Difference between revisions of "2001 AMC 10 Problems/Problem 24"
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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>. | 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>. | ||
Revision as of 17:28, 1 December 2015
Problem
In trapezoid 
, 
and 
are perpendicular to 
, with 
, 
, and 
. What is 
?
Solution
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 7.3, ymin = -3.16, ymax = 6.3; /* image dimensions */
/* draw figures */
draw(circle((0.2,4.92), 1.3));
draw(circle((1.04,1.58), 2.14));
draw((-1.1,4.92)--(0.2,4.92));
draw((0.2,4.92)--(1.04,1.58));
draw((1.04,1.58)--(-1.1,1.58));
draw((-1.1,1.58)--(-1.1,4.92));
/* dots and labels */
dot((-1.1,4.92),dotstyle);
label("
", (-1.02,5.12), NE * labelscalefactor);
dot((0.2,4.92),dotstyle);
label("
", (0.28,5.12), NE * labelscalefactor);
dot((-1.1,1.58),dotstyle);
label("
", (-1.02,1.78), NE * labelscalefactor);
dot((1.04,1.58),dotstyle);
label("
", (1.12,1.78), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
