2005 AMC 10A Problem 8

The big square is made of the inner square plus four matching right triangles, one against each side of the big square. One of those triangles holds all the information, so work with it.

1. Pick the triangle on side $AB$. It is triangle $ABH$, with the right angle at $H$ and hypotenuse $AB = \sqrt{50}$.
2. Find its short leg. The four corner triangles are identical, and in the next triangle over, the short leg is $BE = 1$. So the short leg here is $AH = 1$ too.
3. Find its long leg. The long leg goes from $B$ through $E$ all the way to $H$. The piece $EH$ is a side of the inner square. Call that side $s$. So the long leg is $BH = 1 + s$.
4. Use the Pythagorean theorem.
$$1^2 + (s + 1)^2 = (\sqrt{50})^2 = 50.$$
So $(s + 1)^2 = 49$, which means $s + 1 = 7$ and $s = 6$.
5. Area of the inner square. $6 \times 6 = 36$.

Check: legs $1$ and $7$ give $1 + 49 = 50$, matching $AB^2$.

The answer is $\boxed{\textbf{(C)}\ 36}$.

A tilted square inside a square always makes four identical right triangles. The side of the outer square is the hypotenuse, and the side of the inner square is the long leg minus the short leg. Once you spot that, one Pythagorean equation solves the problem. The number $50$ was chosen so that $50 - 1 = 49$ is a perfect square.

Use areas. The big square has area $50$. It equals the inner square, $s^2$, plus four triangles, each with area $\frac{1}{2} \cdot 1 \cdot (s + 1)$. So $s^2 + 2(s + 1) = 50$, or $s^2 + 2s - 48 = 0$. This factors as $(s + 8)(s - 6) = 0$, so $s = 6$ and the area is $36$.