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inscribed coordinate geometry (m06q32) : Retired Discussions [Locked]

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

1

This post was BOOKMARKED

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 ?

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

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)

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

Re: inscribed coordinate geometry (m06q32) [#permalink]
26 Oct 2012, 04:14

1

This post received KUDOS

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 ?

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 3851 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).

Re: inscribed coordinate geometry (m06q32) [#permalink]
28 Oct 2012, 07:32

2

This post received KUDOS

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) _________________