2005 AMC 10A Problem 8

The four segments from the corners cut the big square into the inner square plus four congruent right triangles, one at each corner. Let $s$ be the side of the inner square.

Look at the triangle with hypotenuse $AB$. Its short leg is $BE = 1$. Its long leg runs from $B$ through $E$ to $H$, and $EH$ is a side of the inner square, so the long leg is $BH = 1 + s$.

Pythagoras on this triangle:
$$1^2 + (s+1)^2 = (\sqrt{50})^2 = 50 \quad\Longrightarrow\quad (s+1)^2 = 49 \quad\Longrightarrow\quad s = 6.$$

The inner square has area $6^2 = 36$.

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

A square with a tilted square inside is the classic "four congruent right triangles" picture; the outer side is a hypotenuse and the inner side is the difference of the two legs. The figure tells you the short leg, so a single Pythagorean equation finds everything. The number $\sqrt{50}$ is chosen so that $50 - 1 = 49$ is a perfect square.

Area bookkeeping: the big square's area $50$ equals $s^2$ plus four triangles of area $\frac{1}{2}(1)(s+1)$, so $s^2 + 2s + 2 = 50$, giving $s^2 + 2s - 48 = (s+8)(s-6) = 0$ and $s = 6$.

• Taking the longer leg to be s rather than s + 1, which gives s = 7 and area 49, or reporting s = 6 instead of the area 36.
• Assuming the triangles are $45^\circ$-$45^\circ$-$90^\circ$; they are not, since one leg is $1$ and the other is $7$.