Difference between revisions of "2001 AMC 10 Problems/Problem 24"
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==Solution== | ==Solution== | ||
| − | If <math> AB=x </math> and <math> CD=y </math>, | + | If <math> AB=x </math> and <math> CD=y </math>,then <math> BC=x+y </math>. By the [[Pythagorean theorem]], we have <math> (x+y)^2=(y-x)^2+49 </math> Solving the equation, we get <math> 4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25} </math>. |
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| − | By the [[Pythagorean theorem]], we have <math> (x+y)^2=(y-x)^2+49 </math> | ||
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| − | Solving the equation, we get <math> 4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25} </math>. | ||
==See Also== | ==See Also== | ||
Revision as of 11:05, 2 August 2015
Problem
In trapezoid 
, 
and 
are perpendicular to 
, with 
, 
, and 
. What is 
?
Solution
If 
and 
,then 
. By the Pythagorean theorem, we have 
Solving the equation, we get 
.
See Also
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
