Question

Two triangular walls of a flyover have been used for advertisements from both sides. The sides of each wall are 120 m, 110 m and 20 m. The advertisements yield an earning of R.s 100 per $m^2$ per year. Find the amount of revenue earned in one year. (Take $\sqrt{7}=2.65$)

• A. Rs. 3,97,500
• B. Rs. 3,47,500
• C. Rs. 5,73,300
• D. Rs. 4,73,500

Answer 1

The correct option is A Rs. 3,97,500

Given a = 120, b = 20, c = 110
$s=\frac{a+b+c}{2}=\frac{120+20+110}{2}=125$
$\therefore Areaofthewall=\sqrt{\left(s(s-a)(s-b)(s-c)\right)}$
$\Rightarrow \sqrt{\left(125(125-120)(125-20)(125-110)\right)}$
$\Rightarrow \sqrt{\left(125\left(5\right)\left(105\right)\left(15\right)\right)}$
$\Rightarrow \sqrt{(25\times 5\left(5\right)(15\times 7)\left(15\right))}$
$\Rightarrow 375\sqrt{7}$
$NowtheAreaofthetwowall=2\times 375\sqrt{7}=750\sqrt{7}=750\times 2.65=1987.5$
Rent of the both side of the walls $=2\times 1987.5\times 100=397500$