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# inscribed coordinate geometry (m06q32)

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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
D

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might be a typo/mistake. this is taken from Challenges
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12 Nov 2007, 12:17
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image
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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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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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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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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]

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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]

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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]

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26 Oct 2012, 04:14
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Expert's post
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 [ 9.99 KiB | Viewed 4265 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 $$\frac{1}{4}(6\pi-12)=3(\frac{\pi}{2} - 1)$$.

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Re: inscribed coordinate geometry (m06q32) [#permalink]

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26 Oct 2012, 04:43
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]

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26 Oct 2012, 04:47
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]

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26 Oct 2012, 04:54
The four parts of circle and 1 part inside the square?
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Re: inscribed coordinate geometry (m06q32) [#permalink]

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26 Oct 2012, 04:57
Marcab wrote:
The four parts of circle and 1 part inside the square?

Check the diagram below:
Attachment:

regions.png [ 10.45 KiB | Viewed 4206 times ]
Hope it's clear.
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Re: inscribed coordinate geometry (m06q32) [#permalink]

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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]

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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]

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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]

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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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# inscribed coordinate geometry (m06q32)

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