Square EFGH has one vertex on each side of square ABCD. Point E is on \overlineAB with AE=7× EB. What is the ratio of the area of EFGH to the area of ABCD?

2011 AMC 10A Problem 11 — Solution

2011 AMC 10A Problem 11

Attempt Problem 11 of the 2011 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10A solutions, or check the answer key.

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11.

Square $EFGH$ has one vertex on each side of square $ABCD.$ Point $E$ is on $\overline{AB}$ with $AE=7\cdot EB.$ What is the ratio of the area of $EFGH$ to the area of $ABCD?$

$\dfrac{49}{64}$

$\dfrac{25}{32}$

$\dfrac78$

$\dfrac{5\sqrt{2}}{8}$

$\dfrac{\sqrt{14}}{4}$

Answer: B

Concepts:square (geometry)Pythagorean Theorem area ratio

Difficulty rating: 1420

Solution:

Let $x = EB.$ Then $AB = 8x.$ Applying the Pythagorean Theorem to a side of $EFGH,$ we get $\sqrt{(7x)^2 + x^2} = \sqrt{50x^2}$

The desired ratio is then $\dfrac{\sqrt{50x^2}^2}{(8x)^2} = \dfrac{50}{64} = \dfrac{25}{32}.$

Thus, B is the correct answer.

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