Three sides of a triangular field are 20, m,2 | vedclass

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(C) Let the sides of the triangle be $a = 20\, m$,$b = 21\, m$,and $c = 29\, m$.First,calculate the semi-perimeter $(s)$:$s = \frac{a + b + c}{2} = \f

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(C) Let the sides of the triangle be $a = 20\, m$,$b = 21\, m$,and $c = 29\, m$.First,calculate the semi-perimeter $(s)$:$s = \frac{a + b + c}{2} = \frac{20 + 21 + 29}{2} = \frac{70}{2} = 35\, m$.Using Heron's formula,the area of the triangle is given by:Area $= \sqrt{s(s - a)(s - b)(s - c)}$.Substituting the values:Area $= \sqrt{35(35 - 20)(35 - 21)(35 - 29)}$Area $= \sqrt{35 	imes 15 	imes 14 	imes 6}$Area $= \sqrt{(5 	imes 7) 	imes (3 	imes 5) 	imes (2 	imes 7) 	imes (2 	imes 3)}$Area $= \sqrt{5^2 	imes 7^2 	imes 3^2 	imes 2^2}$Area $= 5 	imes 7 	imes 3 	imes 2 = 210\, m^2$.

Three sides of a triangular field are $20\, m$,$21\, m$,and $29\, m$ long,respectively. The area of the field is.......$m^2$.

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