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 $=\sqrt{s(s-a)(s-b)(s-c)}$
$=[\sqrt{15(15-12)(15-12)(15-6)}] \mathrm{cm}^{2}$
$=[\sqrt{15(3)(3)(9)}] \mathrm{cm}^{2}$
$=9 \sqrt{15} \mathrm{~cm}^{2}$
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 $=\sqrt{s(s-a)(s-b)(s-c)}$
$=[\sqrt{15(15-12)(15-12)(15-6)}] \mathrm{cm}^{2}$
$=[\sqrt{15(3)(3)(9)}] \mathrm{cm}^{2}$
$=9 \sqrt{15} \mathrm{~cm}^{2}$