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Difference between revisions of "2005 AMC 12A Problems/Problem 7"

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Notice the [[right triangle]] (in blue) with legs <math>1, x+1</math> and [[hypotenuse]] <math>\sqrt{50}</math>. By the [[Pythagorean Theorem]], we have <math>1^2 + (x+1)^2 = (\sqrt{50})^2 \Longrightarrow (x+1)^2 = 49 \Longrightarrow x = 6</math>. Thus, <math>[EFGH] = x^2 = 36\ \mathrm{(C)}</math>
 
Notice the [[right triangle]] (in blue) with legs <math>1, x+1</math> and [[hypotenuse]] <math>\sqrt{50}</math>. By the [[Pythagorean Theorem]], we have <math>1^2 + (x+1)^2 = (\sqrt{50})^2 \Longrightarrow (x+1)^2 = 49 \Longrightarrow x = 6</math>. Thus, <math>[EFGH] = x^2 = 36\ \mathrm{(C)}</math>
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== Solution ==
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You can also notice that the four triangles <math>\triangle ABH,\triangle BCE,\triangle CDF,\triangle DAG</math> are congruent because the right angles of square <math>ABCD</math> cause them to be similar, and the hypotenuse of the triangles are the same because they are the sides of the square. Then you have <math>[ABCD]-4*7=50-28=22\Rightarrow\boxed{C}.</math>
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2005|num-b=6|num-a=8|ab=A}}
 
{{AMC12 box|year=2005|num-b=6|num-a=8|ab=A}}

Revision as of 22:32, 28 September 2022

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, $[EFGH] = x^2 = 36\ \mathrm{(C)}$

Solution

You can also notice that the four triangles $\triangle ABH,\triangle BCE,\triangle CDF,\triangle DAG$ are congruent because the right angles of square $ABCD$ cause them to be similar, and the hypotenuse of the triangles are the same because they are the sides of the square. Then you have $[ABCD]-4*7=50-28=22\Rightarrow\boxed{C}.$

See also

2005 AMC 12A (Problems • Answer Key • Resources)
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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