AMC 8 · 2010 · #19

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference

area-circlespythagorean-theorem area-differenceidentify-subproblems ↑ Prerequisites: area-circlespythagorean-theorem

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Two circles share the same center C. A chord AD of the outer circle is tangent to the inner circle at B. We know AC = 10 (a radius of the outer circle) and AD = 16. Find the area of the ring-shaped region between the two circles.

Pick an answer.

(A)

$36\pi$

(B)

$49\pi$

(C)

$64\pi$

(D)

$81\pi$

(E)

$100\pi$

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The figure is already given, but the key move is to add to it: draw the inner radius CB to the tangent point. Tool #1 (Draw a Diagram) says label everything and add helper segments — once CB ⊥ AD is marked, a right triangle △ CBA pops out with hypotenuse R = 10 and leg AB. Tool #5 (Look for a Pattern) then catches a shortcut: by the Pythagorean Theorem R² - r² = AB², so the annulus area π(R² - r²) is just π · AB² — no need to compute r separately.

1STEP 1

Draw the inner radius CB. Since AD is tangent at B, the radius is perpendicular to it: CB ⊥ AD, so △CBA is right-angled at B.

∠ CBA = 90°, CA = R = 10, CB = r

2STEP 2

A radius perpendicular to a chord bisects it, so B is the midpoint of AD and AB = 8 (half of 16).

AB =

\frac{1}{2}

· AD =

\frac{1}{2}

· 16 = 8

3STEP 3

Pythagoras on △CBA gives AB² + r² = R², which rearranges to exactly what the area needs: R² - r² = 64.

R² - r² = AB² = 8² = 64

4STEP 4

The ring's area is the outer disk minus the inner disk, π(R² - r²). With R² - r² = 64, the area is 64π → (C).

Area = π R² - π r² = π(R² - r²) = 64π → (C)

Answer

64π

The outer disk has area π · 10² = 100π, so any answer must be smaller than 100π. Our answer 64π fits. We can also back-check the inner radius: r² = 100 - 64 = 36, so r = 6, and indeed 6 ＜ 10. The half-chord AB = 8 is the leg of a 6-8-10 right triangle — a classic Pythagorean triple — which confirms the geometry is consistent.

💡Key takeaway

Add one helper line (the radius to the tangent point) and the picture hands you a right triangle — then the Pythagorean Theorem you learn in Grade 8 finishes the problem in one line.

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