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Difference between revisions of "2004 AMC 12B Problems/Problem 10"

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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{\mathrm{(A)\ }\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{\mathrm{(A)\ }\pi a^2}</math>.
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==Solution 2==
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Set <math>c=0,</math> then the area is just <math>\boxed{\pi a^2}.</math>
  
 
== See also ==
 
== See also ==

Revision as of 13:34, 23 April 2023

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 $b$ and $c$, with $b>c$. Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$, and let $OY$ be the radius of the larger circle that contains $Z$. Let $a=XZ$, $d=YZ$, and $e=XY$. 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]

$\mathrm{(A) \ } \pi a^2 \qquad \mathrm{(B) \ } \pi b^2 \qquad \mathrm{(C) \ } \pi c^2 \qquad \mathrm{(D) \ } \pi d^2 \qquad \mathrm{(E) \ } \pi e^2$

Solution

The area of the large circle is $\pi b^2$, the area of the small one is $\pi c^2$, hence the shaded area is $\pi(b^2-c^2)$.

From the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$, hence $b^2-c^2=a^2$ and thus the shaded area is $\boxed{\mathrm{(A)\ }\pi a^2}$.

Solution 2

Set $c=0,$ then the area is just $\boxed{\pi a^2}.$

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

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