Question

Find the length of altitude AD of an isosceles ∆ABC in which AB = AC = 2a units and BC = a units.

Answer 1

In isosceles ∆ ABC, we have:
AB = AC = 2a units and BC = a units
Let AD be the altitude drawn from A that meets BC at D.
Then, D is the midpoint of BC.
BD = BC = $\frac{a}{2}\mathrm{units}$
Applying Pythagoras theorem in right-angled ∆ABD, we have:

$$\begin{gathered} A{B}^2=A{D}^2+B{D}^2 \\ A{D}^2=A{B}^2-B{D}^2=(2a{)}^2-{\left(\frac{a}{2}\right)}^2 \\ A{D}^2=4a^2-\frac{a^2}{4}=\frac{15a^2}{4} \\ AD=\sqrt{\frac{15a^2}{4}}=\frac{a\sqrt{15}}{2}\mathrm{units} \end{gathered}$$