Squares $A B C D$ and $E F G H$ are equal in area. Vertices $B, E, C$, and $H$ lie on the same straight line. Diagonal $A C$ is extended to $J$, the midpoint of $G H$. The fraction of the two squares that is shaded is
(A) $\frac{5}{8}$
(B) $\frac{1}{3}$
(C) $\frac{2}{5}$
(D) $\frac{5}{16}$
(E) $\frac{3}{8}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-052.jpg?height=449&width=374&top_left_y=1662&top_left_x=1339)
Assume that each square has side length 2 , and thus has area 4.

In square $A B C D$, diagonal $A C$ divides the square into 2 equal areas.

Thus, the area of $\triangle A C D$ is one half of the area of square $A B C D$ or 2.

Since $A C$ is the diagonal of square $A B C D, \angle A C D=\angle A C B=90^{\circ} \div 2=45^{\circ}$.

Also, $\angle D C H=\angle B C H-\angle D C B=180^{\circ}-90^{\circ}=90^{\circ}$.

Since $A C J$ is a straight line segment, $\angle A C J=\angle A C D+\angle D C H+\angle H C J=180^{\circ}$.

Thus, $\angle H C J=180^{\circ}-\angle A C D-\angle D C H=180^{\circ}-45^{\circ}-90^{\circ}=45^{\circ}$.

In $\triangle C H J, \angle H C J=45^{\circ}$ and $\angle C H J=90^{\circ}$, so $\angle H J C=180^{\circ}-90^{\circ}-45=45^{\circ}$.

Therefore $\triangle C H J$ is isosceles, so $C H=H J=1$, since $J$ is the midpoint of $G H$.

Therefore $\triangle C H J$ has area $\frac{1}{2} \times 1 \times 1$ or $\frac{1}{2}$.

Since $\triangle A C D$ has area 2 , and $\triangle C H J$ has area $\frac{1}{2}$, then the combined areas of the two shaded regions is $2+\frac{1}{2}$ or $\frac{5}{2}$.

Since the total area of the two large squares is 8 , the fraction of the two squares that is shaded is $\frac{5}{2} \div 8=\frac{5}{2} \times \frac{1}{8}=\frac{5}{16}$.

ANSWER: (D)