AMC 10 · 2002 · #5
Grade 7 geometry-2dCircles of radius 2 and 3 are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.
Pick an answer.
AMC 10 2002 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: A circle of radius $3$ and a circle of radius $2$ sit side by side, touching each other, and both fit exactly inside a larger circle that touches each of them. Find the area inside the big circle but outside the two small circles.
Givens: The small circles have radii $3$ and $2$; The two small circles are externally tangent (they touch at one point); The big circle passes through the far edge of each small circle (circumscribes them); The shaded region is the part of the big circle not covered by either small circle; Answer choices: (A) $3\pi$, (B) $4\pi$, (C) $6\pi$, (D) $9\pi$, (E) $12\pi$
Unknowns: The area of the shaded region
Understand
Restated: A circle of radius $3$ and a circle of radius $2$ sit side by side, touching each other, and both fit exactly inside a larger circle that touches each of them. Find the area inside the big circle but outside the two small circles.
Givens: The small circles have radii $3$ and $2$; The two small circles are externally tangent (they touch at one point); The big circle passes through the far edge of each small circle (circumscribes them); The shaded region is the part of the big circle not covered by either small circle; Answer choices: (A) $3\pi$, (B) $4\pi$, (C) $6\pi$, (D) $9\pi$, (E) $12\pi$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The whole problem turns on one picture fact: the flat line drawn through both small centers is actually a diameter of the big circle, so Tool #1 (Draw a Diagram) unlocks the size of the big circle by reading the lengths straight off that line. Once the big radius is known, Tool #7 (Identify Subproblems) splits the shaded area into three easy circle-area calculations that combine by subtraction — big circle minus the two small ones.
Execute — Answer: E
7.G.B.4 Step 1 Read the big diameter off the line
- The horizontal line runs through the center of each small circle, so it is a straight path across the whole big circle — a diameter.
- Walking along it from one side of the big circle to the other, you cross the radius-$3$ circle twice ($3+3$) and the radius-$2$ circle twice ($2+2$), because the two small circles touch with no gap.
- So the big diameter is $3+3+2+2=10$.
💡 Lining the small circles up edge to edge stretches them across the full width of the big circle.
7.G.B.4 Step 2 Get the big radius
The radius is half the diameter, so the big circle has radius $10\div2=5$.
💡 Radius is always half the distance straight across.
7.G.B.4 Step 3 Area of the big circle
Use the circle-area formula $\pi r^2$ with $r=5$: the big circle has area $\pi\cdot 5^2 = 25\pi$.
💡 Squaring the radius and multiplying by $\pi$ gives the space a circle covers.
7.G.B.4 Step 4 Area of the two small circles
- The same formula gives each small area.
- The radius-$3$ circle covers $\pi(3)^2=9\pi$, and the radius-$2$ circle covers $\pi(2)^2=4\pi$.
- Together they cover $9\pi+4\pi=13\pi$.
💡 Each white circle is just its own $\pi r^2$.
6.EE.A.3 Step 5 Subtract to get the shaded area
- The shaded region is what is left of the big circle after removing both small circles.
- Since every area is a number of $\pi$'s, subtract the counts: $25\pi-13\pi=(25-13)\pi=12\pi$.
- That is choice (E).
💡 Take away the covered part and the leftover $\pi$'s are the shaded area.
7.G.B.4 The horizontal line runs through the center of each small circle, so it is a str 7.G.B.4 The radius is half the diameter, so the big circle has radius $10\div2=5$. 7.G.B.4 Use the circle-area formula $\pi r^2$ with $r=5$: the big circle has area $\pi\c 7.G.B.4 The same formula gives each small area. The radius-$3$ circle covers $\pi(3)^2=9 6.EE.A.3 The shaded region is what is left of the big circle after removing both small ci Review
Reasonableness: The big circle ($25\pi$) is much larger than the two small ones combined ($13\pi$), so a shaded region of $12\pi$ — a little less than half the big circle — looks right for two chunks bitten out of it. A quick sanity check on the smaller choices: $3\pi$ or $4\pi$ would mean the small circles cover almost the whole big circle, which the picture clearly contradicts. The key move that must be right is the diameter: mistakenly using $R=5$ as a diameter or forgetting to double the radii would throw the whole answer off.
Alternative: Work with the diameter fact algebraically instead of by walking the line: the big diameter equals the sum of the two small diameters, $2\cdot3+2\cdot2=10$, so $R=5$ and shaded $=\pi R^2-\pi(3)^2-\pi(2)^2=25\pi-9\pi-4\pi=12\pi$. Same result in one line.
CCSS standards used (min grade 7)
7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Relating the small radii to the big circle's diameter ($10$) and radius ($5$), then computing the areas $25\pi$, $9\pi$, and $4\pi$ with $\pi r^2$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring out the common $\pi$ to subtract $25\pi-13\pi=(25-13)\pi=12\pi$.)
⭐ Line the two circles up edge to edge and their widths add to the big circle's diameter; then the shaded part is just the big circle's area minus the two smaller ones.
⭐ Line the two circles up edge to edge and their widths add to the big circle's diameter; then the shaded part is just the big circle's area minus the two smaller ones.
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