Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length.

Solution:

Let :

$a=18 \mathrm{~cm}, b=24 \mathrm{~cm}$ and $c=30 \mathrm{~cm}$

$\therefore s=\frac{a+b+c}{2}=\frac{18+24+30}{2}=36 \mathrm{~cm}$

By Heron's formula, we have :

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

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

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

$=\sqrt{12 \times 3 \times 6 \times 3 \times 12 \times 6}$

$=12 \times 3 \times 6$

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

We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:

Area of triangle $=216 \mathrm{~cm}^{2}$

$\Rightarrow \frac{1}{2} \times$ Base $\times$ Height $=216$

$\Rightarrow$ Height $=\frac{216 \times 2}{18}=24 \mathrm{~cm}$