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inscribed coordinate geometry (m06q32)

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inscribed coordinate geometry (m06q32) [#permalink] New post 12 Nov 2007, 10:56
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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}

[Reveal] Spoiler: OA
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might be a typo/mistake. this is taken from Challenges
[Reveal] Spoiler: OA
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 [#permalink] New post 12 Nov 2007, 12:17
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 [#permalink] New post 12 Nov 2007, 22:03
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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 [#permalink] New post 13 Nov 2007, 05:17
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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 [#permalink] New post 13 Nov 2007, 05:21
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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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 [#permalink] New post 23 Dec 2007, 17:31
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] New post 23 Oct 2010, 16:01
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] New post 29 Nov 2011, 14:28
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] New post 26 Oct 2012, 04:14
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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}

[Reveal] Spoiler: OA
D

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might be a typo/mistake. this is taken from Challenges


BELOW IS REVISED VERSION OF THIS QUESTION:

The figure below shows a square inscribed in a circle with a radius of \sqrt{6}. What is the area of the shaded region?
Attachment:
Square in a circle.png
Square in a circle.png [ 9.99 KiB | Viewed 3626 times ]

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}

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] New post 26 Oct 2012, 04:43
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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] New post 26 Oct 2012, 04:47
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Marcab wrote:
Bunuel...could you please let me know what does the question means by "cutting into 5 regions"?


Look at the image here: inscribed-coordinate-geometry-m06q32-55450.html#p1135920

You can see that when a square is inscribed in a circle we get 5 regions.

Hope it's clear.
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Re: inscribed coordinate geometry (m06q32) [#permalink] New post 26 Oct 2012, 04:54
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Re: inscribed coordinate geometry (m06q32) [#permalink] New post 26 Oct 2012, 04:57
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Re: inscribed coordinate geometry (m06q32) [#permalink] New post 26 Oct 2012, 21:17
Area of sector BOC - area of triangle BOC

pi*sq(sqrt(6))*[90/360] - [1/2]*sqrt(6)*sqrt(6)

= [3pi/2] - 3
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Re: inscribed coordinate geometry (m06q32) [#permalink] New post 28 Oct 2012, 03:51
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There is a slightly more convoluted approach.
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Re: inscribed coordinate geometry (m06q32) [#permalink] New post 28 Oct 2012, 07:32
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Great problem Bunuel! This one was simpler - Thanks to your MATH NOTE on polygons.

1. As circle is circumscribed to the square - Area of circle = \frac{Pi}{2}*(Area of square). So this means we need to know area of sqaure first.
2. Area of sqaure = \frac{(diagonal)^2}{2} and diagonal (in this case) = 2 * radius i.e 2\sqrt{6}.Therefore, Area of square = 12 and Area of circle (using point 1) = \frac{Pi}{2}*12.
3. Now, area of all 4 sectors = area of circle - area of sqaure and area of 1 sector (say CD) = \frac{1}{4}*(AREA of 4 sectors)
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Re: inscribed coordinate geometry (m06q32) [#permalink] New post 24 Oct 2013, 06:51
We know the diagonal length is 2 sq6

Then we use similar triangles (we know that it's a 45-45-90 triangle) to find the length if the sides

Finally we know that the inscribed square equally divides the chords.

Final formula: (area of circle - area of square) / 4

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Re: inscribed coordinate geometry (m06q32)   [#permalink] 24 Oct 2013, 06:51
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