2005 AMC 12A Problems/Problem 7

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Problem

Square $EFGH$ is inside the square $ABCD$ so that each side of $EFGH$ can be extended to pass through a vertex of $ABCD$. Square $ABCD$ has side length $\sqrt {50}$ and $BE = 1$. What is the area of the inner square $EFGH$?

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

Solution

2005 12A AMC-7b.png

Arguable the hardest part of this question is to visualize the diagram. Since each side of $EFGH$ can be extended to pass through a vertex of $ABCD$, we realize that $EFGH$ must be tilted in such a fashion. Let a side of $EFGH$ be $x$.

2005 12A AMC-7.png

Notice the right triangle (in blue) with legs $1, x+1$ and hypotenuse $\sqrt{50}$. By the Pythagorean Theorem, we have $1^2 + (x+1)^2 = (\sqrt{50})^2 \Longrightarrow (x+1)^2 = 49 \Longrightarrow x = 6$. Thus, $[EFGH] = x^2 = 36\ \mathrm{(C)}$

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 12 Problems and Solutions

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