Since the same region $(A E H D)$ is unshaded inside each rectangle, then the two shaded regions have equal area, since the rectangles have equal area.

Thus, the total shaded area is twice the area of $A E H C B$.

Draw a horizontal line through $E$, meeting $A B$ at $X$ and $H C$ at $Y$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-190.jpg?height=543&width=596&top_left_y=154&top_left_x=819)

We know $\angle B A E=30^{\circ}$ and $A E=12$, so $E X=\frac{1}{2} A E=6$ and $A X=\sqrt{3} E X=6 \sqrt{3}$, by the ratios in a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Thus, the area of $\triangle E X A$ is $\frac{1}{2}(6)(6 \sqrt{3})=18 \sqrt{3}$.

Since $A B=12$ and $A X=6 \sqrt{3}$, then $X B=12-6 \sqrt{3}$.

Therefore, rectangle $B X Y C$ has height $12-6 \sqrt{3}$ and width 18 , so has area $18(12-6 \sqrt{3})$ or $216-108 \sqrt{3}$.

Since $X Y=B C=18$ and $E X=6$, then $E Y=12$.

Since $\angle A E B=60^{\circ}$ and $\angle A E H=90^{\circ}$, then $\angle H E Y=30^{\circ}$.

Since $E Y=12$ and $\angle H E Y=30^{\circ}$, then $H Y=\frac{12}{\sqrt{3}}=4 \sqrt{3}$.

Thus, $\triangle E Y H$ has area $\frac{1}{2}(12)(4 \sqrt{3})=24 \sqrt{3}$.

So the total area of $A E H C B$ is $18 \sqrt{3}+(216-108 \sqrt{3})+24 \sqrt{3}=216-66 \sqrt{3}$.

Therefore, the total shaded area is $2(216-66 \sqrt{3})=432-132 \sqrt{3} \approx 203.369$.

ANSWER: (C)
In the diagram, $A B C D$ is rectangle with $A B=12$ and $B C=18$. Rectangle $A E F G$ is formed by rotating $A B C D$ about $A$ through an angle of $30^{\circ}$. The total area of the shaded regions is closest to
(A) 202.8
(B) 203.1
(C) 203.4
(D) 203.7
(E) 204.0

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-100.jpg?height=387&width=591&top_left_y=2170&top_left_x=1144)