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inscribed coordinate geometry (m06q32) [#permalink]
 12 Nov 2007, 11:56
Question Stats:
77% (02:34) correct
22% (02:02) wrong based on 2 sessions
On the coordinate graph, a circle is centered at the point (3, 3). If the radius of the circle is \sqrt{6} , and there is a square ABCD inscribed into the circle cutting it into 5 regions, what is the area of the segment CD ? (A) \pi(B) 6\pi - \sqrt{12}(C) 6\pi - 2\sqrt{6}(D) 3(\frac{\pi}{2} - 1)(E) 3 \frac{\pi}{4} - \frac{3}{2} Source: GMAT Club Tests - hardest GMAT questions might be a typo/mistake. this is taken from Challenges
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Ravshonbek wrote: D for me
Can you please explain why we cannot simply use 90/360 * pi (r^2) to solve this? Thanks.
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Ravshonbek wrote: image
how did u get the side of the square?
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GK_Gmat wrote: Ravshonbek wrote: D for me Can you please explain why we cannot simply use 90/360 * pi (r^2) to solve this? Thanks. Yeh.. GK once u figure the image out, then you can use that formulae as well. it yields the same answer
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bmwhype2 wrote: Ravshonbek wrote: image how did u get the side of the square? Imagine, AC is a dioganal of the circle and at the same time of the square. then ACD is an isoscales right traingle. thus AD^2+CD^2=AC^2 since AD=CD=x, 2x^2=(2sqrt6)^2 so x^2=12 or x=2sqrt3 that's the side
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Ravshonbek wrote: bmwhype2 wrote: Ravshonbek wrote: image how did u get the side of the square? Imagine, AC is a dioganal of the circle and at the same time of the square. then ACD is an isoscales right traingle. thus AD^2+CD^2=AC^2 since AD=CD=x, 2x^2=(2sqrt6)^2 so x^2=12 or x=2sqrt3 that's the side thanks very helpful 
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excellent question.
It took me some time, but then Got D
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Ravshonbek wrote: bmwhype2 wrote: Ravshonbek wrote: image how did u get the side of the square? Imagine, AC is a dioganal of the circle and at the same time of the square. then ACD is an isoscales right traingle. thus AD^2+CD^2=AC^2 since AD=CD=x, 2x^2=(2sqrt6)^2 so x^2=12 or x=2sqrt3 that's the side Area of sector = 1/4 area of circle (radius is known)
In the figure, assume a center O of circle.
COD is an isosceles right triangle, with 2 sides as radius. You can compute the area of triangle.
I wonder why the question asks "area of arch" CD. It ought to be "SEGMENT" CD
Another thing that baffles me is the redundant info - center (3,3)
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Re: inscribed coordinate geometry (m06q32) [#permalink]
22 Oct 2010, 12:33
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D is the answer. I took me some time to figure out. Is this a 600+level sum?
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Re: inscribed coordinate geometry (m06q32) [#permalink]
22 Oct 2010, 20:12
Good One. My ans is D. This looks like 650+.
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Re: inscribed coordinate geometry (m06q32) [#permalink]
23 Oct 2010, 17:01
D 2r = diagonal of square => side = 2 sqrt(3) area of circle = area of square + 4 * area of segments => area of segment = (pi * sqrt(6)^2 - [2 sqrt(3)]^2 ) / 4
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Re: inscribed coordinate geometry (m06q32) [#permalink]
24 Oct 2010, 21:56
d for sure
it takes 40 secs to solve
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Re: inscribed coordinate geometry (m06q32) [#permalink]
26 Oct 2011, 13:26
I picked D...Ravshonbek explained it best.
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Re: inscribed coordinate geometry (m06q32) [#permalink]
29 Nov 2011, 15:28
D for me. Here is my logic.
Step 1 - tells us that the radius is Sqrt(6)
Step 2 - Diameter = 2Sqrt(6)
Step 3- Since its an embedded square its an equilateral triangle so each side equals = sqrt(12)
Step 4 - Find area of square = 12
Step 5 - Area of Circle = 6π
Step 6 - Area of Circle - Square divided by 4 (to get area of DC) = (6π - 12)/4
Step 7 - Look to see what answer choice matches this.
Hence D.
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Re: inscribed coordinate geometry (m06q32) [#permalink]
26 Oct 2012, 05:14
bmwhype2 wrote: On the coordinate graph, a circle is centered at the point (3, 3). If the radius of the circle is \sqrt{6} , and there is a square ABCD inscribed into the circle cutting it into 5 regions, what is the area of the segment CD ? (A) \pi(B) 6\pi - \sqrt{12}(C) 6\pi - 2\sqrt{6}(D) 3(\frac{\pi}{2} - 1)(E) 3 \frac{\pi}{4} - \frac{3}{2} Source: GMAT Club Tests - hardest GMAT questions might be a typo/mistake. this is taken from Challenges Attachment: 
Square in a circle.png [ 9.99 KiB | Viewed 1621 times ]
A. \piB. 6\pi - \sqrt{12}C. 6\pi - 2\sqrt{6}D. 3(\frac{\pi}{2} - 1)E. 3 \frac{\pi}{4} - \frac{3}{2}The area of the shaded region is 1/4th of the area of the circle minus the area of the square. The are of the circle is \pi{r^2}=6\pi; Since the diagonal of the square equals to the diameter of the circle then diagonal=2\sqrt{6}. The are of a square equals to \frac{diagonal^2}{2}=12; The are of the shaded region equals to \frac{1}{4}(6\pi-12)=3(\frac{\pi}{2} - 1). Answer: D.
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Re: inscribed coordinate geometry (m06q32) [#permalink]
26 Oct 2012, 05:43
Bunuel...could you please let me know what does the question means by "cutting into 5 regions"?
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Re: inscribed coordinate geometry (m06q32) [#permalink]
26 Oct 2012, 05:47
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Re: inscribed coordinate geometry (m06q32) [#permalink]
26 Oct 2012, 05:54
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Re: inscribed coordinate geometry (m06q32) [#permalink]
26 Oct 2012, 05:57
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Re: inscribed coordinate geometry (m06q32)
[#permalink]
26 Oct 2012, 05:57
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