An isosceles triangle has perimeter $ 30 \mathrm{~cm} $ and each of the equal sides is $ 12 \mathrm{~cm} $. Find the area of the triangle.

Given:
 An isosceles triangle has perimeter $30\ cm$ and each of the equal sides is $12\ cm$.

To do:

Let us assume the third side of the triangle as $x$

We have,

Two sides with equal length of $12\ cm$

and perimeter as $30\ cm$

We know that,

Perimeter $P$ of a triangle with sides of length $a\ units, b\ units$ and $c\ units$ 

$P=(a+b+c)\ units$.

This implies,

$30\ cm=12\ cm+12\ cm+x\ cm$

$30\ cm=24\ cm+x\ cm$

This implies,

$x\ cm=30\ cm-24\ cm$

$x\ cm=6\ cm$

By Heron's formula:

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Since,

$S=\frac{a+b+c}{2}$

$S=\frac{12+12+6}{2}$

$S=\frac{30}{2}$

$S=15\ cm$

This implies,

$A=\sqrt{15(15-12)(15-12)(15-06)}$

$A=\sqrt{15(3)(3)(9)}$

$A=9\sqrt{15}\ cm^2$

Therefore,

The area of the triangle is $9\sqrt{15}\ cm^2$.

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Updated on: 10-Oct-2022

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