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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] New post   Updated on: 26 Dec 2017, 08:44
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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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C

Originally posted by ulm on 17 Jul 2010, 10:33.
Last edited by Bunuel on 26 Dec 2017, 08:44, 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] New post   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 \(c=\frac{24*3}{2}=36=\pi{d}\) --> \(d=\frac{36}{\pi}\approx{\frac{36}{3.14}}\approx{11.5}\).

Answer: C.
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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   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.)


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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   25 Aug 2010, 20:30
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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] New post   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.
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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   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...
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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   10 May 2013, 15:40
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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?
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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   11 May 2013, 03:21
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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.
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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   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?

Thanks a bunch!
RCA
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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   01 Nov 2014, 04:57
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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 in the [#permalink] New post   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
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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   21 Dec 2014, 04:43
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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 in the [#permalink] New post   21 Dec 2014, 06:45
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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.
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Ques.png
Ques.png [ 11.56 KiB | Viewed 131841 times ]

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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   21 Dec 2014, 08:44
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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
Untitled.png [ 12.63 KiB | Viewed 131584 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 in the [#permalink] New post   21 Dec 2014, 21:56
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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
Ques3.jpg [ 15.73 KiB | Viewed 111225 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.
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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   08 Jun 2017, 16:41
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Attached is a visual that should help.
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Screen Shot 2017-06-08 at 4.31.56 PM.png
Screen Shot 2017-06-08 at 4.31.56 PM.png [ 225.94 KiB | Viewed 100251 times ]

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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   26 Dec 2017, 08:38
Can someone help me out here?

everything makes sense up to C=36.

Since the circumference is 36 and Circumference=D*pi or 2*r*pi.

Therefore, shouldnt the diameter just be 36? please help.
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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   26 Dec 2017, 08:48
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teamryan15 wrote:
Can someone help me out here?

everything makes sense up to C=36.

Since the circumference is 36 and Circumference=D*pi or 2*r*pi.

Therefore, shouldnt the diameter just be 36? please help.


\(c=\frac{24*3}{2}=36=\pi{d}\);

\(d=\frac{36}{\pi}\approx{\frac{36}{3.14}}\approx{11.5}\).
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Re: In the figure above, equilateral triangle ABC is inscribed in the [#permalink] New post   21 May 2019, 19:33
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ulm 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

Show Spoiler
Attachment:
Circle.JPG


We are given the length of arc ABC is equal to 24. Since arc AB, BC, and CA each represent 1/3 of the circumference, arc ABC represents 2/3 of the circumference. Thus, we can create the following equation.

2/3(circumference) = 24

2/3(πd) = 24

πd = 24 x 3/2

πd = 36

d = 36/π

d ≈ 11

Answer: C
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Re: Imagine an equilateral triangle ABC inscribed in a circle. If the leng [#permalink] New post   23 Jan 2020, 20:51
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naju wrote:
Imagine an equilateral triangle ABC inscribed in a 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



The vertices of an equilateral triangle divide the circle into 3 equal parts. Arc ABC will make up two of those parts so the circumference of the circle will be 36.

\(\pi * d = 36\)
d = 36/3.14 = About 11
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Re: Imagine an equilateral triangle ABC inscribed in a circle. If the leng [#permalink]
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