Let points $A = (0 ,0 ,0)$, $B = (1, 0, 0)$, $C = (0, 2, 0)$, and $D = (0, 0, 3)$. Points $E$, $F$, $G$, and $H$ are midpoints of line segments $\overline{BD},\text{ } \overline{AB}, \text{ } \overline {AC},$ and $\overline{DC}$ respectively. What is the area of $EFGH$? $\textbf{(A)}\ \sqrt{2}\qquad$ $\textbf{(B)}\ \frac{2\sqrt{5}}{3}\qquad$ $\textbf{(C)}\ \frac{3\sqrt{5}}{4}\qquad$ $\textbf{(D)}\ \sqrt{3}\qquad$ $\textbf{(E)}\ \frac{2\sqrt{7}}{3}$
Solution
With $A=(0,0,0),B=(1,0,0),C=(0,2,0),D=(0,0,3)$, the midpoints are $E=(\tfrac12,0,\tfrac32),F=(\tfrac12,0,0),G=(0,1,0),H=(0,1,\tfrac32)$.
$\overrightarrow{FE}=\overrightarrow{GH}=(0,0,\frac32)$ and $\overrightarrow{FG}=(-\frac12,1,0)$, which are perpendicular. So $EFGH$ is a rectangle with sides $\frac32$ and $\frac{\sqrt5}2$, and its area is $\frac{3\sqrt5}4$. The answer is (\textbf{C}).
Let $A=(0,0,0)$, $B=(4,0,0)$, $C=(0,3,0)$, and $D=(0,0,12)$. Points $E$, $F$, $G$, $H$ are the midpoints of $\overline{BD}$, $\overline{AB}$, $\overline{AC}$, $\overline{DC}$ respectively. What is the area of quadrilateral $EFGH$?
$\textbf{(A)}\ \sqrt{2}\qquad$ $\textbf{(B)}\ \frac{2\sqrt{5}}{3}\qquad$ $\textbf{(C)}\ \frac{3\sqrt{5}}{4}\qquad$ $\textbf{(D)}\ \sqrt{3}\qquad$ $\textbf{(E)}\ \frac{2\sqrt{7}}{3}$ 