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17 Jul 2022, 05:18
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
50% (02:58) correct
50%
(02:46)
wrong
based on 22
sessions
History
Date
Time
Result
Not Attempted Yet
if the radius of the Circle with center O is \(\sqrt{2}\), then Find the area of the green shaded region in \(unit^2\).
a) \(\frac{2 \sqrt{2}-1}{4}\)
b) \(\frac{2 \sqrt{2}-1}{2}\)
c) \(\frac{2 \sqrt{2}-1}{\sqrt{2}}\)
d) \(\frac{\pi \sqrt{2}}{2}\)\( -1\)
e) \(\frac{\pi \sqrt{2}}{2}\)
a) \(\frac{2 \sqrt{2}-1}{4}\)
b) \(\frac{2 \sqrt{2}-1}{2}\)
c) \(\frac{2 \sqrt{2}-1}{\sqrt{2}}\)
d) \(\frac{\pi \sqrt{2}}{2}\)\( -1\)
e) \(\frac{\pi \sqrt{2}}{2}\)
Attachments
Circle Que.png [ 96.16 KiB | Viewed 1889 times ]
17 Jul 2022, 17:42
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Note that the radius of the quarter circle is nothing but the hypotenuse of square FOEB plus radius of circle O, so: 2 + sqrt(2)
Area of circle O = pi*(sqrt(2))^2 = 2pi
Area of curved portion FBE = 2 - 2pi/4 = 2-pi/2
Therefore area of green region = (pi/4*(2+sqrt(2))^2-(2-pi/2)-2pi)/2 = pi/2*sqrt(2) - 1
Answer is D
Posted from my mobile device
Area of circle O = pi*(sqrt(2))^2 = 2pi
Area of curved portion FBE = 2 - 2pi/4 = 2-pi/2
Therefore area of green region = (pi/4*(2+sqrt(2))^2-(2-pi/2)-2pi)/2 = pi/2*sqrt(2) - 1
Answer is D
Posted from my mobile device
17 Sep 2024, 17:56
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