The area of a triangle whose sides are 15 , m | vedclass

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(C) Let the sides of the triangle be $a = 15 \, m$,$b = 16 \, m$,and $c = 17 \, m$.First,calculate the semi-perimeter $s$:$s = \frac{a + b + c}{2} = \

Vedclass · Accepted Answer

$24 \sqrt{21} \, m^{2}$

Vedclass · Answer

(C) Let the sides of the triangle be $a = 15 \, m$,$b = 16 \, m$,and $c = 17 \, m$.First,calculate the semi-perimeter $s$:$s = \frac{a + b + c}{2} = \frac{15 + 16 + 17}{2} = \frac{48}{2} = 24 \, m$.Using Heron's formula,the area of the triangle is given by:$	ext{Area} = \sqrt{s(s - a)(s - b)(s - c)}$Substitute the values:$	ext{Area} = \sqrt{24(24 - 15)(24 - 16)(24 - 17)}$$	ext{Area} = \sqrt{24 	imes 9 	imes 8 	imes 7}$$	ext{Area} = \sqrt{12096} = \sqrt{576 	imes 21} = 24 \sqrt{21} \, m^{2}$.

The area of a triangle whose sides are $15 \, m$,$16 \, m$,and $17 \, m$ is

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