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

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17 Jul 2010, 10:33

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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?

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

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17 Jul 2010, 11:02

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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 \(c=\frac{24*3}{2}=36=\pi{d}\) --> \(d\approx{11.5}\).

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

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21 Jul 2010, 10:18

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]

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08 Sep 2010, 07:28

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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.

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

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18 Nov 2010, 20:15

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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... _________________

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

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20 Nov 2010, 02:51

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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]

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10 May 2013, 15:40

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]

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11 May 2013, 03:21

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]

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11 May 2013, 10:08

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!

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

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22 Jul 2014, 21:30

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

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26 Oct 2014, 10:10

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.

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

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01 Nov 2014, 00:15

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?

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

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01 Nov 2014, 04:57

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. _________________

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

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21 Dec 2014, 00:29

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 _________________

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

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21 Dec 2014, 04:43

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. _________________

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

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21 Dec 2014, 06:45

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.

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

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21 Dec 2014, 08:44

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 8239 times ]

The red arc is 24 and its central angle is 240 degrees.

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