AMC 8 · 2010 · #19

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

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

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

The two circles pictured have the same center $C$ . Chord $\overline{AD}$ is tangent to the inner circle at $B$ , $AC$ is $10$ , and chord $\overline{AD}$ has length $16$ . What is the area between the two circles?

Pick an answer.

(A)

$36\pi$

(B)

$49\pi$

(C)

$64\pi$

(D)

$81\pi$

(E)

$100\pi$

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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

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

#1 Draw a Diagram 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 $B$.

$$\angle CBA = 90^\circ,\quad CA = R = 10,\quad CB = r$$

💡 Drawing the radius to the tangent point and marking the right angle is the standard Grade 4 "points, lines, perpendicular lines" vocabulary in action.

#1 Draw a Diagram 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 $AB = \tfrac{1}{2} \cdot 16 = 8$.

$$AB = \tfrac{1}{2} \cdot AD = \tfrac{1}{2} \cdot 16 = 8$$

💡 Marking $B$ as the midpoint on the picture is what makes the right-triangle legs concrete.

#5 Look for a Pattern 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.

$$R^2 - r^2 = AB^2 = 8^2 = 64$$

💡 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.

#5 Look for a 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:

$$\text{Area} = \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = 64\pi \;\Rightarrow\; \textbf{(C)}$$

💡 Knowing the area of a circle is $\pi r^2$ and subtracting to get a ring is a direct Grade 7 circle-area application.

[1] #1 4.G.A.1 Add the inner radius $\overline{CB}$ to the diagram. Because $\overline{AD}$ is

[2] #1 4.G.A.1 A radius that is perpendicular to a chord bisects the chord. Since $\overline{CB

[3] #5 8.G.B.7 Apply the Pythagorean Theorem to $\triangle CBA$: $AB^2 + CB^2 = CA^2$, that is

[4] #5 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.1 Draw 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.4 Know 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.7 Apply 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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