2001 AMC 10 Problems/Problem 24
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(11.6cm); 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 |
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