If the sides of a triangular field are 290 , | vedclass

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(C) The sides of the triangle are $a = 290 \, m$,$b = 290 \, m$,and $c = 400 \, m$.First,calculate the semi-perimeter $(s)$:$s = \frac{a + b + c}{2} =

Vedclass · Accepted Answer

$42000$

Vedclass · Answer

(C) The sides of the triangle are $a = 290 \, m$,$b = 290 \, m$,and $c = 400 \, m$.First,calculate the semi-perimeter $(s)$:$s = \frac{a + b + c}{2} = \frac{290 + 290 + 400}{2} = \frac{980}{2} = 490 \, m$.Using Heron's formula,Area = $\sqrt{s(s-a)(s-b)(s-c)}$.Area = $\sqrt{490(490-290)(490-290)(490-400)}$.Area = $\sqrt{490 	imes 200 	imes 200 	imes 90}$.Area = $\sqrt{490 	imes 90 	imes 200^2}$.Area = $200 	imes \sqrt{44100}$.Area = $200 	imes 210 = 42000 \, m^2$.

If the sides of a triangular field are $290 \, m$,$290 \, m$,and $400 \, m$,find the area of the field in $m^2$.

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