Question

In a ∆ABC, AD is the altitude from A such that AD = 12 cm, BD = 9 cm and DC = 16 cm. Examine if ∆ABC is right angled at A.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{ADC}, \\ \angle \mathrm{ADC}=90°(\mathrm{AD}\mathrm{is}\mathrm{an}\mathrm{altitude}\mathrm{on}\mathrm{BC}.) \\ \mathrm{Using}\mathrm{the}\mathrm{Pythagoras}\mathrm{theorem},\mathrm{we}\mathrm{get}: \\ {12}^2+{16}^2={\mathrm{AC}}^2 \\ {\mathrm{AC}}^2=144+256=400 \\ \mathrm{AC}=20\mathrm{cm} \\ \mathrm{In}△\mathrm{ADB}, \\ \angle \mathrm{ADB}=90°(\mathrm{AD}\mathrm{is}\mathrm{an}\mathrm{altitude}\mathrm{on}\mathrm{BC}.) \\ \mathrm{Using}\mathrm{the}\mathrm{Pythagoras}\mathrm{theorem},\mathrm{we}\mathrm{get}: \\ {12}^2+9^2={\mathrm{AB}}^2 \\ {\mathrm{AB}}^2=144+81=225 \\ \mathrm{AB}=15\mathrm{cm} \\ \mathrm{In}△\mathrm{ABC}, \\ {\mathrm{BC}}^2={25}^2=625 \\ {\mathrm{AB}}^2+{\mathrm{AC}}^2={15}^2+{20}^2=625 \\ {\mathrm{AB}}^2+{\mathrm{AC}}^2={\mathrm{BC}}^2 \\ \mathrm{Because}\mathrm{it}\mathrm{satisfies}\mathrm{the}\mathrm{Pythagoras}\mathrm{theorem},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}△\mathrm{ABC}\mathrm{is}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{A}. \end{gathered}$$