Problem 19 · 2010 AMC 8
Medium
Geometry & Measurement
tangent-radiusannulus-area
The two circles pictured have the same center C. Chord AD is tangent to the inner circle at B, AC is 10, and chord AD has length 16. What is the area between the two circles?
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Answer: C — 64π.
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Hint 1 of 3
You're never told the inner radius — and you don't need it. The ring's area is π(AC2 − CB2), and that difference of squares is exactly what a right triangle hands you.
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Hint 2 of 3
Tangent line ⇒ the radius to the touch point is perpendicular to it. That right angle at B sets up Pythagoras: AC2 − CB2 = AB2.
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Hint 3 of 3
The perpendicular from the center to a chord also bisects it, so B is the midpoint of AD and AB = 8.
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Approach: the missing inner radius cancels via Pythagoras
- AD is tangent to the inner circle at B, so radius CB ⊥ AD. A perpendicular from the center bisects the chord, so AB = AD/2 = 8.
- The ring (annulus) area is πAC2 − πCB2 = π(AC2 − CB2). In right triangle ABC, that bracket is AB2 by Pythagoras.
- So the area = π · 82 = 64π — the unknown inner radius never had to be found.
- Why this transfers: when an answer depends only on a difference of squared radii, look for a right triangle whose legs are those radii. The annulus area becomes π · (half-chord)2 — a recurring ‘ring + tangent chord’ shortcut.
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