Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ a=18\mathrm{cm},b=24\mathrm{cm}\mathrm{and}c=30\mathrm{cm} \\ \therefore s=\frac{a+b+c}{2}=\frac{18+24+30}{2}=36\mathrm{cm} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{36(36-18)(36-24)(36-30)} \\ =\sqrt{36\times 18\times 12\times 6} \\ =\sqrt{12\times 3\times 6\times 3\times 12\times 6} \\ =12\times 3\times 6 \\ =216{\mathrm{cm}}^2 \end{gathered}$$

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:
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=216{\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}=216 \\ \Rightarrow \mathrm{Height}=\frac{216\times 2}{18}=24\mathrm{cm} \end{gathered}$$