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Question

Difficulty: Very hardCircles, Arc Lengths, and Sector Areas

In a circle centered at point OO with radius RR, radii OAOA and OBOB bound a sector AOBAOB with a central angle of 120∘120^\circ. Point PP lies on minor arc ABAB such that the ratio of the length of arc APAP to the length of arc PBPB is 1:31:3. Segment PQPQ is drawn perpendicular to radius OAOA, intersecting OAOA at point QQ. What is the ratio of the area of the region bounded by line segment PQPQ, line segment AQAQ, and minor arc APAP to the area of sector AOBAOB?

  1. 4π−3316π\frac{4\pi - 3\sqrt{3}}{16\pi}Answer
  2. B
    4π−334π\frac{4\pi - 3\sqrt{3}}{4\pi}
  3. C
    4π−338π\frac{4\pi - 3\sqrt{3}}{8\pi}
  4. D
    π−34π\frac{\pi - 3}{4\pi}
  5. E
    14\frac{1}{4}

Answer

4π−3316π\frac{4\pi - 3\sqrt{3}}{16\pi}
The correct answer 4π−3316π\frac{4\pi - 3\sqrt{3}}{16\pi} is obtained by finding the central angle of sector AOPAOP (30∘30^\circ), subtracting the area of right triangle OQPOQP (316R2\frac{\sqrt{3}}{16}R^2) from the area of sector AOPAOP (π12R2\frac{\pi}{12}R^2), and dividing this bounded area by the total sector AOBAOB area (π3R2\frac{\pi}{3}R^2).

Step-by-Step Solution

1
Determine the central angle ∠AOP\angle AOP of sector AOPAOP
∠AOP=11+3×120∘=30∘\angle AOP = \frac{1}{1+3} \times 120^\circ = 30^\circ
Arc length is directly proportional to central angle. A 1:31:3 ratio means sector AOPAOP represents 14\frac{1}{4} of the 120∘120^\circ central angle.
2
Calculate the area of sector AOPAOP and the total area of sector AOBAOB
Area(Sector AOP)=30∘360∘πR2=π12R2\text{Area(Sector } AOP) = \frac{30^\circ}{360^\circ} \pi R^2 = \frac{\pi}{12} R^2, and Area(Sector AOB)=120∘360∘πR2=π3R2\text{Area(Sector } AOB) = \frac{120^\circ}{360^\circ} \pi R^2 = \frac{\pi}{3} R^2
The area of a sector with central angle θ\theta is θ360∘πR2\frac{\theta}{360^\circ} \pi R^2.
3
Calculate the area of right triangle OQPOQP
Area(△OQP)=316R2\text{Area}(\triangle OQP) = \frac{\sqrt{3}}{16} R^2
In △OQP\triangle OQP, ∠QOP=30∘\angle QOP = 30^\circ and hypotenuse OP=ROP = R. Thus OQ=Rcos⁡(30∘)=R32OQ = R \cos(30^\circ) = \frac{R\sqrt{3}}{2} and PQ=Rsin⁡(30∘)=R2PQ = R \sin(30^\circ) = \frac{R}{2}. Area=12×OQ×PQ=12(R32)(R2)=316R2\text{Area} = \frac{1}{2} \times OQ \times PQ = \frac{1}{2} \left(\frac{R\sqrt{3}}{2}\right)\left(\frac{R}{2}\right) = \frac{\sqrt{3}}{16} R^2.
4
Calculate the area of the bounded region and take the ratio to the area of sector AOBAOB
Ratio=π12R2−316R2π3R2=4π−3316π\text{Ratio} = \frac{\frac{\pi}{12}R^2 - \frac{\sqrt{3}}{16}R^2}{\frac{\pi}{3}R^2} = \frac{4\pi - 3\sqrt{3}}{16\pi}
Subtract the triangle area from sector AOPAOP area to get the bounded region area, then divide by sector AOBAOB area.

Key Concept

Sector Area, Arc Length Proportions, and Geometric Region Subdivision
Estimated Time:3m 0s
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Topics

Circles, Arc Lengths, and Sector Areas

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