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2015 AMC 10B Problem 9

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2015 AMC 10B Problem 9AMC 10 Step by Step
Name:                             Date: September 13, 2026
Circles · difficulty 2 of 5 · about 2 min · position P1-10

The shaded region below is called a shark's fin falcata, a figure studied by Leonardo da Vinci. It is bounded by the portion of the circle of radius 33 and center (0,0)(0,0) that lies in the first quadrant, the portion of the circle with radius 32\tfrac{3}{2} and center (0,32)(0,\tfrac{3}{2}) that lies in the first quadrant, and the line segment from (0,0)(0,0) to (3,0)(3,0) . What is the area of the shark's fin falcata?

figure

  1. A)

    4π5\dfrac{4\pi}{5}

  2. B)

    9π8\dfrac{9\pi}{8}

  3. C)

    4π3\dfrac{4\pi}{3}

  4. D)

    7π5\dfrac{7\pi}{5}

  5. E)

    3π2\dfrac{3\pi}{2}

Answer

B

Key insight

The falcata is a quarter disk of radius 3 with a half disk of radius 3/2 removed: 9pi/4 - 9pi/8 = 9pi/8.

Solution

The large circle, centered at the origin with radius 33, contributes the quarter disk in the first quadrant:
14⋅π⋅32=9π4.\frac14 \cdot \pi \cdot 3^2 = \frac{9\pi}{4}.

The small circle has center (0,32)(0, \tfrac32) and radius 32\tfrac32, so it passes through (0,0)(0,0) and (0,3)(0,3) and its center sits on the yy-axis. Exactly half of it lies in the first quadrant (the right half), and that half disk is cut out of the quarter disk:
12⋅π(32)2=9π8.\frac12 \cdot \pi \left(\frac32\right)^2 = \frac{9\pi}{8}.

Falcata area:
9π4−9π8=9π8.\frac{9\pi}{4} - \frac{9\pi}{8} = \frac{9\pi}{8}.

The answer is (B) 9π8\boxed{\textbf{(B)}\ \frac{9\pi}{8}}.

Why this works

Curved regions bounded by circular arcs are almost always "big sector minus small sector or disk." Identify each arc's circle, decide what fraction of that circle the arc encloses (using where the center sits relative to the axes), and subtract. Here the small circle is tangent to the xx-axis at the origin and bisected by the yy-axis, so the piece removed is precisely a semicircle.

The trap

Subtracting a quarter of the small circle instead of half of it; the small circle is centered on the y-axis, so exactly half of it lies in the first quadrant.

Common mistakes

  • Subtracting a quarter of the small circle instead of half of it; the small circle is centered on the yy-axis, so exactly half of it lies in the first quadrant.
  • Using diameter 32\tfrac32 (radius 34\tfrac34) for the small circle, or subtracting the whole small disk.

Techniques

Cut the figure into known shapes (triangles, rectangles, sectors)