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Zorluk: ZorCircles, Arc Lengths, and Sector Areas

In circle OO, sector AOBAOB has a central angle of 60∘60^\circ and a radius of 1212. A smaller circle CC is inscribed within sector AOBAOB such that it is tangent to radii OAOA and OBOB, as well as to arc ABAB. What is the area of the region inside sector AOBAOB that lies outside circle CC?

  1. A
    4π4\pi
  2. 8π8\piCevap
  3. C
    12π12\pi
  4. D
    16π16\pi
  5. E
    18π18\pi

Cevap

The area of the region inside sector AOBAOB outside circle CC is 8π8\pi.
The correct answer is derived by first finding the area of sector AOBAOB using 60360π(122)=24π\frac{60}{360} \pi (12^2) = 24\pi. Then, analyzing the geometry of the inscribed circle reveals that the line from OO to the center of circle CC bisects the 60∘60^\circ angle. In the resulting 30∘−60∘−90∘30^\circ-60^\circ-90^\circ right triangle, the hypotenuse length is 2r2r, making the total radius of sector AOBAOB equal to 2r+r=3r=122r + r = 3r = 12, which yields r=4r = 4. The area of circle CC is π(42)=16π\pi (4^2) = 16\pi. Subtracting the circle area from the sector area gives 24π−16π=8π24\pi - 16\pi = 8\pi.

Adım Adım Çözüm

1
Calculate the area of sector AOBAOB
Sector area =60∘360∘×π(122)=16×144π=24π= \frac{60^\circ}{360^\circ} \times \pi (12^2) = \frac{1}{6} \times 144\pi = 24\pi
The area of a sector with central angle θ\theta and radius RR is θ360∘πR2\frac{\theta}{360^\circ} \pi R^2.
2
Find the radius rr of the inscribed circle CC
Distance from OO to center of circle CC is OP=2rOP = 2r, so total radius R=OP+r=3r=12  ⟹  r=4R = OP + r = 3r = 12 \implies r = 4
The line segment connecting center OO to center PP of circle CC bisects the 60∘60^\circ angle into two 30∘30^\circ angles. A perpendicular dropped from PP to radius OAOA forms a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ right triangle where sin⁡(30∘)=rOP=12\sin(30^\circ) = \frac{r}{OP} = \frac{1}{2}, giving OP=2rOP = 2r.
3
Calculate the area of circle CC
Area of circle C=πr2=π(42)=16πC = \pi r^2 = \pi (4^2) = 16\pi
The area of a circle with radius rr is πr2\pi r^2.
4
Subtract the area of circle CC from the area of sector AOBAOB
24π−16π=8π24\pi - 16\pi = 8\pi
The desired region is the difference between the full sector area and the enclosed circle's area.

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Inscribed circles in sectors and central angle sector area calculations
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