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Difference between revisions of "2011 AMC 10A Problems/Problem 11"

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== Solution ==
 
== Solution ==
  
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Let <math>8s</math> be the length of the sides of square <math>ABCD</math>, then the length of one of the sides of square <math>EFGH</math> is <math>\sqrt{(7s)^2+s^2}=\sqrt{50s^2}</math>, and hence the ratio in the areas is <math>\frac{\sqrt{50s^2}^2}{(8s)^2}=\frac{50}{64} = \boxed{\frac{25}{32} \ \mathbf{(B)}}</math>.
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Let <math>8</math> be the length of the sides of square <math>ABCD</math>. Then, the length of one of the sides of square <math>EFGH</math> is <math>\sqrt{7^2+1^2}=\sqrt{50}</math>, and hence the ratio in the areas is <math>\frac{\sqrt{50}^2}{8^2}=\frac{50}{64} = \boxed{\frac{25}{32} \ \mathbf{(B)}}</math>.
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Latest revision as of 08:02, 7 September 2025

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 $8$ be the length of the sides of square $ABCD$. Then, the length of one of the sides of square $EFGH$ is $\sqrt{7^2+1^2}=\sqrt{50}$, and hence the ratio in the areas is $\frac{\sqrt{50}^2}{8^2}=\frac{50}{64} = \boxed{\frac{25}{32} \ \mathbf{(B)}}$.

Video Solution by OmegaLearn

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

~ pi_is_3.14

See Also

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 10 Problems and Solutions

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