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:
A circle is inscribed in equilateral triangle ABC such that [#permalink]
01 Oct 2006, 10:43
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
1. A circle is inscribed in equilateral triangle ABC such that point D lies on the circle and on line segment AC, point E lies on the circle and on line segment AB and point F lies on the cirlce and on line segment BC. If line segment AB = 6, what is the area of the figure created by line segments AD, AE and the minor arc DE?
a) 3*underroot(3) - 9/4pi
b) 3*underroot(3) - pi
c) 6*underroot(3) - pi
d) 9*underroot(3) - 3pi
e)cannot be determined from the info given
Great explanation jainan but can you explain this point that you made....
"All three perpendicualr bisectors meet at the center of circle and divide themselves in ratio of 2:1"
Can you please explain this point....
Thanks for much.....
Since the triangle formed by the radius of the circle (perpendicular),
half the base of original equilateral triangle (base) and distance from one of the vertices to the center of circle (hypotenuse) form a 30-60-90 triangle and we also know that sum of base and hypotenuse of that triangle by symmetry is 3sqrt(3), thus you can find hypotenuse = 2sqrt(3) and perpendicualr (radius) = sqrt(3).
If you draw the figure accurately, you can see the symmetry