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In the figure above, equilateral triangle ABC is inscribed [#permalink]
17 Jul 2010, 09:33

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A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

65% (01:55) correct
34% (01:08) wrong based on 104 sessions

Attachment:

Circle.JPG [ 5.57 KiB | Viewed 4053 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]
17 Jul 2010, 10:02

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Expert's post

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]
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.
_________________

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

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

Re: Geometry: circle & triangle from mba.com [#permalink]
18 Nov 2010, 19:15

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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]
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.)

COUNTER ARGUMENTS WELCOME, SORRY I CUDN'T ATTACH PICTURE

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}.

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]
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}.

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]
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!

gmatclubot

Re: Geometry: circle & triangle from mba.com
[#permalink]
11 May 2013, 09:08