In rectangle $ABCD$, $AB=1$, $BC=2$, and points $E$, $F$, and $G$ are midpoints of $\overline{BC}$, $\overline{CD}$, and $\overline{AD}$, respectively. Point $H$ is the midpoint of $\overline{GE}$. What is the area of the shaded region? $\textbf{(A)}\ \dfrac1{12}\qquad$ $\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad$ $\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad$ $\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad$ $\textbf{(E)}\ \dfrac16$
Solution
$H$ is the midpoint of $GE$, which is the center of the rectangle, so $A,H,C$ are collinear on the diagonal $AC$. Let $K$ be the intersection of $AC$ and $BF$.
Since $AB\parallel CF$, $\triangle AKB\sim \triangle CKF$, so $BK:FK=AB:CF=2:1$.
$[CFB]=\frac12\cdot\frac12\cdot2=\frac12$, so $[CFK]=\frac13[CFB]=\frac16$; and $[CHE]=\frac12\cdot CE\cdot\frac12=\frac14$.
The shaded area=$[CDGE]-2[CKF]-2[CHE]=1-\frac13-\frac12=\frac16$.
In rectangle $ABCD$, $AB=2$, $BC=6$, and points $E,F,G$ are midpoints of $\overline{BC},\overline{CD},\overline{AD}$, respectively. Point $H$ is the midpoint of $\overline{GE}$. What is the area of the shaded region?
$\textbf{(A)}\ \dfrac1{12}\qquad$ $\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad$ $\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad$ $\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad$ $\textbf{(E)}\ \dfrac16$ 
