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(C) Let the sides of the triangle be $a = 18 \, cm$,$b = 10 \, cm$,and $c$ be the third side.Given that the perimeter $(2s) = 42 \, cm$.Therefore,the

Vedclass · Accepted Answer

$21 \sqrt{11} \, cm^2$

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(C) Let the sides of the triangle be $a = 18 \, cm$,$b = 10 \, cm$,and $c$ be the third side.Given that the perimeter $(2s) = 42 \, cm$.Therefore,the semi-perimeter $s = \frac{42}{2} = 21 \, cm$.The third side $c = 42 - (18 + 10) = 42 - 28 = 14 \, cm$.Using Heron's formula,the area of a triangle $= \sqrt{s(s-a)(s-b)(s-c)}$.Substituting the values: Area $= \sqrt{21(21-18)(21-10)(21-14)} \, cm^2$.Area $= \sqrt{21 	imes 3 	imes 11 	imes 7} \, cm^2$.Area $= \sqrt{(3 	imes 7) 	imes 3 	imes 11 	imes 7} \, cm^2 = \sqrt{3^2 	imes 7^2 	imes 11} \, cm^2$.Area $= 3 	imes 7 \sqrt{11} \, cm^2 = 21 \sqrt{11} \, cm^2$.

Find the area of a triangle two sides of which are $18 \, cm$ and $10 \, cm$ and the perimeter is $42 \, cm$.

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