2011 AMC 10A Problem 11

Only the ratio matters, so let the side of $ABCD$ be $8$; then $AE = 7$ and $EB = 1$.

The four corner triangles cut off by the inner square are congruent: each is a right triangle whose hypotenuse is a side of $EFGH$, and the two acute angles at each vertex of $EFGH$ add to $90^\circ$, so the angles match and the hypotenuses are equal. Thus every corner triangle has legs $7$ and $1$, and in particular $BF = 7$.

By the Pythagorean theorem,
$$EF^2 = EB^2 + BF^2 = 1^2 + 7^2 = 50.$$
So the area of $EFGH$ is $50$, and the ratio is
$$\frac{50}{64} = \frac{25}{32}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{25}{32}}$.

A square inscribed in a square always cuts off four congruent right triangles, so knowing one leg split ($7$ and $1$) determines everything. The inner area is then a hypotenuse squared, which equals the sum of the leg squares, no square roots needed. Choosing side length $8$ turns the ratio $7:1$ into integers.

Subtract the corners: the four right triangles each have area $\frac{1}{2}\cdot 7 \cdot 1 = \frac{7}{2}$, total $14$. The inner square has area $64 - 14 = 50$, giving the same ratio $\frac{50}{64} = \frac{25}{32}$.

• Assuming the inner square's side is 7 (giving 49/64), rather than the hypotenuse of a 7-1 right triangle.
• Taking $AE = 7$ and $EB = 1$ to mean the side of $ABCD$ is $7$ instead of $8$.