Competition · AMC preparation · step 4 of 4
AMC 8 · 2010 · #19
Grade 8 geometry-2d
Pick an answer.
AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Draw the inner radius
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.
Drawing the radius to the tangent point and marking the right angle is the standard Grade 4 "points, lines, perpendicular lines" vocabulary in action.
4.G.A.1Draw A DiagramUse the bisected chord
A radius perpendicular to a chord bisects it, so B is the midpoint of AD and AB = 8 (half of 16).
Marking B as the midpoint on the picture is what makes the right-triangle legs concrete.
4.G.A.1Draw A DiagramApply the Pythagorean Theorem
Pythagoras on △CBA gives AB² + r² = R², which rearranges to exactly what the area needs: R² - r² = 64.
Instead of solving for r then computing R² - r², spotting that the Pythagorean relation is the formula's R² - r² is a Grade 8 Pythagorean Theorem pattern.
In the right triangle whose hypotenuse is the outer radius CA=R and whose legs are the inner radius CB=r and the half-chord AB, the two squared radii differ by exactly the squared half-chord: R² - r² = AB².
▸ Why?
The points C, B, A form a right triangle with legs AB and CB and hypotenuse CA, and in a right triangle the squares on the two legs add up to the square on the hypotenuse, so AB² + CB² = CA²; putting CB=r and CA=R and moving r² to the other side leaves R² - r² = AB².
▸ Why?
The corner at B is a right angle: CB is the inner radius that reaches the chord AD exactly at B, the single point where AD grazes the inner circle, and a radius drawn to that point of contact meets the tangent line square-on.
▸ Why?
Once △ CBA is known to be right-angled at B, its two legs AB and CB and its hypotenuse CA obey the leg-leg-hypotenuse square law, giving AB² + CB² = CA² directly.
Subtract the two disk areas
The ring's area is the outer disk minus the inner disk, π(R² - r²). With R² - r² = 64, the area is 64π → (C).
Knowing the area of a circle is π r² and subtracting to get a ring is a direct Grade 7 circle-area application.
7.G.B.4Look For A PatternAdd 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.
- Draw the inner radius
- Use the bisected chord
- Apply the Pythagorean Theorem
- Subtract the two disk areas
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