Find the area of a triangular field,two sides | vedclass

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(C) Let the sides of the triangle be $a = 510\,m$,$b = 520\,m$,and $c$. The perimeter $P = a + b + c = 1560\,m$. $510 + 520 + c = 1560 \implies 1030 +

Vedclass · Accepted Answer

$1,17,000\,m^2$,Rs. $24,57,000$

Vedclass · Answer

(C) Let the sides of the triangle be $a = 510\,m$,$b = 520\,m$,and $c$. The perimeter $P = a + b + c = 1560\,m$. $510 + 520 + c = 1560 \implies 1030 + c = 1560 \implies c = 530\,m$. The semi-perimeter $s = P/2 = 1560/2 = 780\,m$. Using Heron's Formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$. Area $= \sqrt{780(780-510)(780-520)(780-530)} = \sqrt{780 	imes 270 	imes 260 	imes 250}$. Area $= \sqrt{(78 	imes 10) 	imes (27 	imes 10) 	imes (26 	imes 10) 	imes (25 	imes 10)} = \sqrt{78 	imes 27 	imes 26 	imes 25 	imes 10^4}$. Area $= \sqrt{(3 	imes 26) 	imes (27) 	imes (26) 	imes (25) 	imes 10^4} = \sqrt{26^2 	imes 81 	imes 25 	imes 10^4} = 26 	imes 9 	imes 5 	imes 100 = 1,17,000\,m^2$. Cost $= 	ext{Area} 	imes 	ext{Rate} = 1,17,000 	imes 21 = 	ext{Rs. } 24,57,000$.

Find the area of a triangular field,two sides of which are $510\,m$ and $520\,m$ and the perimeter is $1560\,m$. Also,find the cost of the field at the rate of Rs. $21$ per $m^2$.

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