AMC 10 · 2025 · #10
Grade 8 geometry-2dA semicircle has diameter AB and chord CD of length 16 parallel to AB. A smaller semicircle
with diameter on AB and tangent to CD is cut from the larger semicircle, as shown below.
What is the area of the resulting figure, shown shaded?
Pick an answer.
AMC 10 2025 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 large semicircle has diameter $\overline{AB}$. Inside it a chord $\overline{CD}$ of length $16$ runs parallel to $\overline{AB}$. A smaller semicircle, with its flat diameter lying on $\overline{AB}$ and its curved top just touching $\overline{CD}$, is removed from the large one. Find the area of the shaded region that is left.
Givens: The large region is a semicircle with diameter $\overline{AB}$; Chord $\overline{CD}$ has length $16$ and is parallel to $\overline{AB}$; The removed small semicircle has its diameter on $\overline{AB}$ and is tangent to $\overline{CD}$ from below; Answer choices: (A) $16\pi$, (B) $24\pi$, (C) $32\pi$, (D) $48\pi$, (E) $64\pi$
Unknowns: The area of the shaded figure (large semicircle with the small semicircle cut out)
Understand
Restated: A large semicircle has diameter $\overline{AB}$. Inside it a chord $\overline{CD}$ of length $16$ runs parallel to $\overline{AB}$. A smaller semicircle, with its flat diameter lying on $\overline{AB}$ and its curved top just touching $\overline{CD}$, is removed from the large one. Find the area of the shaded region that is left.
Givens: The large region is a semicircle with diameter $\overline{AB}$; Chord $\overline{CD}$ has length $16$ and is parallel to $\overline{AB}$; The removed small semicircle has its diameter on $\overline{AB}$ and is tangent to $\overline{CD}$ from below; Answer choices: (A) $16\pi$, (B) $24\pi$, (C) $32\pi$, (D) $48\pi$, (E) $64\pi$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #1 Draw a Diagram
No radius is given a number, so I cannot compute either semicircle's area on its own. Tool #4 (Introduce a Variable) is the move: call the two radii $R$ and $r$, write the shaded area in terms of them, and watch what the problem actually forces you to know. It turns out you never need $R$ or $r$ alone — only the single combination $R^2 - r^2$ — and that is exactly what one right triangle hands you. Tool #7 (Identify Subproblems) splits the shaded region into big-semicircle-minus-small-semicircle, and Tool #1 (Draw a Diagram) supplies the radius-to-chord right triangle that pins down $R^2 - r^2$.
Execute — Answer: C
7.G.B.4 Step 1 Split the shaded area
- The shaded figure is what remains after cutting the small semicircle out of the large one, so its area is the large semicircle's area minus the small semicircle's area.
- A semicircle is half a full circle, so each area is half of $\pi \times (\text{radius})^2$.
💡 Cutting one shape out of another means subtract the areas — nothing more.
6.EE.A.3 Step 2 Name the two radii
- Let $R$ be the radius of the large semicircle and $r$ the radius of the small one.
- Writing the areas and factoring out $\tfrac{\pi}{2}$ shows the shaded area depends only on $R^2 - r^2$, not on $R$ and $r$ by themselves.
- That is the whole point: I only have to find this one combined quantity.
💡 Pulling out the common factor $\tfrac{\pi}{2}$ reveals that only $R^2 - r^2$ matters.
7.G.B.4 Step 3 Tangency fixes the height of CD
- The small semicircle rests flat on $\overline{AB}$ and curves upward, so its highest point is exactly $r$ above $\overline{AB}$.
- Being tangent to $\overline{CD}$ means $\overline{CD}$ just grazes that top point.
- Therefore $\overline{CD}$ sits at height exactly $r$ above the diameter $\overline{AB}$.
💡 A semicircle of radius $r$ tops out at height $r$, and a tangent line kisses it right there.
8.G.B.7 Step 4 Right triangle from center to C
- Let $O$ be the center of $\overline{AB}$.
- Drop the perpendicular from $O$ to $\overline{CD}$; a perpendicular from the center bisects a chord, so it meets $\overline{CD}$ at its midpoint and each half is $8$.
- That foot is at height $r$ (the height of $\overline{CD}$), so the perpendicular leg is $r$.
- The segment $OC$ is a radius of the large circle, length $R$.
- The horizontal leg $8$, the vertical leg $r$, and the hypotenuse $R$ form a right triangle, so Pythagoras gives $R^2 - r^2 = 64$.
💡 Half the chord, the drop from the center, and the radius always make a right triangle.
6.EE.A.2 Step 5 Substitute and finish
- Put $R^2 - r^2 = 64$ into the shaded-area expression from Step 2.
- The unknown radii vanish and a single number comes out.
💡 Once the combined quantity $R^2 - r^2$ is known, the answer is one substitution away.
7.G.B.4 The shaded figure is what remains after cutting the small semicircle out of the 6.EE.A.3 Let $R$ be the radius of the large semicircle and $r$ the radius of the small on 7.G.B.4 The small semicircle rests flat on $\overline{AB}$ and curves upward, so its hig 8.G.B.7 Let $O$ be the center of $\overline{AB}$. Drop the perpendicular from $O$ to $\o 6.EE.A.2 Put $R^2 - r^2 = 64$ into the shaded-area expression from Step 2. The unknown ra Review
Reasonableness: The final area $32\pi$ never used a specific value of $R$ or $r$, which matches the problem never telling us where the small semicircle sits — good sign. Test the extreme case where the small semicircle shrinks to nothing ($r \to 0$): then $\overline{CD}$ slides down onto $\overline{AB}$, the chord of length $16$ becomes the diameter, so $R = 8$ and the plain semicircle area is $\tfrac{1}{2}\pi (8)^2 = 32\pi$. Same number, so the subtraction is consistent. And $32\pi$ is choice (C), squarely in the middle of the options.
Alternative: Tool #14 (Extreme Principle) skips most of the algebra. Since the problem's answer cannot depend on the small semicircle's size (its position and radius are never specified), just push it to the limiting case $r = 0$. Then $\overline{CD}$ coincides with $\overline{AB}$, making the length-$16$ chord the diameter of the large semicircle, so $R = 8$ and the area is $\tfrac{1}{2}\pi(8)^2 = 32\pi$ directly. The full derivation confirms this shortcut is legitimate.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for area and circumference of a circle (Writing each semicircle's area as half of $\pi r^2$, and using the circle fact that a semicircle of radius $r$ rises to height $r$ where a tangent line touches it.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring $\tfrac{1}{2}\pi R^2 - \tfrac{1}{2}\pi r^2$ into $\tfrac{\pi}{2}(R^2 - r^2)$ to expose that only $R^2 - r^2$ is needed.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Using the right triangle with legs $8$ and $r$ and hypotenuse $R$ (radius to endpoint $C$) to get $R^2 - r^2 = 64$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting $R^2 - r^2 = 64$ into $\tfrac{\pi}{2}(R^2 - r^2)$ to evaluate the shaded area as $32\pi$.)
⭐ When no radius is given, name them with letters and watch for a combination like $R^2 - r^2$ that a single right triangle can hand you — you often never need the radii by themselves.
⭐ When no radius is given, name them with letters and watch for a combination like $R^2 - r^2$ that a single right triangle can hand you — you often never need the radii by themselves.
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