An isosceles triangle has a perimeter of 30 , | vedclass

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Question

(A) The equal sides of the triangle are $12 \, cm$ each.Let the third side be $x \, cm$.Since the perimeter is $30 \, cm$,we have:$12 \, cm + 12 \, cm

Vedclass · Accepted Answer

$9 \sqrt{15} \, cm^2$

Vedclass · Answer

(A) The equal sides of the triangle are $12 \, cm$ each.Let the third side be $x \, cm$.Since the perimeter is $30 \, cm$,we have:$12 \, cm + 12 \, cm + x \, cm = 30 \, cm$$24 \, cm + x = 30 \, cm$$x = 30 - 24 = 6 \, cm$.Now,the semi-perimeter $s = \frac{	ext{Perimeter}}{2} = \frac{30}{2} = 15 \, cm$.Using Heron's formula,the area of the triangle is:$	ext{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$	ext{Area} = \sqrt{15(15-12)(15-12)(15-6)} \, cm^2$$	ext{Area} = \sqrt{15 	imes 3 	imes 3 	imes 9} \, cm^2$$	ext{Area} = \sqrt{15 	imes 9 	imes 9} \, cm^2 = 9 \sqrt{15} \, cm^2$.Thus,the area of the triangle is $9 \sqrt{15} \, cm^2$.

An isosceles triangle has a perimeter of $30 \, cm$ and each of the equal sides is $12 \, cm$. Find the area of the triangle.

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