2011 AMC 10A Problem 11

Only a ratio is asked for, so we can choose the size of $ABCD$. The split $AE = 7 \cdot EB$ cuts side $AB$ into $8$ equal pieces, so make the side $8$. Then $AE = 7$ and $EB = 1$.

1. Look at the corners. The inner square $EFGH$ cuts four right triangles off the corners of $ABCD$. Each triangle's longest side is a side of $EFGH$, so all four have the same longest side.
2. The four corner triangles are identical. Stand at $E$: the angle at $A$'s triangle, the $90^\circ$ corner of the inner square, and the angle at $B$'s triangle make a straight line. So those two triangle angles add to $90^\circ$. That means the triangles have the same angles, and with the same longest side, they are identical copies turned around the square.
3. So every corner triangle has short sides $7$ and $1$. For example, triangle $EBF$ has $EB = 1$ and $BF = 7$.
4. Area of the inner square. Its side is $EF$, and by the Pythagorean theorem
$$EF^2 = 1^2 + 7^2 = 1 + 49 = 50.$$
The area of a square is its side squared, so the area of $EFGH$ is $50$. (No square roots needed.)
5. The ratio. The big square has area $8^2 = 64$, so
$$\frac{50}{64} = \frac{25}{32}.$$

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

A square drawn inside another square with one corner on each side always cuts off four identical right triangles. So knowing how one side is split tells you everything. The inner square's area is its side squared, and the Pythagorean theorem hands you that square directly. Choosing side $8$ turns the $7$-to-$1$ split into whole numbers.

Subtract the corners instead. Each corner triangle has area $\frac{1}{2} \times 7 \times 1 = \frac{7}{2}$, so the four together have area $14$. The inner square is what is left: $64 - 14 = 50$. The ratio is again $\frac{50}{64} = \frac{25}{32}$.