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 $EFGH$?

AMC102005Aq.png

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

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 won’t 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 $6^2=\boxed{36} \Rightarrow \mathrm{(C)}$.

Video Solution

CHECK OUT Video Solution: https://youtu.be/z6T9Zm6Jvk0

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 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

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