 2.

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Before, finding the answer. Let’s find out on how we can find the answer.[tex]\sf \sqrt{s (s-a) (s-b) (s-c)}[/tex]

Where,

S = Semi-perimeter

A = Side 1

B = Side 2

C = Side 3

___________________

Given :To find :Solution :Semi-perimeter = [tex]\sf \dfrac{a + b + c}{2}[/tex]

= [tex]\sf \dfrac{122 + 22 + 120}{2}[/tex]

= [tex]\sf \dfrac{264}{2}[/tex]

= 132 m

Area of Triangular side walls = [tex]\sf \sqrt{s (s-a) (s-b) (s-c)}[/tex]

= [tex]\sf \sqrt{132 (132-122) (132-22) (132-120)}[/tex]

= [tex]\sf \sqrt{132 \times (10) \times (110) \times (12)}[/tex]

= [tex]\sf \sqrt{1742400}[/tex]

= 1320 m²

∴ Area of the Wall will be 1320 m².

⇒ Then total earning will be = Total area × per m² cost.

Total Earning = 5000 × 1320

→ Rs. 66,00,000

⇒ So, further it is said that the company hired one of its walls for 3 months. Its rent for 3 months will be :

= Total earnings × 3/12

= 66,00,000 × 3/12

→ Rs. 16,50,000