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Difficulty: Very hardCircle Geometry: Arc Length and Sector Area

A circle is inscribed inside a sector of a larger circle. The larger circle has a radius of 18 inches18\text{ inches} and the sector has a central angle of 60∘60^\circ, as shown in the figure. What is the area, in square inches, of the region that is inside the sector but outside the inscribed circle?

  1. A
    27π27\pi
  2. B
    45π45\pi
  3. 18π18\piAnswer
  4. D
    72π72\pi
  5. E
    288π288\pi

Answer

The correct area is 18π18\pi square inches.
The correct answer is 18π18\pi square inches. First, the sector area is calculated as 60360×π×182=54π\frac{60}{360} \times \pi \times 18^2 = 54\pi. Second, using the right triangle formed by the sector's center, the inscribed circle's center, and the point of tangency, we set up sin⁡(30∘)=r18−r\sin(30^\circ) = \frac{r}{18-r}. Solving for rr gives r=6r = 6, so the area of the inscribed circle is π×62=36π\pi \times 6^2 = 36\pi. The difference between the two areas is 54π−36π=18π54\pi - 36\pi = 18\pi.

Step-by-Step Solution

1
Calculate the area of the 60∘60^\circ sector of the larger circle.
Sector Area = 54π54\pi square inches
The sector has a radius of R=18R = 18 and a central angle of 60∘60^\circ. The sector's area is a fraction of the total circle's area: Areasector=60360×πR2=16×π×182=324π6=54π\text{Area}_{\text{sector}} = \frac{60}{360} \times \pi R^2 = \frac{1}{6} \times \pi \times 18^2 = \frac{324\pi}{6} = 54\pi.
2
Determine the radius rr of the inscribed circle using trigonometry.
Inscribed Radius r=6r = 6 inches
The center of the inscribed circle, II, lies on the angle bisector of the sector. The line segment from the center of the sector OO to II bisects the 60∘60^\circ angle, forming a 30∘30^\circ angle. The distance from OO to the outer boundary of the sector is R=18R = 18, and the distance from II to the boundary is rr, so the hypotenuse OI=18−rOI = 18 - r. Drawing a perpendicular from II to one of the straight edges of the sector creates a right triangle with opposite side rr (the radius) and hypotenuse 18−r18 - r. Applying the sine ratio: sin⁡(30∘)=r18−r\sin(30^\circ) = \frac{r}{18 - r}. Since sin⁡(30∘)=0.5\sin(30^\circ) = 0.5, we solve 0.5=r18−r⇒18−r=2r⇒3r=18⇒r=60.5 = \frac{r}{18 - r} \Rightarrow 18 - r = 2r \Rightarrow 3r = 18 \Rightarrow r = 6.
3
Calculate the area of the inscribed circle.
Inscribed Circle Area = 36π36\pi square inches
The area of the inscribed circle with radius r=6r = 6 is Areacircle=πr2=π×62=36π\text{Area}_{\text{circle}} = \pi r^2 = \pi \times 6^2 = 36\pi.
4
Subtract the area of the inscribed circle from the area of the sector.
Remaining Area = 18π18\pi square inches
The area of the region inside the sector but outside the circle is found by subtracting the inscribed circle area from the sector area: 54π−36π=18π54\pi - 36\pi = 18\pi.

Key Concept

Using trigonometry on angle bisectors to determine the radius of a circle inscribed inside a sector, and applying sector and circle area formulas.
Estimated Time:3m 0s
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Topics

Circle Geometry: Arc Length and Sector Area

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