Difference between revisions of "2002 AMC 10B Problems/Problem 22"

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(Redirected page to 2002 AMC 12B Problems/Problem 20)
 
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== Problem==
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#REDIRECT[[2002 AMC 12B Problems/Problem 20]]
 
 
Let <math>\triangle{XOY}</math> be a right-triangle with <math>m\angle{XOY}=90^\circ</math>. Let <math>M</math> and <math>N</math> be the midpoints of the legs <math>OX</math> and <math>OY</math>, respectively. Given <math>XN=19</math> and <math>YM=22</math>, find <math>XY</math>.
 
 
 
<math> \mathrm{(A) \ } 24\qquad \mathrm{(B) \ } 26\qquad \mathrm{(C) \ } 28\qquad \mathrm{(D) \ } 30\qquad \mathrm{(E) \ } 32 </math>
 
 
 
== Solution ==
 
 
 
Let <math>OM=MX=x</math> and <math>ON=NY=y</math>. By the Pythagorean Theorem, <math>x^2+4y^2=484</math> and <math>4x^2+y^2=361</math> We wish to find <math>\sqrt{4x^2+4y^2}</math>. So, we add the two equations, multiply by <math>\frac{4}{5}</math>, and take the square root.
 
<cmath>\begin{align*}
 
5x^2+5y^2&=845\\
 
4x^2+4y^2&=676\\
 
\sqrt{4x^2+4y^2}&=\boxed{\mathrm{(B) \ } 26}
 
\end{align*}</cmath>
 
 
 
==See Also==
 
 
 
{{AMC10 box|year=2002|ab=B|num-b=21|num-a=23}}
 

Latest revision as of 15:28, 29 July 2011

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