AMC 8 · 2010 · #19
Grade 8 geometry-2dProblem
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?
Pick an answer.
AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two circles share the same center $C$. A chord $\overline{AD}$ 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.
Givens: Both circles have the same center $C$; $\overline{AD}$ is a chord of the outer circle and is tangent to the inner circle at $B$; $AC = 10$, so the outer radius is $R = 10$; $AD = 16$; Answer choices: (A) $36\pi$, (B) $49\pi$, (C) $64\pi$, (D) $81\pi$, (E) $100\pi$
Unknowns: The area of the annulus (region between the two circles)
Understand
Restated: Two circles share the same center $C$. A chord $\overline{AD}$ 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.
Givens: Both circles have the same center $C$; $\overline{AD}$ is a chord of the outer circle and is tangent to the inner circle at $B$; $AC = 10$, so the outer radius is $R = 10$; $AD = 16$; Answer choices: (A) $36\pi$, (B) $49\pi$, (C) $64\pi$, (D) $81\pi$, (E) $100\pi$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #5 Look for a Pattern
The figure is already given, but the key move is to *add to it*: draw the inner radius $\overline{CB}$ to the tangent point. Tool #1 (Draw a Diagram) says label everything and add helper segments — once $\overline{CB} \perp \overline{AD}$ is marked, a right triangle $\triangle CBA$ pops out with hypotenuse $R = 10$ and leg $AB$. Tool #5 (Look for a Pattern) then catches a shortcut: by the Pythagorean Theorem $R^2 - r^2 = AB^2$, so the annulus area $\pi(R^2 - r^2)$ is just $\pi \cdot AB^2$ — no need to compute $r$ separately.
Execute — Answer: C
4.G.A.1 Step 1 - Add the inner radius $\overline{CB}$ to the diagram.
- Because $\overline{AD}$ is tangent to the inner circle at $B$, the radius to the point of tangency is perpendicular to the tangent: $\overline{CB} \perp \overline{AD}$.
- So $\triangle CBA$ is a right triangle with the right angle 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.1 Step 2 - A radius that is perpendicular to a chord bisects the chord.
- Since $\overline{CB} \perp \overline{AD}$ and $AD = 16$, point $B$ is the midpoint of $\overline{AD}$, so $AB = \tfrac{1}{2} \cdot 16 = 8$.
💡 Marking $B$ as the midpoint on the picture is what makes the right-triangle legs concrete.
8.G.B.7 Step 3 - Apply the Pythagorean Theorem to $\triangle CBA$: $AB^2 + CB^2 = CA^2$, that is $AB^2 + r^2 = R^2$.
- Rearrange to isolate $R^2 - r^2$ — the very quantity inside the annulus area formula.
💡 Instead of solving for $r$ then computing $R^2 - r^2$, spotting that the Pythagorean relation *is* the formula's $R^2 - r^2$ is a Grade 8 Pythagorean Theorem pattern.
7.G.B.4 Step 4 - The annulus area is the outer disk minus the inner disk.
- Using $R^2 - r^2 = 64$ from the previous step:
💡 Knowing the area of a circle is $\pi r^2$ and subtracting to get a ring is a direct Grade 7 circle-area application.
4.G.A.1 Add the inner radius $\overline{CB}$ to the diagram. Because $\overline{AD}$ is 4.G.A.1 A radius that is perpendicular to a chord bisects the chord. Since $\overline{CB 8.G.B.7 Apply the Pythagorean Theorem to $\triangle CBA$: $AB^2 + CB^2 = CA^2$, that is 7.G.B.4 The annulus area is the outer disk minus the inner disk. Using $R^2 - r^2 = 64$ Review
Reasonableness: The outer disk has area $\pi \cdot 10^2 = 100\pi$, so any answer must be smaller than $100\pi$. Our answer $64\pi$ fits. We can also back-check the inner radius: $r^2 = 100 - 64 = 36$, 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.
Alternative: Tool #7 (Identify Subproblems): solve for $r$ first, then plug in. From $8^2 + r^2 = 10^2$ we get $r^2 = 36$, so $r = 6$. Then the annulus area is $\pi(10)^2 - \pi(6)^2 = 100\pi - 36\pi = 64\pi$. Same answer (C), but it does one extra step (computing $r$) that the pattern-spotting route avoided.
CCSS standards used (min grade 8)
4.G.A.1Draw and identify points, lines, line segments, rays, angles, and perpendicular lines (Adding the inner radius $\overline{CB}$ to the figure and marking the right angle where it meets the tangent chord $\overline{AD}$, plus identifying $B$ as the midpoint of $\overline{AD}$.)7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Computing the annulus area as $\pi R^2 - \pi r^2$, a direct difference of two circle areas.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Relating the outer radius, inner radius, and half-chord via $AB^2 + r^2 = R^2$ in right triangle $\triangle CBA$, giving $R^2 - r^2 = AB^2 = 64$.)
⭐ 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.
⭐ 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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