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| − | == Problem==
| + | #REDIRECT[[2002 AMC 12B Problems/Problem 20]] |
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| − | 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>.
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| − | <math> \mathrm{(A) \ } 24\qquad \mathrm{(B) \ } 26\qquad \mathrm{(C) \ } 28\qquad \mathrm{(D) \ } 30\qquad \mathrm{(E) \ } 32 </math>
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| − | == Solution ==
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| − | 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.
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| − | <cmath>\begin{align*}
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| − | 5x^2+5y^2&=845\\
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| − | 4x^2+4y^2&=676\\
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| − | \sqrt{4x^2+4y^2}&=\boxed{\mathrm{(B) \ } 26}
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| − | \end{align*}</cmath>
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| − | ==See Also==
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| − | {{AMC10 box|year=2002|ab=B|num-b=21|num-a=23}}
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