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:

Circle A, center X. XB is the radius. There is a chord which [#permalink]
21 Mar 2008, 23:36

1

This post received KUDOS

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 1 sessions

Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area?

Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area?

My computer does not support drawing, pls get it by yoursefl 1. XDA = 90 degr. so XB perpendicular with AC, and so D is mid point of AC, that is AD=DC =6

2 Triangle XDA: (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36 ==> XD=8 --> r = 8+2 = 10 --> Area of cirle = 100pi

EXPECT A LESS TIME-CONSUMING APPROACH! _________________

the first step is the most important in my approach:

the triangles DBA and ABY are right triangles, because angles ADB and YAB are right angles. the triangles DBA and ABY have the same angle ABY=ABD and, therefore, third angles in triangles are equal: AYB=DAB So, 1. ABY is geometrically similar to DBA

Next step is easer, because I only use the formula for similar triangles:

If anyone is wondering how we know that angle BAY is right, look up Thales' theorem. Until I looked at walker's method, I had no idea this theorem existed. I really don't remember this from high school. Did I simply forget it, or is it not taught in US public schools?

Also, if anyone is confused by the first (simple) method, the trick is to recognize that XB = XA = XD + DB

Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area?

My computer does not support drawing, pls get it by yoursefl 1. XDA = 90 degr. so XB perpendicular with AC, and so D is mid point of AC, that is AD=DC =6

2 Triangle XDA: (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36 ==> XD=8 --> r = 8+2 = 10 --> Area of cirle = 100pi

Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area?

My computer does not support drawing, pls get it by yoursefl 1. XDA = 90 degr. so XB perpendicular with AC, and so D is mid point of AC, that is AD=DC =6

2 Triangle XDA: (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36 ==> XD=8 --> r = 8+2 = 10 --> Area of cirle = 100pi

(XA)^2 = (XD +2)^2 (Because XA and (XD + 2) are both equal to the radius of the circle)

(XD)^2 + (DA)^2 (Imagine a triangle ADX, with angle ADX a right angle. This is the pythagorean theorem. Leg AD squared plus leg XD squared equals hypotenuse AX squared. This equals XD squared plus 36.

So AX^2 equals both (XD +2)^2 and (XD)^2 + 36, and since (XD +2)^2 = (XD)^2 + 36, we can solve for XD

Essay B for Stanford GSB will essentially ask you to explain why you’re doing what you’re doing. Namely, the essay wants to know, A) why you’re seeking...

The following pictures perfectly describe what I’ve been up to these days. MBA is an extremely valuable tool in your career, no doubt, just that it is also...