In the figure above, equilateral triangle ABC is inscribed : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 23 Jan 2017, 07:13

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In the figure above, equilateral triangle ABC is inscribed

Author Message
TAGS:

### Hide Tags

Manager
Joined: 03 Jun 2010
Posts: 183
Location: United States (MI)
Concentration: Marketing, General Management
Followers: 6

Kudos [?]: 57 [4] , given: 40

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

### Show Tags

17 Jul 2010, 09:33
4
KUDOS
9
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

66% (02:03) correct 34% (01:14) wrong based on 356 sessions

### HideShow timer Statistics

Attachment:

Circle.JPG [ 5.57 KiB | Viewed 34189 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?

A. 5
B. 8
C. 11
D. 15
E. 19
[Reveal] Spoiler: OA

Last edited by Bunuel on 18 Oct 2012, 09:13, edited 2 times in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93523 [5] , given: 10568

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

### Show Tags

17 Jul 2010, 10:02
5
KUDOS
Expert's post
8
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 $$c=\frac{24*3}{2}=36=\pi{d}$$ --> $$d\approx{11.5}$$.

_________________
Manager
Joined: 24 Jan 2010
Posts: 164
Location: India
Schools: ISB
Followers: 2

Kudos [?]: 53 [0], given: 14

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

### Show Tags

21 Jul 2010, 09:18
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 $$c=\frac{24*3}{2}=36=\pi{d}$$ --> $$d\approx{11.5}$$.

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

_________________

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

Manager
Status: Last few days....Have pressed the throttle
Joined: 20 Jun 2010
Posts: 70
WE 1: 6 years - Consulting
Followers: 3

Kudos [?]: 46 [0], given: 27

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

### Show Tags

25 Aug 2010, 19:30
Another approach could be as follows:

Angle C subtends AB arc and angle A subtends BC arc. That means Angle C+ angle A subtends an arc of 24

120 degrees subtends arc of length 24 (since equilteral triangle so one angle = 60)
so 180 degrees subtends 36 (full circular arc or circumference)

2 pi r = 36

Diameter = 36*7/22 = 11.5 (Ans - C)
_________________

Consider giving Kudos if my post helps in some way

Intern
Joined: 13 Aug 2010
Posts: 2
Followers: 0

Kudos [?]: 1 [1] , given: 10

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

### Show Tags

08 Sep 2010, 06:28
1
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.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7130
Location: Pune, India
Followers: 2140

Kudos [?]: 13703 [1] , given: 222

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

### Show Tags

18 Nov 2010, 19:15
1
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
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...
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Senior Manager
Joined: 25 May 2010
Posts: 321
Location: United States
Concentration: Strategy, Finance
Schools: CBS '14 (A)
GMAT 1: 590 Q47 V25
GMAT 2: 560 Q47 V20
GMAT 3: 600 Q47 V25
GMAT 4: 680 Q49 V34
Followers: 6

Kudos [?]: 57 [0], given: 32

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

### Show Tags

19 Nov 2010, 11:50
Pi*D=24(3/2)
This gives us approximately 11. So answer should be C.
_________________

"Whether You Think You Can or Can't, You're Right"--Henry Ford
680 Debrief
600 Debrief
590 Debrief
My GMAT Journey

Intern
Joined: 07 Sep 2010
Posts: 18
Followers: 0

Kudos [?]: 3 [2] , given: 3

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

### Show Tags

20 Nov 2010, 01:51
2
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
SVP
Joined: 16 Nov 2010
Posts: 1672
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 33

Kudos [?]: 514 [0], given: 36

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

### Show Tags

19 Apr 2011, 18:08
(120+ 120)/360 = 24/pi*d (because Angle subtended by Arc AB at Center + Angle subtended by Arc BC at Center = 2*60 + 2*60)

pi*d = 3 * 24/2

d = 36/3.14 ~ 11.something

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Manager
Joined: 14 Dec 2012
Posts: 82
Location: United States
Followers: 1

Kudos [?]: 16 [0], given: 186

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

### Show Tags

10 May 2013, 14:40
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 $$c=\frac{24*3}{2}=36=\pi{d}$$ --> $$d\approx{11.5}$$.

Hi Bunuel,
Can you please explain the 2/3 of circumference part? how do u derive this?
Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93523 [0], given: 10568

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

### Show Tags

11 May 2013, 02:21
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 $$c=\frac{24*3}{2}=36=\pi{d}$$ --> $$d\approx{11.5}$$.

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.
_________________
Manager
Joined: 14 Dec 2012
Posts: 82
Location: United States
Followers: 1

Kudos [?]: 16 [0], given: 186

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

### Show Tags

11 May 2013, 09:08
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 $$c=\frac{24*3}{2}=36=\pi{d}$$ --> $$d\approx{11.5}$$.

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!
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13513
Followers: 577

Kudos [?]: 163 [0], given: 0

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

### Show Tags

22 Jul 2014, 20:30
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.
_________________
Current Student
Joined: 26 Aug 2014
Posts: 827
Concentration: Marketing
GMAT 1: 680 Q V
GPA: 3.4
Followers: 7

Kudos [?]: 159 [0], given: 98

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

### Show Tags

26 Oct 2014, 09:10
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.
Intern
Joined: 12 Dec 2013
Posts: 20
Followers: 1

Kudos [?]: 7 [0], given: 34

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

### Show Tags

31 Oct 2014, 23:15
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?

Thanks a bunch!
RCA
Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93523 [0], given: 10568

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

### Show Tags

01 Nov 2014, 03:57
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.
_________________
Current Student
Joined: 20 Jan 2014
Posts: 186
Location: India
Concentration: Technology, Marketing
Followers: 1

Kudos [?]: 58 [0], given: 120

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

### Show Tags

20 Dec 2014, 23:29
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
_________________

Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93523 [0], given: 10568

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

### Show Tags

21 Dec 2014, 03:43
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.
_________________
Current Student
Joined: 20 Jan 2014
Posts: 186
Location: India
Concentration: Technology, Marketing
Followers: 1

Kudos [?]: 58 [0], given: 120

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

### Show Tags

21 Dec 2014, 05:45
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.
Attachments

Ques.png [ 11.56 KiB | Viewed 13141 times ]

_________________

Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93523 [1] , given: 10568

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

### Show Tags

21 Dec 2014, 07:44
1
KUDOS
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 13079 times ]
The red arc is 24 and its central angle is 240 degrees.

Hope it's clear.
_________________
Re: In the figure above, equilateral triangle ABC is inscribed   [#permalink] 21 Dec 2014, 07:44

Go to page    1   2    Next  [ 24 posts ]

Similar topics Replies Last post
Similar
Topics:
10 A semicircle is inscribed in equilateral triangle ABC, as 5 29 May 2014, 13:51
45 In the figure above, triangle ABC is equilateral, and point 23 03 Dec 2012, 03:36
3 In the figure above, equilateral triangle ABC is inscribed in the circ 4 07 May 2011, 13:26
12 In the figure above, equilateral triangle ABC is inscribed i 17 02 Mar 2009, 08:50
19 The triangles in the figure above are equilateral and the 12 05 Apr 2008, 21:14
Display posts from previous: Sort by