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22 Oct 2022, 07:41
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:
75%
(hard)
Question Stats:
47% (02:48) correct
53%
(02:33)
wrong
based on 36
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
20a9935023dd9de5a53a6fb3809342dd.jpg [ 3.96 KiB | Viewed 2849 times ]
A. \(\frac{5 \pi}{π −1}\)
B. \( \frac{10 \pi}{π −1}\)
C. \(\frac{ 20 \pi}{π −1}\)
D. \(\frac{10 \pi}{π −2}\)
E. \(\frac{20 \pi}{π −2}\)
23 Oct 2022, 21:14
Kudos
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Afroditee
Join OAD, to make a right angled triangle. (In a square, diagonals are perpendicular to each other)
The sides of the triangle are r and hypotenuse \(r\sqrt{2}\), which is also the side of the square.
Area of sector OAD = A(circle)/4 = \(\frac{\pi r^2}{4}\)
Area of triangle OAD = \(\frac{1}{2}*OA*OD=\frac{r^2}{2}\)
Area of shaded portion AD = \(\frac{5\pi}{2}\)
Thus, \(\frac{\pi r^2}{4}-\frac{r^2}{2}=\frac{5\pi}{2}\)
\(\pi r^2-2r^2=10\pi…….r^2=\frac{10\pi}{\pi -2}\)
Now, area of square = \(r\sqrt{2}*r\sqrt{2}=2r^2=2*\frac{10\pi}{\pi -2}\)
E
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