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:

In the figure above, equilateral triangle ABC is inscribed [#permalink]

Show Tags

17 Jul 2010, 10:33

2

This post received KUDOS

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

65% (02:05) correct
35% (01:12) wrong based on 302 sessions

HideShow timer Statistics

Attachment:

Circle.JPG [ 5.57 KiB | Viewed 25284 times ]

In the figure above, equilateral triangle ABC is inscribed in the circle. If the length of arc ABC is 24, what is the approximate diameter of the circle?

Re: Geometry: circle & triangle from mba.com [#permalink]

Show Tags

17 Jul 2010, 11:02

5

This post received KUDOS

Expert's post

7

This post was BOOKMARKED

In the figure above, equilateral triangle ABC is inscribed in the circle. If the length of arc ABC is 24, what is the approximate diameter of the circle?

A. 5 B. 8 C. 11 D. 15 E. 19

Arc ABC is \(\frac{2}{3}\) of the circumference (as ABC is equilateral triangle and thus arc AB=arc BC=arc AC, so arc AB+arc BC=arc ABC = 2/3 of circumference) --> \(24=c*\frac{2}{3}\), hence circumference \(c=\frac{24*3}{2}=36=\pi{d}\) --> \(d\approx{11.5}\).

Re: Geometry: circle & triangle from mba.com [#permalink]

Show Tags

21 Jul 2010, 10:18

Bunuel wrote:

An equilateral triangle ABC is inscribed in the circle. If the length of arc ABC is 24, what is the approximate diameter of the circle?

Arc ABC is \(\frac{2}{3}\) of the circumference (as ABC is equilateral triangle and thus arc AB=arc BC=arc AC, so arc AB+arc BC=arc ABC = 2/3 of circumference) --> \(24=c*\frac{2}{3}\), hence circumference \(c=\frac{24*3}{2}=36=\pi{d}\) --> \(d\approx{11.5}\).

Answer: C.

Thanks Bunuel for the explanation. Is there any other way to solve this type of questions?

Thanks in advance. _________________

_________________ If you like my post, consider giving me a kudos. THANKS!

Re: Geometry: circle & triangle from mba.com [#permalink]

Show Tags

08 Sep 2010, 07:28

1

This post received KUDOS

Not exactly a scientific method, but I figured it out like this:

Triangle ABC = 24, and is an equilateral, so each side is 8. Since AB = 8, and does not go across the radius, the circumference of the circle is greater than 8. So you can then eliminate answer choices A and B. You are left with C (11), D (15) and E (19). Answer choice D, 15, is almost twice the length of AB. That almost certainly makes it too large of a number - thus you can rule out D and E. And you are left with C.

Again, not a very scientific method, but that's how I figured it out in my head in a few seconds.

Re: Geometry: circle & triangle from mba.com [#permalink]

Show Tags

18 Nov 2010, 20:15

1

This post received KUDOS

Expert's post

prashantbacchewar wrote:

It is strange that GMAT is asking for such approximation and calculations

Actually Prashant, you don't need to calculate at all. You see that 2/3rd of circumference is 24, so the circumference of the circle must be 36. The question asks for 2r and we know 2 pi r = 36. So 2r = 36/pi We know pi = 3.14 Answer = 36/(3.14) When I divide 36 by 3, I get 12 so my answer has to be less than 12. When I divide 36 by 4, I get 9 so my answer has to more than 9. It will be closer to 12 than 9. Only possible answer is 11.

I remember, when I took GMAT, I almost never had to put the marker to the scratch pad for calculations. May be to make a little diagram, may be to make an equation to satisfy myself... _________________

Re: Geometry: circle & triangle from mba.com [#permalink]

Show Tags

20 Nov 2010, 02:51

2

This post received KUDOS

Let centre be denoted as O. Draw lines OA & OC. Now Angle ABC=60(equilateral triangle) & hence Angle AOC=120. Length of the arc ABC = (x /360) * circumference, where x = 240 (ext. angle of O) 24 = (240/360)* 2 pi r D = (36*7)/22 = 11.14 (approx.)

COUNTER ARGUMENTS WELCOME, SORRY I CUDN'T ATTACH PICTURE

Re: Geometry: circle & triangle from mba.com [#permalink]

Show Tags

10 May 2013, 15:40

Bunuel wrote:

In the figure above, equilateral triangle ABC is inscribed in the circle. If the length of arc ABC is 24, what is the approximate diameter of the circle?

A. 5 B. 8 C. 11 D. 15 E. 19

Arc ABC is \(\frac{2}{3}\) of the circumference (as ABC is equilateral triangle and thus arc AB=arc BC=arc AC, so arc AB+arc BC=arc ABC = 2/3 of circumference) --> \(24=c*\frac{2}{3}\), hence circumference \(c=\frac{24*3}{2}=36=\pi{d}\) --> \(d\approx{11.5}\).

Answer: C.

Hi Bunuel, Can you please explain the 2/3 of circumference part? how do u derive this?

Re: Geometry: circle & triangle from mba.com [#permalink]

Show Tags

11 May 2013, 03:21

Expert's post

up4gmat wrote:

Bunuel wrote:

In the figure above, equilateral triangle ABC is inscribed in the circle. If the length of arc ABC is 24, what is the approximate diameter of the circle?

A. 5 B. 8 C. 11 D. 15 E. 19

Arc ABC is \(\frac{2}{3}\) of the circumference (as ABC is equilateral triangle and thus arc AB=arc BC=arc AC, so arc AB+arc BC=arc ABC = 2/3 of circumference) --> \(24=c*\frac{2}{3}\), hence circumference \(c=\frac{24*3}{2}=36=\pi{d}\) --> \(d\approx{11.5}\).

Answer: C.

Hi Bunuel, Can you please explain the 2/3 of circumference part? how do u derive this?

Sure. Triangle ABC is equilateral, thus minor arcs AC, AB, and BC are equal and each is 1/3rd of the circumference (since these 3 arcs make the whole circumference). From which it follows that arc ABC is 1/3+1/3=2/3 of the circumference.

Re: Geometry: circle & triangle from mba.com [#permalink]

Show Tags

11 May 2013, 10:08

Bunuel wrote:

up4gmat wrote:

Bunuel wrote:

In the figure above, equilateral triangle ABC is inscribed in the circle. If the length of arc ABC is 24, what is the approximate diameter of the circle?

A. 5 B. 8 C. 11 D. 15 E. 19

Arc ABC is \(\frac{2}{3}\) of the circumference (as ABC is equilateral triangle and thus arc AB=arc BC=arc AC, so arc AB+arc BC=arc ABC = 2/3 of circumference) --> \(24=c*\frac{2}{3}\), hence circumference \(c=\frac{24*3}{2}=36=\pi{d}\) --> \(d\approx{11.5}\).

Answer: C.

Hi Bunuel, Can you please explain the 2/3 of circumference part? how do u derive this?

Sure. Triangle ABC is equilateral, thus minor arcs AC, AB, and BC are equal and each is 1/3rd of the circumference (since these 3 arcs make the whole circumference). From which it follows that arc ABC is 1/3+1/3=2/3 of the circumference.

Hope it's clear.

yes,i got it : is it that we measure the arc length from centre of circle : (120/360)* c = c/3.. thanks bunuel!

Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]

Show Tags

22 Jul 2014, 21:30

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]

Show Tags

26 Oct 2014, 10:10

funny how we all learn differently. I see now the "correct" way to approach this problem. For some reason when I did the test, it didn't register that AB, BC & AC arcs where also equal - I was too focused on the triangle inside and didn't see how knowing anything about the circumference would help me as I didn't have a radius. Here is what I did (might help folks that get stuck on these types of problems make good guesses.)

I said, since ABC is almost the entire circumference of the circle we can assume piD>24 (ie entire circumference is greater than 24) d> 24/3.14 d> 8

I knew ABC was almost the whole circle, so picked the closest number bigger which was 11.

Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]

Show Tags

01 Nov 2014, 00:15

Hi Bunuel,

I was trying to approach this problem with the formula R=sqrt of 3/3 * (side) and then find the diameter. Eventually got it wrong. Is there a possibility to solve this question using this formula? Also Bunuel, could you shed some light on this approach - what sort of problems can be solved using this formula?

Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]

Show Tags

01 Nov 2014, 04:57

Expert's post

Rca wrote:

Hi Bunuel,

I was trying to approach this problem with the formula R=sqrt of 3/3 * (side) and then find the diameter. Eventually got it wrong. Is there a possibility to solve this question using this formula? Also Bunuel, could you shed some light on this approach - what sort of problems can be solved using this formula?

Thanks a bunch! RCA

This formula links the radius of the circumscribed circle to the side of equilateral triangle inscribed in it: \(R=side*\frac{\sqrt{3}}{3}\). We don;t know the side, so this is not a good formula to apply here. _________________

Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]

Show Tags

21 Dec 2014, 00:29

Whats wrong in this approach??

As it is equilateral triangle - So all angles are 60 degree. We know that angle at the center is twice the angle made at circumference. , so angle at centre = 120 degree.

Length of arc = n/360 * 2 pi R = 120/360 * Pi * R =24 => (1/3)*3.14 * R = 24, but by this calculation R is coming approx 24

Can you please tell that where i am wrong _________________

Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]

Show Tags

21 Dec 2014, 04:43

Expert's post

him1985 wrote:

Whats wrong in this approach??

As it is equilateral triangle - So all angles are 60 degree. We know that angle at the center is twice the angle made at circumference. , so angle at centre = 120 degree.

Length of arc = n/360 * 2 pi R = 120/360 * Pi * R =24 => (1/3)*3.14 * R = 24, but by this calculation R is coming approx 24

Can you please tell that where i am wrong

The central angle of the arc ABC is 120 + 120 = 240 degrees. _________________

Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]

Show Tags

21 Dec 2014, 06:45

Bunuel wrote:

him1985 wrote:

Whats wrong in this approach??

As it is equilateral triangle - So all angles are 60 degree. We know that angle at the center is twice the angle made at circumference. , so angle at centre = 120 degree.

Length of arc = n/360 * 2 pi R = 120/360 * Pi * R =24 => (1/3)*3.14 * R = 24, but by this calculation R is coming approx 24

Can you please tell that where i am wrong

The central angle of the arc ABC is 120 + 120 = 240 degrees.

I still can not understand how it becomes 240+240 . I have attached the figure. Can you please clear my doubt.

Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]

Show Tags

21 Dec 2014, 08:44

Expert's post

him1985 wrote:

Bunuel wrote:

him1985 wrote:

Whats wrong in this approach??

As it is equilateral triangle - So all angles are 60 degree. We know that angle at the center is twice the angle made at circumference. , so angle at centre = 120 degree.

Length of arc = n/360 * 2 pi R = 120/360 * Pi * R =24 => (1/3)*3.14 * R = 24, but by this calculation R is coming approx 24

Can you please tell that where i am wrong

The central angle of the arc ABC is 120 + 120 = 240 degrees.

I still can not understand how it becomes 240+240 . I have attached the figure. Can you please clear my doubt.

Attachment:

Untitled.png [ 12.63 KiB | Viewed 8734 times ]

The red arc is 24 and its central angle is 240 degrees.

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...