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:
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
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
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