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01 Mar 2023, 10:53
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (03:23) correct
26%
(02:51)
wrong
based on 23
sessions
History
Date
Time
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Not Attempted Yet
In the figure, ARD is an arc of a circle centered at point O, ABCD is a rectangle, and CD = a/2. What is the area of the shaded region?
A. \((\frac{\sqrt{2}}{2} - \frac{1}{2})a^2\)
B. \((\frac{\sqrt{2}}{2} - \frac{\pi}{4})a^2\)
C. \((\frac{\sqrt{2}}{2} - \frac{\pi}{4} - \frac{1}{2})a^2\)
D. \((\frac{\sqrt{2}}{2} + \frac{\pi}{4} - \frac{1}{2})a^2\)
E.\((\frac{\sqrt{2}}{2} - \frac{\pi}{4} + \frac{1}{2})a^2\)
Attachment:
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01 Mar 2023, 23:11
Kudos
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Area of the sector = (pi*a*a) /4
Area of the triangle = 0.5*a*a
area of the rectangle = (a*a)/1.4
Shaded area = Area of the rectangle- (Area of the sector- area of the triangle)
Option E
Area of the triangle = 0.5*a*a
area of the rectangle = (a*a)/1.4
Shaded area = Area of the rectangle- (Area of the sector- area of the triangle)
Option E
02 Mar 2023, 00:15
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Bunuel
Area of shaded region = Area of rectangle - Area of segment ARDA
Area of rectangle = lb = \((\frac{a}{2}*\sqrt{2}a) = \frac{\sqrt{2}}{2}*a^2\)
Area of segment ARDA = Area of sector OARDO - Area of triangle OAD
Area of segment ARDA = \(\frac{1}{4}*\pi*a^2 - \frac{1}{2}*a^2\)
Area of shaded region = \(\frac{\sqrt{2}}{2}*a^2 - (\frac{1}{4}*\pi*a^2 - \frac{1}{2}*a^2)\)
Area of shaded region =\(a^2 * (\frac{\sqrt{2}}{2} - \frac{1}{4}*\pi + \frac{1}{2})\)
Answer (E)
