Points $A,B,C,D$ and $E$ are located in 3-dimensional space with $AB=BC=CD=DE=EA=2$ and $\angle ABC=\angle CDE=\angle DEA=90^o$. The plane of $\triangle ABC$ is parallel to $\overline{DE}$. What is the area of $\triangle BDE$? $\mathrm {(A)} \sqrt{2}\qquad$ $ \mathrm {(B)} \sqrt{3}\qquad$ $ \mathrm {(C)} 2\qquad$ $ \mathrm {(D)} \sqrt{5}\qquad$ $ \mathrm {(E)} \sqrt{6}$
Solution
Place $B=(0,0,0), A=(2,0,0), C=(0,2,0)$ (right angle at $B$, legs $2$). Requiring $CD=DE=EA=2$, $\angle CDE=\angle DEA=90^\circ$, and $D,E$ at a common height (so $\overline{DE}\parallel$ plane $ABC$) gives (one valid solution) $D=(0,1,\sqrt3),E=(2,1,\sqrt3)$.
$\vec{BD}=(0,1,\sqrt3)$ and $\vec{BE}=(2,1,\sqrt3)$, so $\vec{BD}\times\vec{BE}=(0,2\sqrt3,-2)$ with length $4$, and $[BDE]=\frac12\cdot4=2$. The answer is (\textbf{C}).
Let $ABCDEFGH$ be a cube of edge length $4$ with bottom face $ABCD$ and top face $EFGH$ ($E$ above $A$, $F$ above $B$, and so on). Let $P$, $Q$, $R$ be the midpoints of $\overline{AB}$, $\overline{CG}$, and $\overline{EH}$. What is the area of $\triangle PQR$?
$\mathrm {(A)} \sqrt{2}\qquad$ $ \mathrm {(B)} \sqrt{3}\qquad$ $ \mathrm {(C)} 2\qquad$ $ \mathrm {(D)} \sqrt{5}\qquad$ $ \mathrm {(E)} \sqrt{6}$ 