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 500,000 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:

In the figure above, point O is the center of the circle [#permalink]

Show Tags

05 Mar 2013, 11:38

1

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

66% (03:44) correct
34% (03:53) wrong based on 120 sessions

HideShow timer Statistics

Attachment:

File comment: Image of circle

photo.JPG [ 175.07 KiB | Viewed 4103 times ]

In the figure above, point O is the center of the circle and line segment BD is tangent to the to the circle at point C. If BC = 4, OB = 8, and OC = CD, then what is the area of the region whose perimeter is radius OA, arc ACE, and radius OE?

Re: In the figure above, point O is the center of the circle [#permalink]

Show Tags

05 Mar 2013, 14:13

2

This post received KUDOS

My question is how did you know where to put point A and point E? Was the image included in the problem or did you draw it?

Angle OCD is 90, and given OC=CD, we know angle COD and CDO =45 degrees. OB=8, CB=4, and properities of the special 30, 60, 90 triangle. x, 2x, \(x\sqrt{3}\), we find out the radius is \(4\sqrt{3}\). We also know angle COB, which is opposite of x in the triangle = 30 degrees Radius= \(4\sqrt{3}\) Area=48pie Angle AOE =45+30=75 degree

Re: In the figure above, point O is the center of the circle [#permalink]

Show Tags

05 Mar 2013, 14:14

DoItRight wrote:

My question is how did you know where to put point A and point E? Was the image included in the problem or did you draw it?

Angle OCD is 90, and given OC=CD, we know angle COD and CDO =45 degrees. OB=8, CB=4, and properities of the special 30, 60, 90 triangle. x, 2x, \(x\sqrt{3}\), we find out the radius is \(4\sqrt{3}\). We also know angle COB, which is opposite of x in the triangle = 30 degrees Radius= \(4\sqrt{3}\) Area=48pie Angle AOE =45+30=75 degree

Re: In the figure above, point O is the center of the circle [#permalink]

Show Tags

09 Mar 2015, 10:12

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

The GMAT is based heavily on patterns, so building up your 'pattern-matching' skills is a valuable part of your training. When complex-looking questions appear, they are almost certainly going to be based on a series of overlapping patterns (since it's difficult to make a question complex by accident), so you should be on the lookout for "little" patterns, then think about how they 'connect' to one another.

Here, the first rule that you need to know is that lines that are TANGENT to a circle always form 90 degree angles. This means that triangles OBC and ODC are both RIGHT triangles.

With triangle OBC, we're given two of the sides: one of the legs is 4 and the hypotenuse is 8. You should be thinking....."what type of right triangle has a hypotenuse that is exactly DOUBLE one of its legs.....?" The pattern is that it's a 30/60/90 right triangle.

Next, with triangle ODC, we're told that the two legs of that right triangle are equal. What type of right triangle has two legs that are EQUAL....? The pattern is that it's an ISOSCELES right triangle, so we're dealing with a 45/45/90 right triangle.

From here, it's just a few more steps to figure out the central angle of the circle and the sector area of that piece of circle.

As you continue to study, remember that you're not expected to do every step of a question 'all at once.' Break prompts into small pieces, look for patterns and do the work on the pad.

Re: In the figure above, point O is the center of the circle [#permalink]

Show Tags

02 Jun 2016, 08:33

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: In the figure above, point O is the center of the circle [#permalink]

Show Tags

02 Jun 2016, 08:57

OB = 8 = 2*4 CB = 4 Using pythagoras therom we can deduce OC = 4√3 As we can see this triangle follows 30-60-90 Right angled triangle hence angle COB = 30 degrees. We know from the question OC = DC hence Computing OD will be 4√6 which means triangle ODC is a 45-45-90 right triangle hence angle DOC = 45degrees. Finally, OE=OC=OA = 4√3 as all are the radii of the circle. The area covering OE,OC,OA arcs EC and CA is to be calculated using the proportion of central angle as follows: [[(75°)/(360°)]* π*r^(2 ) =>[5/24]* π*(4√3)^2=>10π

Re: In the figure above, point O is the center of the circle [#permalink]

Show Tags

02 Jun 2016, 09:48

alex1233 wrote:

Attachment:

photo.JPG

In the figure above, point O is the center of the circle and line segment BD is tangent to the to the circle at point C. If BC = 4, OB = 8, and OC = CD, then what is the area of the region whose perimeter is radius OA, arc ACE, and radius OE?

A. 8π B. 10π C. 12π D. 16π E. 20π

OB= 8, BC = 4

Since DB is tangent on point C, it makes 90 degrees angle at C

So OC = \sqrt{OB^2 - CB^2}= 4\sqrt{3}

Sides are in ratio 1:2:\sqrt{3}, Hence angle COB= 30 degrees

OC=DC; that means ODC is an isosceles triangle with angle DOC= 45 degrees

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

As you leave central, bustling Tokyo and head Southwest the scenery gradually changes from urban to farmland. You go through a tunnel and on the other side all semblance...

Ghibli studio’s Princess Mononoke was my first exposure to Japan. I saw it at a sleepover with a neighborhood friend after playing some video games and I was...