Question

∆ABC is an equilateral triangle of a side 2a units. Find each of its altitudes.

Answer 1

Let AD, BE and CF be the altitudes of ∆ABC meeting BC, AC and AB at D, E and F, respectively.
Then, D, E and F are the midpoints of BC, AC and AB, respectively.

In right-angled ∆ABD, we have:
AB = 2a and BD = a
Applying Pythagoras theorem, we get:
$$\begin{gathered} A{B}^2=A{D}^2+B{D}^2 \\ A{D}^2=A{B}^2-B{D}^2=(2a{)}^2-a^2 \\ A{D}^2=4a^2-a^2=3a^2 \\ AD=\sqrt{3}a\mathrm{units} \end{gathered}$$

Similarly,
BE = $a\sqrt{3}$ units and CF = $a\sqrt{3}$ units