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Difference between revisions of "2005 AMC 10A Problems/Problem 8"
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==Solution==
 
==Solution==
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'''(C)''' 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'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:
  
 
<math>1^2 + (HE+1)^2=\sqrt{50}^2</math>
 
<math>1^2 + (HE+1)^2=\sqrt{50}^2</math>
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<math>HE+1=7</math>
 
<math>HE+1=7</math>
  
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<math>HE=6</math> So, the area of the square is <math>6^2=\boxed{36}</math>.
+
<math>HE=6</math> So, the area of the square is <math>6^2=\boxed{36} \rightarrow (C)</math>.
  
 
==See Also==
 
==See Also==

Revision as of 21:35, 3 January 2016

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 $EFGH$?

AMC102005Aq.png

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

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'll not 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 $6^2=\boxed{36} \rightarrow (C)$.

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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