Examkin

Question

Difficulty: HardCircles, Arc Lengths, and Sector Areas

In a circle with center OO and radius 1212, radii OAOA and OBOB are perpendicular. Point CC lies on segment OAOA such that CC is the midpoint of OAOA. A line segment perpendicular to OAOA is drawn from point CC to intersect minor arc ABAB at point DD. What is the area of the region bounded by line segment CDCD, line segment CACA, and minor arc ADAD?

  1. A
    12π12\pi
  2. B
    24π−1824\pi - 18
  3. 24π−18324\pi - 18\sqrt{3}Answer
  4. D
    36π−18336\pi - 18\sqrt{3}
  5. E
    24π−36324\pi - 36\sqrt{3}

Answer

24π−18324\pi - 18\sqrt{3}
The area of the region bounded by line segment CDCD, segment CACA, and minor arc ADAD is obtained by subtracting the area of right triangle OCDOCD from the area of sector OADOAD. With OC=6OC = 6 and radius OD=12OD = 12, triangle OCDOCD is a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ right triangle, giving CD=63CD = 6\sqrt{3} and central angle ∠AOD=60∘\angle AOD = 60^\circ. The sector area is 60∘360∘⋅π(122)=24π\frac{60^\circ}{360^\circ} \cdot \pi (12^2) = 24\pi, and the triangle area is 12⋅6⋅63=183\frac{1}{2} \cdot 6 \cdot 6\sqrt{3} = 18\sqrt{3}. Subtracting the triangle area from the sector area yields 24π−18324\pi - 18\sqrt{3}.

Step-by-Step Solution

1
Determine the length of OCOC and the height CDCD
OC=6OC = 6 and CD=63CD = 6\sqrt{3}
Since CC is the midpoint of radius OA=12OA = 12, OC=6OC = 6. Triangle OCDOCD is a right triangle at CC with hypotenuse OD=12OD = 12 (radius of circle). By the Pythagorean theorem, CD=122−62=108=63CD = \sqrt{12^2 - 6^2} = \sqrt{108} = 6\sqrt{3}.
2
Find the central angle ∠AOD\angle AOD
∠AOD=60∘\angle AOD = 60^\circ
In right triangle OCDOCD, cos⁡(∠AOD)=OCOD=612=12\cos(\angle AOD) = \frac{OC}{OD} = \frac{6}{12} = \frac{1}{2}, which implies ∠AOD=60∘\angle AOD = 60^\circ.
3
Calculate the area of sector OADOAD
Area(Sector OAD)=24π\text{Area(Sector } OAD) = 24\pi$
The area of a sector with central angle 60∘60^\circ and radius 1212 is 60∘360∘⋅π(122)=16⋅144π=24π\frac{60^\circ}{360^\circ} \cdot \pi (12^2) = \frac{1}{6} \cdot 144\pi = 24\pi.
4
Calculate the area of right triangle OCDOCD
Area(Triangle OCD)=183\text{Area(Triangle } OCD) = 18\sqrt{3}
The area of right triangle OCDOCD is 12⋅base⋅height=12⋅6⋅63=183\frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 6 \cdot 6\sqrt{3} = 18\sqrt{3}.
5
Subtract the area of triangle OCDOCD from the area of sector OADOAD
Area of shaded region=24π−183\text{Area of shaded region} = 24\pi - 18\sqrt{3}
The bounded region is formed by removing triangle OCDOCD from sector OADOAD.

Key Concept

Calculating the area of a region bounded by a circle arc and line segments by subtracting a right triangle area from a sector area.
Rate this question

Topics

Circles, Arc Lengths, and Sector Areas

More questions from this topic