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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] New post 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] New post 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}.

Answer: C.
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Re: Geometry: circle & triangle from mba.com [#permalink] New post 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}.

Answer: C.


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

Thanks in advance.
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Re: Geometry: circle & triangle from mba.com [#permalink] New post 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] New post 08 Sep 2010, 06:28
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] New post 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
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...
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Re: Geometry: circle & triangle from mba.com [#permalink] New post 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] New post 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] New post 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
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Re: Geometry: circle & triangle from mba.com [#permalink] New post 10 May 2013, 14:40
Bunuel wrote:
Image
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?
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Re: Geometry: circle & triangle from mba.com [#permalink] New post 11 May 2013, 02:21
Expert's post
up4gmat wrote:
Bunuel wrote:
Image
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.
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Re: Geometry: circle & triangle from mba.com [#permalink] New post 11 May 2013, 09:08
Bunuel wrote:
up4gmat wrote:
Bunuel wrote:
Image
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!
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Re: In the figure above, equilateral triangle ABC is inscribed [#permalink] New post 22 Jul 2014, 20:30
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Re: In the figure above, equilateral triangle ABC is inscribed   [#permalink] 22 Jul 2014, 20:30
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