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(D) Given,the three sides of the triangle are $a = 56 \ cm$,$b = 60 \ cm$,and $c = 52 \ cm$.First,calculate the semi-perimeter $(s)$ of the triangle:$

Vedclass · Accepted Answer

$1344$

Vedclass · Answer

(D) Given,the three sides of the triangle are $a = 56 \ cm$,$b = 60 \ cm$,and $c = 52 \ cm$.First,calculate the semi-perimeter $(s)$ of the triangle:$s = \frac{a + b + c}{2} = \frac{56 + 60 + 52}{2} = \frac{168}{2} = 84 \ cm$.Using Heron's formula,the area of the triangle is given by $\sqrt{s(s-a)(s-b)(s-c)}$.Area $= \sqrt{84(84 - 56)(84 - 60)(84 - 52)}$Area $= \sqrt{84 	imes 28 	imes 24 	imes 32}$Area $= \sqrt{(4 	imes 3 	imes 7) 	imes (4 	imes 7) 	imes (8 	imes 3) 	imes (8 	imes 4)}$Area $= \sqrt{4^4 	imes 7^2 	imes 3^2 	imes 8^2} = 4^2 	imes 7 	imes 3 	imes 8 = 16 	imes 168 = 1344 \ cm^2$.Thus,the area of the triangle is $1344 \ cm^2$.

The sides of a triangle are $56 \ cm$,$60 \ cm$,and $52 \ cm$ long. Then the area of the triangle is (in $cm^2$)

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