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 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 $6^2=\boxed{\textbf{(C) }36}$

2005 AMC 10A (Problems • Answer Key • Resources)
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
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2005 AMC 10A Problems/Problem 8

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