# 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 $6^2=\boxed{36} \Rightarrow \mathrm{(C)}$.

## See also

 2005 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS