2011 AMC 10A Problems/Problem 11

Problem 11

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

$\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4}$

Solution

Let $8s$ be the length of the sides of square $ABCD$ , then the length of one of the sides of square $EFGH$ is $\sqrt{(7s)^2+s^2}=\sqrt{50s^2}$ , and hence the ratio in the areas is $\frac{\sqrt{50s^2}^2}{(8s)^2}=\frac{50}{64} = \boxed{\frac{25}{32} \ \mathbf{(B)}}$ .

Video Solution by OmegaLearn

https://youtu.be/GrCtzL0S-Uo?t=292

~ pi_is_3.14

2011 AMC 10A (Problems • Answer Key • Resources)
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2011 AMC 10A Problems/Problem 11

Contents

Problem 11

Solution

Video Solution by OmegaLearn

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