Question

Find the area of ∆ABC in which BC = 8 cm, AC = 15 cm and AB = 17 cm. Find the length of altitude drawn on AB.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ a=8\mathrm{cm},b=15\mathrm{cm}\mathrm{and}c=17\mathrm{cm} \\ s=\frac{a+b+c}{2}=\frac{8+15+17}{2}=20\mathrm{cm} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{20(20-8)(20-15)(20-17)} \\ =\sqrt{20\times 12\times 5\times 3} \\ =\sqrt{4\times 5\times 4\times 3\times 5\times 3} \\ =4\times 5\times 3 \\ =60{\mathrm{cm}}^2 \end{gathered}$$

Altitude of the triangle corresponding to AB:
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=60{\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}=60 \\ \Rightarrow \mathrm{Height}=\frac{60\times 2}{17}=7.06\mathrm{cm} \end{gathered}$$