2004 AMC 12B Problem 10

Attempt Problem 10 of the 2004 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

An annulus is the region between two concentric circles. The concentric circles in the figure have radii bb and c,c, with b>c.b \gt c. Let OX‾\overline{OX} be a radius of the larger circle, let XZ‾\overline{XZ} be tangent to the smaller circle at Z,Z, and let OY‾\overline{OY} be the radius of the larger circle that contains Z.Z. Let a=XZ,a = XZ, d=YZ,d = YZ, and e=XY.e = XY. What is the area of the annulus?
πa2\pi a^2
πb2\pi b^2
πc2\pi c^2
πd2\pi d^2
πe2\pi e^2
Answer: A
Concepts:annulustangent linePythagorean Theorem
Small hint:
The annulus area is πb2−πc2\pi b^2 - \pi c^2
Big hint:
Since XZ‾\overline{XZ} is tangent, △OZX\triangle OZX is right-angled at Z,Z, so b2−c2=a2b^2 - c^2 = a^2
Solution:
The annulus area is πb2−πc2.\pi b^2 - \pi c^2. Because XZ‾\overline{XZ} is tangent to the smaller circle at Z,Z, it is perpendicular to radius OZ‾,\overline{OZ}, so △OZX\triangle OZX is right-angled at Z.Z. Then b2=c2+a2,b^2 = c^2 + a^2, giving b2−c2=a2.b^2 - c^2 = a^2. The area is πa2.\pi a^2.
Thus, the correct answer is A.
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