2005 AMC 10B Problem 14

Take $CD$ as the base of $\triangle CDM$. Since $C$ is the midpoint of $BD$, $CD = BC = 2$.

For the height, measure the distance from $M$ to line $BD$. Point $A$ is at height $2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$ above line $BC$ (the altitude of an equilateral triangle of side $2$), and $C$ is on the line at height $0$. Because $M$ is the midpoint of $AC$, its height is the average, $\frac{\sqrt{3}}{2}$.

Therefore
$$[\triangle CDM] = \frac{1}{2} \cdot 2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}$.

Choosing the base that lies along a known line makes the height a single vertical distance, and a midpoint sits at the average of the heights of its endpoints. Equivalently, $\triangle CDM$ has the same base length as $\triangle ABC$ and half its height, so its area is half of $\sqrt{3}$.

Trigonometry: $\angle MCD = 180^\circ - 60^\circ = 120^\circ$, with $CM = 1$ and $CD = 2$, so the area is $\frac{1}{2} \cdot 1 \cdot 2 \cdot \sin 120^\circ = \frac{\sqrt{3}}{2}$.

• Treating MC as the height of triangle CDM, or using the full equilateral height sqrt3 instead of half of it.
• Using $\sin 60^\circ$ correctly but a wrong pair of sides, e.g. $CM = 1$ and $BD = 4$, which doubles the area.