Find the area of a triangle whose sides are 3 | vedclass

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Question

(B) Let the sides of the triangle be $a = 33 \, cm$,$b = 34 \, cm$,and $c = 65 \, cm$.First,calculate the semi-perimeter $s$:$s = \frac{a + b + c}{2}

Vedclass · Accepted Answer

$264$

Vedclass · Answer

(B) Let the sides of the triangle be $a = 33 \, cm$,$b = 34 \, cm$,and $c = 65 \, cm$.First,calculate the semi-perimeter $s$:$s = \frac{a + b + c}{2} = \frac{33 + 34 + 65}{2} = \frac{132}{2} = 66 \, cm$.Using Heron's formula,the area of the triangle is $\sqrt{s(s - a)(s - b)(s - c)}$.Area $= \sqrt{66(66 - 33)(66 - 34)(66 - 65)}$Area $= \sqrt{66 	imes 33 	imes 32 	imes 1}$Area $= \sqrt{(2 	imes 33) 	imes 33 	imes (16 	imes 2) 	imes 1}$Area $= \sqrt{33^2 	imes 2^2 	imes 4^2} = 33 	imes 2 	imes 4 = 264 \, cm^2$.

Find the area of a triangle whose sides are $33 \, cm$,$34 \, cm$,and $65 \, cm$. (in $, cm^2$)

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