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Question

(A) Given,the perimeter of the triangle is $32 \, cm$. Let the sides be $a = 8 \, cm$ and $b = 11 \, cm$.The third side $c$ is calculated as:$c = 32 \

Vedclass · Accepted Answer

$8 \sqrt{30} \, cm^2$

Vedclass · Answer

(A) Given,the perimeter of the triangle is $32 \, cm$. Let the sides be $a = 8 \, cm$ and $b = 11 \, cm$.The third side $c$ is calculated as:$c = 32 \, cm - (8 + 11) \, cm = 32 \, cm - 19 \, cm = 13 \, cm$.The semi-perimeter $s$ is:$s = \frac{	ext{Perimeter}}{2} = \frac{32}{2} \, cm = 16 \, cm$.Now,calculate $(s - a)$,$(s - b)$,and $(s - c)$:$s - a = 16 - 8 = 8 \, cm$$s - b = 16 - 11 = 5 \, cm$$s - c = 16 - 13 = 3 \, cm$Using Heron's formula,the area of the triangle is:$	ext{Area} = \sqrt{s(s - a)(s - b)(s - c)}$$	ext{Area} = \sqrt{16 	imes 8 	imes 5 	imes 3} \, cm^2$$	ext{Area} = \sqrt{16 	imes 120} \, cm^2$$	ext{Area} = 4 	imes \sqrt{4 	imes 30} \, cm^2$$	ext{Area} = 4 	imes 2 \sqrt{30} \, cm^2 = 8 \sqrt{30} \, cm^2$.

Find the area of a triangle,two sides of which are $8 \, cm$ and $11 \, cm$ and the perimeter is $32 \, cm$.

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