# Difference between revisions of "2005 AMC 10A Problems/Problem 8"

## Problem

In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE$=1. What is the area of the inner square $EFGH$?

$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42$

## Solution

We see that side $BE$, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, $AH = 1$. Then $HB = HE + BE = HE + 1$, and $HE$ is one of the sides of the square whose area we want to find. So:

$$1^2 + (HE+1)^2=\sqrt{50}^2$$

$$1 + (HE+1)^2=50$$

$$(HE+1)^2=49$$

$$HE+1=7$$

$$HE=6$$ So, the area of the square is $6^2=\boxed{36} \Rightarrow \mathrm{(C)}$.