Difference between revisions of "2008 AMC 12B Problems/Problem 4"
(→See Also) |
|||
| Line 31: | Line 31: | ||
Since a circle has <math>360^\circ</math>, the desired ratio is <math>\frac{105^\circ}{360^\circ}=\frac{7}{24} \Rightarrow D</math>. | Since a circle has <math>360^\circ</math>, the desired ratio is <math>\frac{105^\circ}{360^\circ}=\frac{7}{24} \Rightarrow D</math>. | ||
| − | + | See Also== | |
{{AMC12 box|year=2008|ab=B|num-b=3|num-a=5}} | {{AMC12 box|year=2008|ab=B|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:35, 10 June 2017
Problem
On circle 
, points 
and 
are on the same side of diameter 
, 
, and 
. What is the ratio of the area of the smaller sector 
to the area of the circle?
![[asy] unitsize(6mm); defaultpen(linewidth(0.7)+fontsize(8pt)); pair C = 3*dir (30); pair D = 3*dir (135); pair A = 3*dir (0); pair B = 3*dir(180); pair O = (0,0); draw (Circle ((0, 0), 3)); label ("\(C\)", C, NE); label ("\(D\)", D, NW); label ("\(B\)", B, W); label ("\(A\)", A, E); label ("\(O\)", O, S); label ("\(45^\circ\)", (-0.3,0.1), WNW); label ("\(30^\circ\)", (0.5,0.1), ENE); draw (A--B); draw (O--D); draw (O--C); [/asy]](http://latex.artofproblemsolving.com/6/e/f/6ef568ce45b5abd28fca3d7565af0a5d7af4c6f2.png)
Solution
.
Since a circle has 
, the desired ratio is 
.
See Also==
| 2008 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
