2005 AMC 12A Problem 7

Below is the professionally curated solution for Problem 7 of the 2005 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 12A solutions, or check the answer key.

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Concepts:square (geometry)Pythagorean Theorem congruence (geometry)

Difficulty rating: 1460

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

$25$

$32$

$36$

$40$

$42$

Solution:

By the symmetry of the figure, triangles $ABH,$ $BCE,$ $CDF,$ and $DAG$ are congruent right triangles. Hence $BH = CE = \sqrt{BC^2 - BE^2} = \sqrt{50 - 1} = 7.$

Since $E$ lies between $B$ and $H,$ the side of the inner square is $EH = BH - BE = 7 - 1 = 6.$

Therefore the area of $EFGH$ is $6^2 = 36.$

Thus, the correct answer is C.

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