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:

Re: Inscribed circle in an Equilateral triangle [#permalink]
31 Mar 2012, 04:01

2

This post received KUDOS

Expert's post

enigma123 wrote:

In the fi gure below, if the radius of circle O is r and triangle ABC is equilateral, what is the length of AB, in terms of r?

(A) r \(\sqrt{2}\) (B) r \(\sqrt{3}\) (C) 2r \(\sqrt{3}\) (D) \(\frac{3}{2}\) r (E) 2r

Any idea how to solve?

There are many ways to solve this question including using the direct formula saying that the radius of the inscribed circle in an equilateral triangle is \(r=a*\frac{\sqrt{3}}{6}\).

One can also do the following, consider the diagram below:

Attachment:

Circle.gif [ 6.02 KiB | Viewed 2484 times ]

Angle DCO is 60/2=30 degrees, hence triangle DOC is 30°-60°-90° right triangle (angle DOC is 60°). Next, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\) and the leg opposite 30° (OD=r) corresponds with \(1\) and the leg opposite 60° (DC) corresponds with \(\sqrt{3}\), so \(\frac{r}{DC}=\frac{1}{\sqrt{3}}\) --> \(DC=r\sqrt{3}\). Now, since DC=AC/2 then \(AC=2DC=2r\sqrt{3}\).

I was checking my phone all day. I wasn’t sure when I would receive the admission decision from Tepper. I received an acceptance from Goizueta in the early morning...

This an overview of my journey till now! I started bloging in August last year. I had a nice list before getting started for GMAT studies. I took my first GMAT prep...

How should business schools teach public policy? On Wednesday, 22nd April 2015, between 2 - 3pm BST , two experts will answer readers' questions on the importance of integrating politics...