Question

Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm. Also, find the length of the altitude corresponding to the smallest side of the triangle.

Answer 1

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.
$$\begin{gathered} s=\frac{1}{2}(a+b+c) \\ s=\frac{1}{2}(18+24+30) \\ s=36\mathrm{cm} \end{gathered}$$


$$\begin{gathered} \mathrm{Area}\mathrm{of}\text{a t}\mathrm{riangle\; =}\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{36(36-18)(36-24)(36-30)} \\ =\sqrt{36\times 18\times 12\times 6} \\ =\sqrt{46656} \\ =216{\mathrm{cm}}^2 \end{gathered}$$
The smallest side is 18 cm long. This is the base.

Now, area of a triangle =$\frac{1}{2}\times b\times h$
$$\begin{gathered} \Rightarrow 216=\frac{1}{2}\times 18\times h \\ \Rightarrow 216=9h \\ \Rightarrow \frac{216}{9}=h \\ \Rightarrow h=24\mathrm{cm} \end{gathered}$$

The length of the altitude corresponding to the smallest side is 24 cm.