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# In the figure above, equilateral triangle ABC is inscribed

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In the figure above, equilateral triangle ABC is inscribed [#permalink]  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
[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.
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Re: Geometry: circle & triangle from mba.com [#permalink]  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 $$c=\frac{24*3}{2}=36=\pi{d}$$ --> $$d\approx{11.5}$$.

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Re: Geometry: circle & triangle from mba.com [#permalink]  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: Geometry: circle & triangle from mba.com [#permalink]  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)
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Re: Geometry: circle & triangle from mba.com [#permalink]  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: Geometry: circle & triangle from mba.com [#permalink]  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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Re: Geometry: circle & triangle from mba.com [#permalink]  19 Nov 2010, 11:50
Pi*D=24(3/2)
This gives us approximately 11. So answer should be C.
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Re: Geometry: circle & triangle from mba.com [#permalink]  20 Nov 2010, 01:51
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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.)

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

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Re: Geometry: circle & triangle from mba.com [#permalink]  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?
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Re: Geometry: circle & triangle from mba.com [#permalink]  11 May 2013, 02:21
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 $$c=\frac{24*3}{2}=36=\pi{d}$$ --> $$d\approx{11.5}$$.

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.
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Re: Geometry: circle & triangle from mba.com [#permalink]  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!
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Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]  22 Jul 2014, 20:30
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Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]  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.
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Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]  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!
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Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]  01 Nov 2014, 03:57
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.
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Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]  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
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Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]  21 Dec 2014, 03:43
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.
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Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]  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 5843 times ]

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Re: In the figure above, equilateral triangle ABC is inscribed [#permalink]  21 Dec 2014, 07:44
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 5807 times ]
The red arc is 24 and its central angle is 240 degrees.

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

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