Problem 28 · 2025 Math Kangaroo
Stretch
Geometry & Measurement
pythagorean-triplearea
In the diagram we see two touching circles and the diameter through their common point. The outer circle has a chord parallel to this diameter with length 16, which touches the inner circle. What is the area of the grey region?
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Answer: C — \(64\pi\)
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Hint 1 of 3
The grey region is the big disk minus the small disk, so its area is \(\pi(R^2-r^2)\) — you never need \(R\) and \(r\) separately.
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Hint 2 of 3
Drop the perpendicular from the centre to the chord: the half-chord, the inner radius, and the outer radius form a right triangle.
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Hint 3 of 3
Tangency makes the centre-to-chord distance equal to \(r\), so \(r^2+8^2=R^2\).
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Approach: annulus area via the chord
- The chord of length 16 (half-length 8) is tangent to the inner circle, so the perpendicular distance from the common centre to the chord equals the inner radius \(r\).
- By the right triangle (radius, half-chord, distance): \(R^2=r^2+8^2\), hence \(R^2-r^2=64\).
- Grey area \(=\pi R^2-\pi r^2=\pi(R^2-r^2)=\) \(64\pi\), answer C.
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