Find the area of the triangle whose sides are 18 cm, 24 cm and 30 cm.

Solution:

Let the sides of triangle be ​a = 18 cm, b = 24 cm and c = 30 cm.
Let s be the semi-perimeter of the triangle.

$s=\frac{1}{2}(a+b+c)$

$s=\frac{1}{2}(18+24+30)$

$s=36 \mathrm{~cm}$

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

$=\sqrt{36(36-18)(36-24)(36-30)}$

$=\sqrt{36 \times 18 \times 12 \times 6}$

$=\sqrt{46656}$

$=216 \mathrm{~cm}^{2}$

The smallest side is 18 cm long. This is the base.

Now, area of a triangle $=\frac{1}{2} \times b \times h$

$\Rightarrow 216=\frac{1}{2} \times 18 \times h$

$\Rightarrow 216=9 h$

$\Rightarrow \frac{216}{9}=h$

$\Rightarrow h=24 \mathrm{~cm}$

The height corresponding to the smallest side is 24 cm.