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01 Sep 2014, 20:46
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D
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Difficulty:
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Question Stats:
77% (02:23) correct
23%
(02:56)
wrong
based on 175
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Circle is inscribed in the square as shown in figure below (Not to scale).
circ.png [ 1.85 KiB | Viewed 5339 times ]
If area of the shaded region is 2, then radius of the circle would be
A: \(\frac{4}{4-\pi}\)
B: \(\frac{4}{\sqrt{8-\pi}}\)
C: \(\frac{16}{4-\pi}\)
D: \(\frac{4}{\sqrt{4-\pi}}\)
E: \(\frac{\sqrt{4+\pi}}{2}\)
Attachment:
circ.png [ 1.85 KiB | Viewed 5339 times ]
A: \(\frac{4}{4-\pi}\)
B: \(\frac{4}{\sqrt{8-\pi}}\)
C: \(\frac{16}{4-\pi}\)
D: \(\frac{4}{\sqrt{4-\pi}}\)
E: \(\frac{\sqrt{4+\pi}}{2}\)
02 Sep 2014, 06:13
Kudos
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Answer is D.
We have area of the square = 4r^2
Area of the circle = \(\pi\)r^2
So the area around the circle but still inside the square is given by: 4r^2 - \(\pi\)r^2
The area we know is 1/8 of the total area outside the circle so now: (4r^2 - \(\pi\)r^2)/8 = 2
Solve this and we get D.
We have area of the square = 4r^2
Area of the circle = \(\pi\)r^2
So the area around the circle but still inside the square is given by: 4r^2 - \(\pi\)r^2
The area we know is 1/8 of the total area outside the circle so now: (4r^2 - \(\pi\)r^2)/8 = 2
Solve this and we get D.
General Discussion
02 Sep 2014, 21:26
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PareshGmat
Circle is inscribed in the square as shown in figure below (Not to scale).
A: \(\frac{4}{4-\pi}\)
B: \(\frac{4}{\sqrt{8-\pi}}\)
C: \(\frac{16}{4-\pi}\)
D: \(\frac{4}{\sqrt{4-\pi}}\)
E: \(\frac{\sqrt{4+\pi}}{2}\)
Attachment:
circ.png
If area of the shaded region is 2, then radius of the circle would beA: \(\frac{4}{4-\pi}\)
B: \(\frac{4}{\sqrt{8-\pi}}\)
C: \(\frac{16}{4-\pi}\)
D: \(\frac{4}{\sqrt{4-\pi}}\)
E: \(\frac{\sqrt{4+\pi}}{2}\)
Let the radius of circle = a
Area of circle \(= \pi a^2\) ............. (1)
As circle is inscribed in square,
Area of Square\(= \frac{4}{\pi}* \pi a^2 = 4a^2\) .............. (2)
Shaded region is \(\frac{1}{8} th\) of the total region outside the circle, but inside the square
So the area outside the circle \(= \frac{2}{\frac{1}{8}} = 16\) .............. (3)
Setting up the equation (2) = (1) + (3)
\(4a^2 = \pi a^2 + 16\)
\(a = \frac{4}{\sqrt{4-\pi}}\)
Answer = D
