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Difficulty: HardCircles, Arc Lengths, and Sector Areas

In circle OO, line segments ABAB and CDCD are perpendicular diameters, each of length 1212. An arc of a second circle, centered at point AA with radius ACAC, is drawn from point CC to point DD through the interior of circle OO. What is the area of the crescent-shaped region bounded by the semicircle CBDCBD of circle OO and arc CDCD of the second circle?

Answer: 36

Answer

36
Circle OO has radius r=6r = 6, giving semicircle CBDCBD an area of 12π(62)=18π\frac{1}{2}\pi(6^2) = 18\pi. The distance AC=62+62=62AC = \sqrt{6^2 + 6^2} = 6\sqrt{2} is the radius of circle AA. Because ∠CAD=90∘\angle CAD = 90^\circ, sector ACDACD has area 90∘360∘π(62)2=18π\frac{90^\circ}{360^\circ}\pi(6\sqrt{2})^2 = 18\pi. Subtracting the area of triangle ACDACD (12×12×6=36\frac{1}{2} \times 12 \times 6 = 36) yields a segment area of 18π−3618\pi - 36. Subtracting this segment area from the semicircle area yields 18π−(18π−36)=3618\pi - (18\pi - 36) = 36.

Step-by-Step Solution

1
Find the radii of circle OO and circle AA.
Radius of circle OO is r=6r = 6. In right triangle AOCAOC, OA=OC=6OA = OC = 6, so radius AC=62+62=62AC = \sqrt{6^2 + 6^2} = 6\sqrt{2}.
Perpendicular diameters ABAB and CDCD intersect at center OO, dividing each diameter into radii of length 66.
2
Calculate the area of sector ACDACD of circle AA and triangle ACDACD.
Sector area =90∘360∘π(62)2=18π= \frac{90^\circ}{360^\circ} \pi (6\sqrt{2})^2 = 18\pi. Triangle area =12×12×6=36= \frac{1}{2} \times 12 \times 6 = 36.
Angle ∠CAD=90∘\angle CAD = 90^\circ because △ACD\triangle ACD is a right isosceles triangle with hypotenuse CD=12CD = 12.
3
Find the area of the circular segment bounded by chord CDCD and arc CDCD of circle AA.
Segment Area =18π−36= 18\pi - 36.
The area of a circular segment is equal to the sector area minus the triangle area.
4
Subtract the segment area from the area of semicircle CBDCBD of circle OO.
Region Area =18π−(18π−36)=36= 18\pi - (18\pi - 36) = 36.
Semicircle CBDCBD has radius 66 and area 12π(62)=18π\frac{1}{2}\pi(6^2) = 18\pi. Subtracting the segment area leaves the crescent region.

Key Concept

Area of circular sectors, segments, and compound regions (Lune of Hippocrates)
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Topics

Circles, Arc Lengths, and Sector Areas

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