Square of the altitude in an equilateral triangle

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Let $ABC$ be an equilateral triangle of side $\ell$ and let $h$ be its altitude.

• (a) Prove that the square of the altitude is triple the square of half the side, that is $h^2=3\left(\dfrac{\ell}{2}\right)^2$.
• (b) Taking $\ell=10$, compute the altitude $h$.

Solution

(a) Proof. In an equilateral triangle the altitude $AH$ to the base $BC$ falls on the midpoint $H$ of $BC$, so $\overline{BH}=\dfrac{\ell}{2}$. Triangle $ABH$ is right-angled at $H$, with hypotenuse $\overline{AB}=\ell$ and legs $\overline{AH}=h$ and $\overline{BH}=\dfrac{\ell}{2}$. By the Pythagorean theorem: $h^2=\ell^2-\left(\frac{\ell}{2}\right)^2=\ell^2-\frac{\ell^2}{4}=\frac{3\ell^2}{4}=3\left(\frac{\ell}{2}\right)^2.$ The square of the altitude is therefore triple the square of half the side. $\;\blacksquare$

(b) Numerical computation. With $\ell=10$ we have $\dfrac{\ell}{2}=5$, hence $h=\sqrt{3\cdot 5^2}=\sqrt{75}=5\sqrt3\approx 8{,}660.$ $\boxed{h=5\sqrt3\approx 8{,}660}$