AMC 10 · 2003 · #19
Grade 8 geometry-2dA semicircle of diameter 1 sits at the top of a semicircle of diameter 2, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.
Pick an answer.
AMC 10 2003 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 small semicircle with diameter $1$ sits on top of a large semicircle with diameter $2$. The flat edge of the small semicircle is a chord of the large semicircle, and both of its endpoints lie on the large arc. The lune is the region inside the small semicircle but outside the large semicircle. Find its area.
Givens: The large semicircle has diameter $2$, so its radius is $1$.; The small semicircle has diameter $1$, so its radius is $\tfrac12$.; The small semicircle's straight edge (its diameter of length $1$) is a chord of the large circle, with both endpoints sitting on the large arc.; Answer choices: (A) $\tfrac16\pi-\tfrac{\sqrt3}{4}$, (B) $\tfrac{\sqrt3}{4}-\tfrac1{12}\pi$, (C) $\tfrac{\sqrt3}{4}-\tfrac1{24}\pi$, (D) $\tfrac{\sqrt3}{4}+\tfrac1{24}\pi$, (E) $\tfrac{\sqrt3}{4}+\tfrac1{12}\pi$.
Unknowns: The area of the lune — the shaded crescent inside the small semicircle and outside the large semicircle.
Understand
Restated: A small semicircle with diameter $1$ sits on top of a large semicircle with diameter $2$. The flat edge of the small semicircle is a chord of the large semicircle, and both of its endpoints lie on the large arc. The lune is the region inside the small semicircle but outside the large semicircle. Find its area.
Givens: The large semicircle has diameter $2$, so its radius is $1$.; The small semicircle has diameter $1$, so its radius is $\tfrac12$.; The small semicircle's straight edge (its diameter of length $1$) is a chord of the large circle, with both endpoints sitting on the large arc.; Answer choices: (A) $\tfrac16\pi-\tfrac{\sqrt3}{4}$, (B) $\tfrac{\sqrt3}{4}-\tfrac1{12}\pi$, (C) $\tfrac{\sqrt3}{4}-\tfrac1{24}\pi$, (D) $\tfrac{\sqrt3}{4}+\tfrac1{24}\pi$, (E) $\tfrac{\sqrt3}{4}+\tfrac1{12}\pi$.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement
The crescent has curved edges on both sides, so there is no single formula for it. Tool #7 (Identify Subproblems) breaks the shape into pieces whose areas we do know: a semicircle, a circular sector, and a triangle. Tool #1 (Draw a Diagram) is what reveals the key fact — the small semicircle's flat edge is a chord of the big circle, and drawing the two big radii to its ends exposes a triangle. Tool #16 (Change Focus / Count the Complement) supplies the framing: instead of chasing the odd crescent directly, take the whole small semicircle and subtract the single sliver of it that dips inside the big circle. That sliver is a plain circular segment we can measure.
Execute — Answer: C
7.G.B.4 Step 1 Fix the two radii
- Diameter and radius differ by a factor of $2$.
- The large semicircle has diameter $2$, so its radius is $R=1$.
- The small semicircle has diameter $1$, so its radius is $r=\tfrac12$.
- The small semicircle's flat edge has length $1$ (its diameter), and its two ends rest on the large arc, so that edge is a chord of the large circle of length $1$.
💡 Halving each diameter gives the two radii you actually compute with.
7.G.B.4 Step 2 Area of the small semicircle
- A semicircle is half a circle, so its area is $\tfrac12\pi r^2$.
- With $r=\tfrac12$, this is $\tfrac12\pi\left(\tfrac12\right)^2=\tfrac12\pi\cdot\tfrac14=\tfrac{\pi}{8}$.
- This whole half-disk is what we start from; the lune is this minus whatever part of it lies inside the big semicircle.
💡 The crescent is the little half-circle with one bite taken out of it, so start with the full half-circle.
7.G.B.6 Step 3 Name the bite: a circular segment
- The part of the small semicircle that lies inside the large semicircle is the region trapped between the chord and the large arc — a circular segment of the big circle sitting above the chord.
- So the lune equals the small semicircle minus this segment.
- Everything now reduces to measuring that one segment.
💡 Rather than measure the strange crescent head-on, subtract the single simple piece that overlaps the big circle.
7.G.B.4 Step 4 The chord equals the big radius, so the angle is 60 degrees
- Draw the two radii of the large circle out to the chord's endpoints.
- Each radius is $1$ and the chord is also $1$, so this triangle has all three sides equal to $1$ — it is equilateral.
- Every angle of an equilateral triangle is $60^\circ$, so the central angle of the big circle over this chord is $60^\circ$.
- The sector (pie slice) it cuts is $\tfrac{60}{360}=\tfrac16$ of the whole circle: $\tfrac16\pi R^2=\tfrac16\pi(1)^2=\tfrac{\pi}{6}$.
💡 When a chord matches the radius, the slice it makes is a tidy sixth of the circle.
8.G.B.7 Step 5 Segment = sector minus triangle
- A circular segment is the sector with its triangle removed.
- The triangle is equilateral with side $1$; its height comes from the Pythagorean theorem, $h=\sqrt{1^2-\left(\tfrac12\right)^2}=\tfrac{\sqrt3}{2}$, so its area is $\tfrac12\cdot 1\cdot\tfrac{\sqrt3}{2}=\tfrac{\sqrt3}{4}$.
- Therefore the segment above the chord is $\tfrac{\pi}{6}-\tfrac{\sqrt3}{4}$.
💡 Slice off the flat triangular base of the pie slice and only the curved sliver is left.
7.G.B.6 Step 6 Assemble the lune
- Subtract the segment from the small semicircle: $\tfrac{\pi}{8}-\left(\tfrac{\pi}{6}-\tfrac{\sqrt3}{4}\right)=\tfrac{\pi}{8}-\tfrac{\pi}{6}+\tfrac{\sqrt3}{4}$.
- Combine the $\pi$ terms over $24$: $\tfrac{3\pi}{24}-\tfrac{4\pi}{24}=-\tfrac{\pi}{24}$.
- So the lune equals $\tfrac{\sqrt3}{4}-\tfrac{\pi}{24}$, which is choice (C).
💡 Full half-circle minus the overlapping sliver leaves exactly the crescent.
7.G.B.4 Diameter and radius differ by a factor of $2$. The large semicircle has diameter 7.G.B.4 A semicircle is half a circle, so its area is $\tfrac12\pi r^2$. With $r=\tfrac1 7.G.B.6 The part of the small semicircle that lies inside the large semicircle is the re 7.G.B.4 Draw the two radii of the large circle out to the chord's endpoints. Each radius 8.G.B.7 A circular segment is the sector with its triangle removed. The triangle is equi 7.G.B.6 Subtract the segment from the small semicircle: $\tfrac{\pi}{8}-\left(\tfrac{\pi Review
Reasonableness: The lune is a thin crescent, so its area should be a bit less than the whole small semicircle and clearly positive. Numerically $\tfrac{\sqrt3}{4}-\tfrac{\pi}{24}\approx 0.433-0.131=0.302$, while the small semicircle is $\tfrac{\pi}{8}\approx 0.393$ — the lune is a little smaller, exactly as expected after cutting off a small segment. Choice (A) is negative ($\approx-0.09$) and impossible for an area; (B) $\approx 0.171$ is too small; (C) $\approx 0.302$ fits.
Alternative: Tool #3, Eliminate Possibilities: every choice shares the $\tfrac{\sqrt3}{4}$ term, so the fight is only over the $\pi$ correction. Since the overlap segment ($\tfrac{\pi}{6}-\tfrac{\sqrt3}{4}$) is subtracted from the half-circle ($\tfrac{\pi}{8}$), the $\pi$ part is $\tfrac{\pi}{8}-\tfrac{\pi}{6}=-\tfrac{\pi}{24}$. Only choice (C) carries $-\tfrac{\pi}{24}$, so it must be the answer without recomputing the triangle.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Computing the area of the small semicircle ($\tfrac12\pi r^2$) and the 60-degree sector of the large circle ($\tfrac16\pi R^2$).)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding the height $\tfrac{\sqrt3}{2}$ of the equilateral triangle of side $1$ to get its area $\tfrac{\sqrt3}{4}$.)7.G.B.6Solve real-world and mathematical problems involving area of two-dimensional objects composed of triangles, quadrilaterals, and polygons (Decomposing the lune into a semicircle minus a circular segment, and the segment into a sector minus a triangle, then combining the areas.)
⭐ The crescent equals the little half-circle minus the sliver of it that overlaps the big circle; since the flat edge equals the big radius, that sliver is a $60^\circ$ pie slice with its equilateral triangle removed, leaving $\tfrac{\sqrt3}{4}-\tfrac{\pi}{24}$.
⭐ The crescent equals the little half-circle minus the sliver of it that overlaps the big circle; since the flat edge equals the big radius, that sliver is a $60^\circ$ pie slice with its equilateral triangle removed, leaving $\tfrac{\sqrt3}{4}-\tfrac{\pi}{24}$.
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