Question

If the perimeter of an isosceles triangle is 64 and its equal sides are 20 . Then, find the area of an
isosceles triangle.

Answer 1

Given

• Perimeter = 64 Unit
• One of the Equal sides = 20 Unit

Explanation:

Let the third side of an isosceles triangle be x

• Two Equal sides 20 Unit and 20 Unit

$\maltese {\large{\pmb{\boxed{\underline{\sf{ Perimeter_{( \Delta )} = a +b+c }}}}}} \\ \\ \\ \colon\implies{\sf{ 64 = 20+20+x }} \\ \\ \\ \colon\implies{\sf{ 64 = 40+x}} \\ \\ \\ \colon\implies{\sf{ 64-40 = x }} \\ \\ \\ \colon\implies{\sf{ x = 24 \ Unit }}$

Therefore, Third side is 24 Unit (base).

Now, We can find the Area of the isosceles triangle as:-

$\maltese {\large{\pmb{\boxed{\underline{\sf{ Area_{( \Delta )} = \left[ \dfrac{Base}{4} \times \sqrt{ 4(Equal \ side)^2 -(Base)^2 } \right] }}}}}} \\ \\ \\ \colon\implies{\sf{ Area_{( \Delta )} = \left[ \cancel{ \dfrac{24}{4} } \times \sqrt{ 4(20)^2 -(24)^2 } \right] }} \\ \\ \\ \colon\implies{\sf{ Area_{( \Delta )} = 6 \times \sqrt{ 4(400) -(576) } }} \\ \\ \\ \colon\implies{\sf{ Area_{( \Delta )} = 6 \times \sqrt{ 1600 -576 } }} \\ \\ \\ \colon\implies{\sf{ Area_{( \Delta )} = 6 \times \sqrt{ 1024 } }} \\ \\ \\ \colon\implies{\sf{ Area_{( \Delta )} = 6 \times 32 }} \\ \\ \\ \colon\implies{\sf{ Area_{( \Delta )} = 192 \ Unit^2 }}$

Hence,

• The Area of the isosceles triangle is 192 unit².