2012 AMC 10A Problem 21

1. Find the midpoints. A midpoint is found by averaging the coordinates of the two ends.
- $E$, middle of $BD$: $B = (1, 0, 0)$ and $D = (0, 0, 3)$ give $E = \left(\tfrac12, 0, \tfrac32\right)$.
- $F$, middle of $AB$: $F = \left(\tfrac12, 0, 0\right)$.
- $G$, middle of $AC$: $G = (0, 1, 0)$.
- $H$, middle of $DC$: $D = (0, 0, 3)$ and $C = (0, 2, 0)$ give $H = \left(0, 1, \tfrac32\right)$.
2. Side $FE$. $F$ and $E$ differ only in the third coordinate, by $\frac32$. So $FE$ points straight up and has length $\frac32$.
3. Side $GH$. $G$ and $H$ also differ only in the third coordinate, by $\frac32$. So $GH$ also points straight up with length $\frac32$. Two opposite sides are equal and parallel, so $EFGH$ is a parallelogram.
4. Side $FG$. Both $F$ and $G$ have third coordinate $0$, so $FG$ lies flat on the floor. It goes $\frac12$ in one direction and $1$ in the other, so its length is $\sqrt{\left(\tfrac12\right)^2 + 1^2} = \sqrt{\tfrac54} = \tfrac{\sqrt5}{2}$.
5. The corner at $F$. Side $FE$ is straight up and side $FG$ is flat, so they meet at a right angle. A parallelogram with a right angle is a rectangle.
6. Area:
$$\frac32 \times \frac{\sqrt5}{2} = \frac{3\sqrt5}{4}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{3\sqrt{5}}{4}}$.

A segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long. In triangle $ABD$, $FE$ is half of $AD$; in triangle $ACD$, $GH$ is half of $AD$; in triangles $ABC$ and $DBC$, $FG$ and $EH$ are half of $BC$. So joining midpoints this way always gives a parallelogram. Here $AD$ points straight up and $BC$ lies flat, so the parallelogram is a rectangle with sides $\frac{AD}{2} = \frac32$ and $\frac{BC}{2} = \frac{\sqrt5}{2}$.