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:

Re: geometry helppppp [#permalink]
08 Aug 2012, 14:41

The answer is C) 10.

I thought this was a very difficult problem. I stared at this for a while before figuring it out. Make point X the point where DB and AF intersect. The angles of the four sided figure in the middle (BCFX), add up to 360. If there are two 90 degree angles, then the remaining angles, FXB and FCB, add up to 180. You will also note that angles FXB and FXD also add up to 180 as well as angles FXB and AXB.

FXB + FCB = 180 FXB + FXD = 180 FXB + AXB = 180

Therefore, Angles FCB, FXD, and AXB are all equal. Lets call their value Y for now. Now you have 3 triangles-- AXB, DXF, AFC, and DBC-- that all have angles that are 90 and Y degrees. They all have the same 3rd angle, which we can call Z for now. We do not need to know the value of Y or Z.

Now, from the above configuration lets extract 2 triangles: CFA and CBD. We can find their proportionality be comparing known values on proportional sides. Since we have values for both CA and CD and they proportional, lets use them. CB:CD = (12):(18) = 2:3. Therefore, the ratio of the sides of triangle CFA to CBD is 2:3. Now lets find AF. Since AF and BD are proportional at 2:3 and BD = 15, then AF = 15(2/3) = 30/3 = 10

Such figures should immediately remind you of similar triangles. The two triangles FCA and BCD have a right angle each and a common angle (angle A) each. Hence, their third corresponding angles would also be equal to each other. By AA rule, the triangles are similar. Therefore, the ratio of their corresponding sides will be equal.