Difference between revisions of "2004 AMC 12B Problems/Problem 10"
(New page: {{duplicate|2004 AMC 12B #10 and 2004 AMC 10B #12}} ==Problem== An annulus is the region between two concentric circles. The concentri...) |
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| − | The area of the large circle is <math>\pi b^2</math>, the area of the small one is <math>\pi | + | The area of the large circle is <math>\pi b^2</math>, the area of the small one is <math>\pi c^2</math>, hence the shaded area is <math>\pi(b^2-c^2)</math>. |
From the [[Pythagorean Theorem]] for the right triangle <math>OXZ</math> we have <math>a^2 + c^2 = b^2</math>, hence <math>b^2-c^2=a^2</math> and thus the shaded area is <math>\boxed{\pi a^2}</math>. | From the [[Pythagorean Theorem]] for the right triangle <math>OXZ</math> we have <math>a^2 + c^2 = b^2</math>, hence <math>b^2-c^2=a^2</math> and thus the shaded area is <math>\boxed{\pi a^2}</math>. | ||
Revision as of 16:20, 19 February 2009
- The following problem is from both the 2004 AMC 12B #10 and 2004 AMC 10B #12, so both problems redirect to this page.
Problem
An annulus is the region between two concentric circles. The concentric circles in the figure have radii 
and 
, with 
. Let 
be a radius of the larger circle, let 
be tangent to the smaller circle at 
, and let 
be the radius of the larger circle that contains . Let 
, 
, and 
. What is the area of the annulus?
![[asy] unitsize(1.5cm); defaultpen(0.8); real r1=1.5, r2=2.5; pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2); pair Y=(0,r2), Z=(0,r1), X=intersectionpoint( Z--(Z+(10,0)), outer ); filldraw(outer,lightgray,black); filldraw(inner,white,black); draw(X--O--Y); draw(Y--X--Z); label("$O$",O,SW); label("$X$",X,E); label("$Y$",Y,N); label("$Z$",Z,SW); label("$a$",X--Z,N); label("$b$",0.25*X,SE); label("$c$",O--Z,E); label("$d$",Y--Z,W); label("$e$",Y*0.65 + X*0.35,SW); defaultpen(0.5); dot(O); dot(X); dot(Z); dot(Y); [/asy]](http://latex.artofproblemsolving.com/0/c/a/0ca2ea60f3f8c72bd6e0bb1dd3335aaf69a85ba9.png)
Solution
The area of the large circle is 
, the area of the small one is 
, hence the shaded area is 
.
From the Pythagorean Theorem for the right triangle 
we have 
, hence 
and thus the shaded area is 
.
See also
| 2004 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 | |
| 2004 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
