Difference between revisions of "2013 AMC 8 Problems/Problem 23"
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==Problem== | ==Problem== | ||
| + | Angle <math>ABC</math> of <math>\triangle ABC</math> is a right angle. The sides of <math>\triangle ABC</math> are the diameters of semicircles as shown. The area of the semicircle on <math>\overline{AB}</math> equals <math>8\pi</math>, and the arc of the semicircle on <math>\overline{AC}</math> has length <math>8.5\pi</math>. What is the radius of the semicircle on <math>\overline{BC}</math>? | ||
| + | [asy] | ||
| + | import graph; | ||
| + | draw((0,8)..(-4,4)..(0,0)--(0,8)); | ||
| + | draw((0,0)..(7.5,-7.5)..(15,0)--(0,0)); | ||
| + | real theta = aTan(8/15); | ||
| + | draw(arc((15/2,4),17/2,-theta,180-theta)); | ||
| + | draw((0,8)--(15,0)); | ||
| + | [/asy] | ||
| + | |||
| + | <math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9</math> | ||
==Solution== | ==Solution== | ||
Revision as of 08:36, 27 November 2013
Problem
Angle 
of 
is a right angle. The sides of are the diameters of semicircles as shown. The area of the semicircle on 
equals 
, and the arc of the semicircle on 
has length 
. What is the radius of the semicircle on 
?
[asy]
import graph;
draw((0,8)..(-4,4)..(0,0)--(0,8));
draw((0,0)..(7.5,-7.5)..(15,0)--(0,0));
real theta = aTan(8/15);
draw(arc((15/2,4),17/2,-theta,180-theta));
draw((0,8)--(15,0));
[/asy]
Solution
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
