# Find the area of a triangle two sides of which are 18 cm

Question. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Solution:

Let the third side of the triangle be x.

Perimeter of the given triangle = 42 cm

$18 \mathrm{~cm}+10 \mathrm{~cm}+x=42$

$x=14 \mathrm{~cm}$

$s=\frac{\text { Perimeter }}{2}=\frac{42 \mathrm{~cm}}{2}=21 \mathrm{~cm}$

By Heron's formula,

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

Area of the given triangle $=(\sqrt{21(21-18)(21-10)(21-14)}) \mathrm{cm}^{2}$

$=(\sqrt{21(3)(11)(7)}) \mathrm{cm}^{2}$

$=21 \sqrt{11} \mathrm{~cm}^{2}$