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28 Jul 2021, 07:15
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
25% (02:17) correct
75%
(02:29)
wrong
based on 24
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Two circles with radius 2R and (√2*R) intersect each other at points A and B. The centers of both the circles are on the same side of AB. O is the center of the bigger circle and ∠AOB is 60°. What is the area of the common region between two circles?
A. \((√3 − π − 1)R^2\)
B. \((√3 − π)R^2\)
C. \((\frac{13π}{6} + 1 − √3)R^2\)
D. \((\frac{13π}{6} + √3)R^2\)
E. None of the above
A. \((√3 − π − 1)R^2\)
B. \((√3 − π)R^2\)
C. \((\frac{13π}{6} + 1 − √3)R^2\)
D. \((\frac{13π}{6} + √3)R^2\)
E. None of the above
28 Jul 2021, 11:40
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Attachment:
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C is the center of smaller circle.
Since AOB is an equilateral triangle, length of AB= 2R. Hence, ∠ACB =90°
Area of Green region= \(\frac{3}{4}(π*(√2R)^2) = \frac{9}{6}*π*R^2\)
Area of Brown region \(= \frac{1}{2}* (√2R)^2 = R^2\)
Area of Red region \(= \frac{1}{6}*(π*(2R)^2) - (\frac{√3}{4} *(2R)^2) = \frac{4}{6}*π*R^2 - √3 * R^2\)
Area of the common region
\(= \frac{9}{6}*π*R^2 + R^2 + \frac{4}{6}*π*R^2 - √3 * R^2 \)
= \((\frac{13π}{6} + 1 − √3)R^2\)
Bunuel
Thelegend2631
28 Jul 2021, 11:50
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nick1816
Welcome back ! Long time!
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