Points $K, L, M,$ and $N$ lie in the plane of the square $ABCD$ such that $AKB$, $BLC$, $CMD$, and $DNA$ are equilateral triangles. If $ABCD$ has an area of 16, find the area of $KLMN$. $\textrm{(A)}\ 32\qquad$ $\textrm{(B)}\ 16+16\sqrt{3}\qquad$ $\textrm{(C)}\ 48\qquad$ $\textrm{(D)}\ 32+16\sqrt{3}\qquad$ $\textrm{(E)}\ 64$
Solution
The square has side $4$. Each equilateral triangle has area $[CDM]=\frac{\sqrt3}4\cdot 16=4\sqrt3$.
At each vertex, e.g. $C$, the angle $\angle MCL=360^\circ-90^\circ-60^\circ-60^\circ=150^\circ$, so $[MCL]=\frac12\cdot 4 \cdot 4 \cdot \sin 150^\circ=4$.
$KLMN$ is made of the square, four equilateral triangles and four such triangles: $16+16\sqrt3+16=32+16\sqrt3$.
Rectangle $ABCD$ has side lengths $AB=8$ and $BC=6$. Equilateral triangles $AKB$, $BLC$, $CMD$, and $DNA$ are constructed outward on its sides. What is the area of $KLMN$?
$\textrm{(A)}\ 32\qquad$ $\textrm{(B)}\ 16+16\sqrt{3}\qquad$ $\textrm{(C)}\ 48\qquad$ $\textrm{(D)}\ 32+16\sqrt{3}\qquad$ $\textrm{(E)}\ 64$ 
