Difference between revisions of "2002 AMC 10B Problems/Problem 22"
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| − | == Problem | + | == Problem== |
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>. | 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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\sqrt{4x^2+4y^2}&=\boxed{\mathrm{(B) \ } 26} | \sqrt{4x^2+4y^2}&=\boxed{\mathrm{(B) \ } 26} | ||
\end{align*}</cmath> | \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}} | ||
Revision as of 02:42, 5 June 2011
Problem
Let 
be a right-triangle with 
. Let 
and 
be the midpoints of the legs 
and 
, respectively. Given 
and 
, find 
.
Solution
Let 
and 
. By the Pythagorean Theorem, 
and 
We wish to find 
. So, we add the two equations, multiply by 
, and take the square root.
See Also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 10 Problems and Solutions | ||
