2014 AMC 10A Problem 9

In a right triangle each leg is the altitude to the other leg, so the missing altitude is the one drawn to the hypotenuse.

The hypotenuse has length
$$\sqrt{(2\sqrt3)^2 + 6^2} = \sqrt{12 + 36} = \sqrt{48} = 4\sqrt3.$$

Compute the area with the legs: $\frac12 \cdot 2\sqrt3 \cdot 6 = 6\sqrt3$. Computing it with the hypotenuse as base and the unknown altitude $h$ gives $\frac12 \cdot 4\sqrt3 \cdot h$. Equating,
$$\frac12 \cdot 4\sqrt3 \cdot h = 6\sqrt3 \quad\Longrightarrow\quad h = 3.$$

The answer is $\boxed{\textbf{(C)}\ 3}$.

The area of a triangle is a single number, so any base-height pair gives the same value. For a right triangle the legs supply the area for free, and "area equals half the hypotenuse times its altitude" then gives $h = \frac{\text{leg}_1 \cdot \text{leg}_2}{\text{hypotenuse}}$. This formula is worth remembering.

The legs are in ratio $2\sqrt3 : 6 = 1 : \sqrt3$, so this is a $30$-$60$-$90$ triangle with hypotenuse $4\sqrt3$. The altitude to the hypotenuse is the short leg times $\sin 60^\circ$: $6 \cdot \frac12 = 3$ from the $6$ leg, or equivalently $2\sqrt3 \cdot \frac{\sqrt3}{2} = 3$ from the other leg.

• Forgetting that the legs of a right triangle are themselves altitudes, and looking for three separate altitudes to compute.
• Using $\frac12 \cdot 4\sqrt3 \cdot h$ but setting it equal to $2\sqrt3 \cdot 6$ (forgetting the $\frac12$ on the other side), which gives $h = 6$.