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In the figure above, equilateral triangle ABC is inscribed in the

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

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17 Jul 2010, 09:33
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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

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Circle.JPG [ 5.57 KiB | Viewed 55240 times ]
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Dec 2017, 07:44, edited 3 times in total.
Renamed the topic and edited the question.

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

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17 Jul 2010, 10:02
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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 $$c=\frac{24*3}{2}=36=\pi{d}$$ --> $$d=\frac{36}{\pi}\approx{\frac{36}{3.14}}\approx{11.5}$$.

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

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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?

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

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25 Aug 2010, 19:30
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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)
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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink]

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08 Sep 2010, 06:28
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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.

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

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18 Nov 2010, 19:15
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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...
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18465 [2], given: 237 Senior Manager Joined: 25 May 2010 Posts: 319 Kudos [?]: 60 [0], given: 32 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 Re: In the figure above, equilateral triangle ABC is inscribed in the [#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 Kudos [?]: 60 [0], given: 32 Intern Joined: 07 Sep 2010 Posts: 18 Kudos [?]: 7 [6], given: 3 Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] Show Tags 20 Nov 2010, 01:51 6 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 Kudos [?]: 7 [6], given: 3 TOEFL Forum Moderator Joined: 16 Nov 2010 Posts: 1588 Kudos [?]: 619 [0], given: 42 Location: United States (IN) Concentration: Strategy, Technology Re: In the figure above, equilateral triangle ABC is inscribed in the [#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 Answer - C _________________ Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant) GMAT Club Premium Membership - big benefits and savings Kudos [?]: 619 [0], given: 42 Manager Joined: 14 Dec 2012 Posts: 78 Kudos [?]: 16 [0], given: 186 Location: United States Re: In the figure above, equilateral triangle ABC is inscribed in the [#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 $$c=\frac{24*3}{2}=36=\pi{d}$$ --> $$d\approx{11.5}$$. Answer: C. Hi Bunuel, Can you please explain the 2/3 of circumference part? how do u derive this? Kudos [?]: 16 [0], given: 186 Math Expert Joined: 02 Sep 2009 Posts: 43312 Kudos [?]: 139329 [0], given: 12783 Re: In the figure above, equilateral triangle ABC is inscribed in the [#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 $$c=\frac{24*3}{2}=36=\pi{d}$$ --> $$d\approx{11.5}$$. 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. _________________ Kudos [?]: 139329 [0], given: 12783 Manager Joined: 14 Dec 2012 Posts: 78 Kudos [?]: 16 [0], given: 186 Location: United States Re: In the figure above, equilateral triangle ABC is inscribed in the [#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 $$c=\frac{24*3}{2}=36=\pi{d}$$ --> $$d\approx{11.5}$$. 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! Kudos [?]: 16 [0], given: 186 Current Student Joined: 26 Aug 2014 Posts: 827 Kudos [?]: 163 [0], given: 98 Concentration: Marketing GPA: 3.4 Re: In the figure above, equilateral triangle ABC is inscribed in the [#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. Kudos [?]: 163 [0], given: 98 Intern Joined: 12 Dec 2013 Posts: 17 Kudos [?]: 9 [0], given: 34 Re: In the figure above, equilateral triangle ABC is inscribed in the [#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 Kudos [?]: 9 [0], given: 34 Math Expert Joined: 02 Sep 2009 Posts: 43312 Kudos [?]: 139329 [0], given: 12783 Re: In the figure above, equilateral triangle ABC is inscribed in the [#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. _________________ Kudos [?]: 139329 [0], given: 12783 Current Student Joined: 20 Jan 2014 Posts: 175 Kudos [?]: 77 [0], given: 120 Location: India Concentration: Technology, Marketing Re: In the figure above, equilateral triangle ABC is inscribed in the [#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 _________________ Consider +1 Kudos Please Kudos [?]: 77 [0], given: 120 Math Expert Joined: 02 Sep 2009 Posts: 43312 Kudos [?]: 139329 [0], given: 12783 Re: In the figure above, equilateral triangle ABC is inscribed in the [#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. _________________ Kudos [?]: 139329 [0], given: 12783 Current Student Joined: 20 Jan 2014 Posts: 175 Kudos [?]: 77 [0], given: 120 Location: India Concentration: Technology, Marketing Re: In the figure above, equilateral triangle ABC is inscribed in the [#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 23956 times ] _________________ Consider +1 Kudos Please Kudos [?]: 77 [0], given: 120 Math Expert Joined: 02 Sep 2009 Posts: 43312 Kudos [?]: 139329 [2], given: 12783 Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] Show Tags 21 Dec 2014, 07:44 2 This post received 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 23863 times ] The red arc is 24 and its central angle is 240 degrees. Hope it's clear. _________________ Kudos [?]: 139329 [2], given: 12783 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7863 Kudos [?]: 18465 [2], given: 237 Location: Pune, India Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] Show Tags 21 Dec 2014, 20:56 2 This post received KUDOS Expert's post him1985 wrote: I still can not understand how it becomes 240+240 . I have attached the figure. Can you please clear my doubt. Let me add to what Bunuel said by giving you a figure (excuse my drawing) that will help you determine central angles. Attachment: Ques3.jpg [ 15.73 KiB | Viewed 3718 times ] The red arc subtends the red central angle. It is the angle formed by the red lines at the center (toward the red arc) The green arc subtends the green angle at the center formed by green lines (toward the arc). So the blue arc will also subtend the blue angle formed by blue lines (toward the arc). This angle will be greater than 180 and will be the central angle subtended by the blue arc which is greater than semi circle. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: In the figure above, equilateral triangle ABC is inscribed in the   [#permalink] 21 Dec 2014, 20:56

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