2004 AMC 12B Problem 10

Below is the professionally curated solution for Problem 10 of the 2004 AMC 12B, 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.

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Concepts:annulustangent linePythagorean Theorem

Difficulty rating: 1460

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

Solution:

The annulus area is πb2πc2.\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 b2c2=a2.b^2 - c^2 = a^2. The area is πa2.\pi a^2.

Thus, the correct answer is A.

Problem 10 in Other Years