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Circle is inscribed in the square as shown in figure

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PareshGmat
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Circle is inscribed in the square as shown in figure [#permalink] New post   01 Sep 2014, 20:46
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Circle is inscribed in the square as shown in figure below (Not to scale).
Attachment:
circ.png
circ.png [ 1.85 KiB | Viewed 5190 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: \(\frac{\sqrt{4+\pi}}{2}\)
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Tobias92
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Re: Circle is inscribed in the square as shown in figure [#permalink] New post   02 Sep 2014, 06:13
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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.
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PareshGmat
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Re: Circle is inscribed in the square as shown in figure [#permalink] New post   02 Sep 2014, 21:26
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PareshGmat wrote:
Circle is inscribed in the square as shown in figure below (Not to scale).
Attachment:
circ.png

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: \(\frac{\sqrt{4+\pi}}{2}\)



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
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Re: Circle is inscribed in the square as shown in figure [#permalink] New post   09 Sep 2014, 19:50
very nice question , it took me sometime to get it , but was involving,

thanks paresh, keep posting please
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Circle is inscribed in the square as shown in figure [#permalink] New post   Updated on: 17 Oct 2022, 01:01
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PareshGmat wrote:
Circle is inscribed in the square as shown in figure below (Not to scale).
Attachment:
circ.png

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: \(\frac{\sqrt{4+\pi}}{2}\)


Check out the blog posts on the link given in my signature below.

Originally posted by KarishmaB on 09 Sep 2014, 21:15.
Last edited by KarishmaB on 17 Oct 2022, 01:01, edited 1 time in total.
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CEdward
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Re: Circle is inscribed in the square as shown in figure [#permalink] New post   27 Mar 2021, 13:54
A minor point. Is the GMAT OK with radicals in the denominator?
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Re: Circle is inscribed in the square as shown in figure [#permalink] New post   01 Apr 2021, 19:52
Tobias92 wrote:
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.



Hi, can you explain why is the area of square 4r^2?
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Abhibarua
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Re: Circle is inscribed in the square as shown in figure [#permalink] New post   26 May 2021, 04:35
Jaya6 wrote:
Tobias92 wrote:
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.



Hi, can you explain why is the area of square 4r^2?


If the radius of the circle is r, each sides of the square will be 2r, which makes the area of the square 4r^2.
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Re: Circle is inscribed in the square as shown in figure [#permalink]
26 May 2021, 04:35
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