Difference between revisions of "2002 AMC 12B Problems/Problem 20"
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+ | {{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #20]] and [[2002 AMC 10B Problems|2002 AMC 10B #22]]}} | ||
== Problem == | == Problem == | ||
Let <math>\triangle XOY</math> be a [[right triangle|right-angled triangle]] with <math>m\angle XOY = 90^{\circ}</math>. Let <math>M</math> and <math>N</math> be the [[midpoint]]s of legs <math>OX</math> and <math>OY</math>, respectively. Given that <math>XN = 19</math> and <math>YM = 22</math>, find <math>XY</math>. | Let <math>\triangle XOY</math> be a [[right triangle|right-angled triangle]] with <math>m\angle XOY = 90^{\circ}</math>. Let <math>M</math> and <math>N</math> be the [[midpoint]]s of legs <math>OX</math> and <math>OY</math>, respectively. Given that <math>XN = 19</math> and <math>YM = 22</math>, find <math>XY</math>. | ||
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By the Pythagorean Theorem again, we have | By the Pythagorean Theorem again, we have | ||
− | <cmath>(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = | + | <cmath>(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \boxed{\mathrm{(B)}\ 26}</cmath> |
== See also == | == See also == | ||
+ | {{AMC10 box|year=2002|ab=B|num-b=21|num-a=23}} | ||
{{AMC12 box|year=2002|ab=B|num-b=19|num-a=21}} | {{AMC12 box|year=2002|ab=B|num-b=19|num-a=21}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Revision as of 15:28, 29 July 2011
- The following problem is from both the 2002 AMC 12B #20 and 2002 AMC 10B #22, so both problems redirect to this page.
Problem
Let be a right-angled triangle with . Let and be the midpoints of legs and , respectively. Given that and , find .
Solution
Let , . By the Pythagorean Theorem on respectively,
Summing these gives .
By the Pythagorean Theorem again, we have
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 |
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |