Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 3821 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) _________________