Bunuel wrote:
In circle O above, if POQ is a right triangle and radius OP = 2, what is the area of the shaded region?
(A) \(4\pi – 2\)
(B) \(4\pi — 4\)
(C) \(2\pi – 2\)
(D) \(2\pi — 4\)
(E) \(\pi - 2\)
Area of shaded region = (Sector area) - (Triangle Area)
• Sector area as fraction of the circle's area
The key to these kinds of problems often is:
"The sector is what fraction of the circle?"
Use the sector's central angle (given, = 90°) to find that fraction
\(\frac{Part}{Whole}=\frac{SectorAngle}{360°}=\frac{90}{360}=\frac{1}{4}=\frac{SectorArea}{CircleArea}\)Area of sector = \(\frac{1}{4}\) area of circle
• Area of circle and area of sector
Circle area, r = 2:
\(πr^2 = 4π\)Sector area:
\(\frac{4π}{4}\)=
\(π\)• Area of triangle
Area of triangle with radii as sides (s = b and h):
\(\frac{s*s}{2} = \frac{4}{2}\) = \(2\)• Area of shaded region
= (Area of sector) - (Area of triangle)
Area of shaded region =
\(π - 2\)Answer E
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