An isosceles triangle has perimeter 30 cm
Question. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.


Solution:

Let the third side of this triangle be x.

Perimeter of triangle = 30 cm

$12 \mathrm{~cm}+12 \mathrm{~cm}+x=30 \mathrm{~cm}$

$x=6 \mathrm{~cm}$

$s=\frac{\text { Perimeter of triangle }}{2}=\frac{30 \mathrm{~cm}}{2}=15 \mathrm{~cm}$

By Heron’s formula,

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$=[\sqrt{15(15-12)(15-12)(15-6)}] \mathrm{cm}^{2}$

$=[\sqrt{15(3)(3)(9)}] \mathrm{cm}^{2}$

$=9 \sqrt{15} \mathrm{~cm}^{2}$
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