Difference between revisions of "2005 AMC 10A Problems/Problem 8"

Revision as of 20:10, 12 November 2019

Problem

In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE=1$ . What is the area of the inner square $EFGH$ ?

$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42$

Solution

We see that side $BE$ , which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, $AH = 1$ . Then $HB = HE + BE = HE + 1$ , and $HE$ is one of the sides of the square whose area we want to find. So:

$1^2 + (HE+1)^2=\sqrt{50}^2$

$1 + (HE+1)^2=50$

$(HE+1)^2=49$

$HE+1=7$

$HE=6$ So, the area of the square is $6^2=\boxed{36} \Rightarrow \mathrm{(C)}$ .

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AMC 10 Problems and Solutions

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@@ Line 7: / Line 7: @@
 ==Solution==
-We see that side <math>BE</math>, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, <math>AH = 1</math>. Then <math>HB = HE + BE = HE + 1</math>, and <math>HE</math> is one of the sides of the square whose area we want to find. So:
+We see that side <math>BE</math>, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, <math>AH = 1</math>. Then <math>HB = HE + BE = HE + 1</math>, and <math>HE</math> is one of the sides of the square whose area we want to find. So:
 <cmath>1^2 + (HE+1)^2=\sqrt{50}^2</cmath>

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Difference between revisions of "2005 AMC 10A Problems/Problem 8"

Revision as of 20:10, 12 November 2019

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Solution

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