2005 AMC 10B Problem 14

For a triangle's area we need a base and a height. Side $CD$ lies along line $BD$, so use it as the base; then the height is just how far $M$ is above that line.

1. Base. $C$ is the midpoint of $BD$, so $CD = BC = 2$.
2. Height of $A$. Drop a perpendicular from $A$ to $BC$. It lands at the midpoint of $BC$, making a right triangle with hypotenuse $2$ and one leg $1$. The height is $\sqrt{2^2 - 1^2} = \sqrt{3}$.
3. Height of $M$. $M$ is halfway from $C$ (height $0$) to $A$ (height $\sqrt{3}$). So $M$ is halfway up: its height is $\frac{\sqrt{3}}{2}$.
4. Area.
$$\frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}.$$

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

Pick a base that lies on a line you know well; then the height is a single up-and-down distance. A midpoint is always at the average height of its two ends. Another way to see it: triangle $CDM$ has the same base length as triangle $ABC$ and half the height, so its area is half of $\sqrt{3}$.

If you know the formula $\text{Area} = \frac{1}{2}ab\sin C$: the angle $MCD$ is $180^\circ - 60^\circ = 120^\circ$, with $CM = 1$ and $CD = 2$. So the area is $\frac{1}{2} \cdot 1 \cdot 2 \cdot \sin 120^\circ = \frac{\sqrt{3}}{2}$, since $\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$.

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