AMC 10 · 2002 · #5
Grade 7 geometry-2dEach of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. 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: Seven unit circles are packed inside one large circle: one circle sits at the center, six more ring it and touch both the center circle and the large outer circle. Find the area of the shaded part of the large circle that none of the seven small circles covers.
Givens: Every small circle has radius $1$; The center circle is tangent to each of the six surrounding circles; Each surrounding circle is tangent to the large circle and to its two small-circle neighbors; The shaded region is the part of the large disk left uncovered by all seven small circles; Answer choices: (A) $\pi$, (B) $1.5\pi$, (C) $2\pi$, (D) $3\pi$, (E) $3.5\pi$
Unknowns: The area of the shaded region
Understand
Restated: Seven unit circles are packed inside one large circle: one circle sits at the center, six more ring it and touch both the center circle and the large outer circle. Find the area of the shaded part of the large circle that none of the seven small circles covers.
Givens: Every small circle has radius $1$; The center circle is tangent to each of the six surrounding circles; Each surrounding circle is tangent to the large circle and to its two small-circle neighbors; The shaded region is the part of the large disk left uncovered by all seven small circles; Answer choices: (A) $\pi$, (B) $1.5\pi$, (C) $2\pi$, (D) $3\pi$, (E) $3.5\pi$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The shaded region has a jagged, six-lobed boundary that would be miserable to measure directly. Tool #16 (Count the Complement) sidesteps that entirely: the shaded area is just the whole large disk minus the seven round holes punched out of it, and a disk's area is easy. Before subtracting, tool #1 (Draw a Diagram) reads the tangency chain off the picture to pin down the one missing number — the large radius. Tool #7 (Identify Subproblems) then splits the job into two clean pieces: the big circle's area and the total area of the seven identical small circles. The main trap is miscounting the small circles as six (the visible ring) and forgetting the center one, which gives the decoy (D) $3\pi$.
Execute — Answer: C
7.G.B.4 Step 1 Build the large radius
- Put the center circle's center at the origin.
- It has radius $1$ and is tangent to a surrounding circle of radius $1$, so their centers are $1+1=2$ apart: each surrounding circle is centered $2$ units out.
- That surrounding circle reaches one more unit outward to touch the large circle, so the large radius runs $2+1=3$ from the center.
- The large circle has radius $3$.
💡 Walking straight out from the center, the radii of the touching circles line up end to end and simply add.
7.G.B.4 Step 2 Area of the large circle
- Using the circle-area formula with radius $3$, the whole large disk covers $\pi R^2 = \pi(3)^2 = 9\pi$.
- This is the total area we will carve the holes out of.
💡 The area of a circle is $\pi$ times its radius squared.
7.G.B.4 Step 3 Total area of the seven holes
- Count the small circles carefully: one at the center plus six in the ring makes $7$, and each has radius $1$, so each covers $\pi(1)^2 = \pi$.
- Seven of them cover $7 \times \pi = 7\pi$ altogether.
💡 Identical circles cover identical amounts, so the total is one circle's area times how many there are.
6.EE.A.3 Step 4 Subtract the holes
- The shaded region is everything in the large disk that the small circles do not cover, so subtract: $9\pi - 7\pi = 2\pi$.
- Since $\pi$ is a common factor, this is just $(9-7)\pi = 2\pi$.
- That is choice (C).
- (Using only the six ring circles by mistake would leave $9\pi - 6\pi = 3\pi$, the decoy (D).)
💡 Taking the round holes out of the big disk leaves exactly the shaded area behind.
7.G.B.4 Put the center circle's center at the origin. It has radius $1$ and is tangent t 7.G.B.4 Using the circle-area formula with radius $3$, the whole large disk covers $\pi 7.G.B.4 Count the small circles carefully: one at the center plus six in the ring makes 6.EE.A.3 The shaded region is everything in the large disk that the small circles do not Review
Reasonableness: The seven small circles cover $7\pi$ out of the large circle's $9\pi$, so only $2\pi$ — a bit over a fifth of the disk — stays shaded. That fits the picture, where the circles crowd nearly the whole interior and leave just thin curved slivers between them. The answer $2\pi$ is smaller than every circle-total in play, which is what an 'uncovered leftover' should be, and it is positive, as any real area must be.
Alternative: Skip the complement and reason with a ratio: the large radius $3$ is exactly three times a small radius $1$, so the large area is $3^2 = 9$ times a small area $\pi$, i.e. $9\pi$. Nine unit-circle areas of room, seven unit circles filling it, leaves $9-7 = 2$ unit-circle areas of shading, or $2\pi$.
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 (Reading the tangency chain to get the large radius $3$, then computing the large area $9\pi$ and each small area $\pi$ with $A=\pi r^2$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring out the common $\pi$ to combine like terms: $9\pi - 7\pi = (9-7)\pi = 2\pi$.)
⭐ When a shape is a big region with holes punched out, find its area by taking the whole area and subtracting the holes — just be sure to count every hole.
⭐ When a shape is a big region with holes punched out, find its area by taking the whole area and subtracting the holes — just be sure to count every hole.
More like this
Same archetype — closest grade level first.
