Theorem 22

For an equilateral triangle with side length $s$, the altitude has length $\frac{\sqrt3}2s$.

\begin{proof}$AC=s, CD=\frac{s}2$, so $AD^2=s^2-\frac14s^2=\frac34s^2$, hence $AD=\frac{\sqrt3}2s$.\end{proof}

← Back to Exercise 108

proof

$AC=s, CD=\frac{s}2$, so $AD^2=s^2-\frac14s^2=\frac34s^2$, hence $AD=\frac{\sqrt3}2s$.