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# In the figure above, showing circle with center O and points

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In the figure above, showing circle with center O and points [#permalink]

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In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?

A. 12
B. 18
C. 24
D. 36
E. 72
[Reveal] Spoiler: OA

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Last edited by Bunuel on 26 Feb 2013, 01:43, edited 1 time in total.
Edited the question.
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Re: In the figure above, showing circle with center O and points [#permalink]

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25 Feb 2013, 04:17
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emmak wrote:
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?

12

18

24

36

72

$$\angle AOB = 120^{\circ}$$

So,

$$\angle$$ ACB = $$60^{\circ}$$

So, the $$\triangle$$ ABC is a 30, 60, 90 triangle and the sides are in the ratio 1 : $$\sqrt{3}$$ : 2

or 12 : 12$$\sqrt{3}$$ : 24

So, AC = 24 and hence AO = 12

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Re: In the figure above, showing circle with center O and points [#permalink]

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26 Feb 2013, 02:09
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In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?

A. 12
B. 18
C. 24
D. 36
E. 72

Good question. It tests three must know properties for the GMAT:

1. A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B.

2. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

According to the above, central angle AOB is twice inscribed angle ACB (notice that both subtend the same arc AB). Hence since AOB=120 degrees (third of 360 degrees), then ACB=60 degrees. So, we have that triangle ABC is 30°-60°-90° right triangle.

3. In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$.

Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

Thus, $$AB:AC=\sqrt{3}: 2$$ --> $$\frac{12\sqrt{3}}{AC}=\frac{\sqrt{3}}{2}$$ --> AC=diameter=24 --> radius=12.

For more on triangles check: math-triangles-87197.html
For more on circles check: math-circles-87957.html

Hope it helps.
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Re: In the figure above, showing circle with center O and points [#permalink]

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22 May 2013, 18:38
can someone explain me why cant we use the formula arc length=2x3.14xr(c/360) we know the arc length AB and C/360 as 120..But am not getting the answer.can anyone explain why we should nt use this formula here
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Re: In the figure above, showing circle with center O and points [#permalink]

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23 May 2013, 02:35
skamal7 wrote:
can someone explain me why cant we use the formula arc length=2x3.14xr(c/360) we know the arc length AB and C/360 as 120..But am not getting the answer.can anyone explain why we should nt use this formula here

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Re: In the figure above, showing circle with center O and points [#permalink]

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24 Nov 2013, 05:44
MacFauz wrote:
emmak wrote:
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?

12

18

24

36

72

$$\angle AOB = 120^{\circ}$$

So,

$$\angle$$ ACB = $$60^{\circ}$$

So, the $$\triangle$$ ABC is a 30, 60, 90 triangle and the sides are in the ratio 1 : $$\sqrt{3}$$ : 2

or 12 : 12$$\sqrt{3}$$ : 24

So, AC = 24 and hence AO = 12

How do you that
since $$\angle AOB = 120^{\circ}$$
so $$\angle$$ ACB = $$60^{\circ}$$ ?

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Re: In the figure above, showing circle with center O and points [#permalink]

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24 Nov 2013, 11:28
Bunuel wrote:

In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?

A. 12
B. 18
C. 24
D. 36
E. 72

Good question. It tests three must know properties for the GMAT:

1. A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B.

2. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

According to the above, central angle AOB is twice inscribed angle ACB (notice that both subtend the same arc AB). Hence since AOB=120 degrees (third of 360 degrees), then ACB=60 degrees. So, we have that triangle ABC is 30°-60°-90° right triangle.

3. In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$.

Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

Thus, $$AB:AC=\sqrt{3}: 2$$ --> $$\frac{12\sqrt{3}}{AC}=\frac{\sqrt{3}}{2}$$ --> AC=diameter=24 --> radius=12.

For more on triangles check: math-triangles-87197.html
For more on circles check: math-circles-87957.html

Hope it helps.

I'm having trouble visualizing your 2nd point; could you draw a rough illustration of it? I'm familiar with the theorem, great illustration here: http://www.mathopenref.com/arccentralangletheorem.html, but I can't visualize it in this problem. I got the answer correct as the sqrt(3) tipped me off to a likely 30-60-90, but I'd like to learn a bit more from this problem
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Re: In the figure above, showing circle with center O and points [#permalink]

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24 Nov 2013, 16:55
AccipiterQ wrote:
Bunuel wrote:

In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?

A. 12
B. 18
C. 24
D. 36
E. 72

Good question. It tests three must know properties for the GMAT:

1. A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B.

2. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

According to the above, central angle AOB is twice inscribed angle ACB (notice that both subtend the same arc AB). Hence since AOB=120 degrees (third of 360 degrees), then ACB=60 degrees. So, we have that triangle ABC is 30°-60°-90° right triangle.

3. In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$.

Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

Thus, $$AB:AC=\sqrt{3}: 2$$ --> $$\frac{12\sqrt{3}}{AC}=\frac{\sqrt{3}}{2}$$ --> AC=diameter=24 --> radius=12.

For more on triangles check: math-triangles-87197.html
For more on circles check: math-circles-87957.html

Hope it helps.

I'm having trouble visualizing your 2nd point; could you draw a rough illustration of it? I'm familiar with the theorem, great illustration here: http://www.mathopenref.com/arccentralangletheorem.html, but I can't visualize it in this problem. I got the answer correct as the sqrt(3) tipped me off to a likely 30-60-90, but I'd like to learn a bit more from this problem

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Re: In the figure above, showing circle with center O and points [#permalink]

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26 Nov 2013, 08:26
AccipiterQ wrote:
Bunuel wrote:

In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?

A. 12
B. 18
C. 24
D. 36
E. 72

Good question. It tests three must know properties for the GMAT:

1. A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B.

2. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

According to the above, central angle AOB is twice inscribed angle ACB (notice that both subtend the same arc AB). Hence since AOB=120 degrees (third of 360 degrees), then ACB=60 degrees. So, we have that triangle ABC is 30°-60°-90° right triangle.

3. In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$.

Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

Thus, $$AB:AC=\sqrt{3}: 2$$ --> $$\frac{12\sqrt{3}}{AC}=\frac{\sqrt{3}}{2}$$ --> AC=diameter=24 --> radius=12.

For more on triangles check: math-triangles-87197.html
For more on circles check: math-circles-87957.html

Hope it helps.

I'm having trouble visualizing your 2nd point; could you draw a rough illustration of it? I'm familiar with the theorem, great illustration here: http://www.mathopenref.com/arccentralangletheorem.html, but I can't visualize it in this problem. I got the answer correct as the sqrt(3) tipped me off to a likely 30-60-90, but I'd like to learn a bit more from this problem

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Re: In the figure above, showing circle with center O and points [#permalink]

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08 Dec 2013, 23:26
emmak wrote:
Attachment:
1.jpg
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?

A. 12
B. 18
C. 24
D. 36
E. 72

Guys I did this question with this approach, can you kindly evaluate the approach
$$AB= \frac{1}{3} Circumference$$
So,$$AB = \frac{2}{3}Pi* R$$
Also $$AB=12\sqrt{3}$$
Therefore, $$12\sqrt{3} = \frac{2}{3}Pi* R$$
Manipulating$$R = 12$$
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Re: In the figure above, showing circle with center O and points [#permalink]

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11 Dec 2013, 14:37
Do we just assume that AC is the diameter?
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Re: In the figure above, showing circle with center O and points [#permalink]

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11 Dec 2013, 15:06
Ok, assuming we know this is a right triangle because we know that the picture is accurate...

If the triangle is a right triangle, we know that angle B is the right angle. Furthermore, we are given that arc AB = 1/3 of the circle's circumference (i.e it = 120 degrees) This means that AOB, the central angle = 120 degrees and because AO and BO are radii, we know that triangle AOB is an isosceles triangle. We also know the angle measure of angle BOC which = 180 - 120 = 60. If angle AOB is the interior angle, it is twice the measure of the "exterior" angle ACB, so ACB = 60. Now we know that triangle BOC is equilateral. More importantly, we know that ABC is a 30:60:90 triangle. With this, we can find the measure of any given side knowing just the measure of one. The ratio of side lengths in a 30:60:90 is (x/2):(3/2*x):(x) in this case, x (the radius) = 12.
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Re: In the figure above, showing circle with center O and points [#permalink]

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12 Dec 2013, 02:20
WholeLottaLove wrote:
Do we just assume that AC is the diameter?

We did not assume that. AC passes through the center of the circle thus it's the diameter.

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B.
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Re: In the figure above, showing circle with center O and points [#permalink]

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12 Dec 2013, 05:21
Bunuel wrote:
WholeLottaLove wrote:
Do we just assume that AC is the diameter?

We did not assume that. AC passes through the center of the circle thus it's the diameter.

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B.

I understand the rule, but in other words, how do we know that the drawing isn't misleading and that AC passes above the center?
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Re: In the figure above, showing circle with center O and points [#permalink]

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12 Dec 2013, 05:31
WholeLottaLove wrote:
Bunuel wrote:
WholeLottaLove wrote:
Do we just assume that AC is the diameter?

We did not assume that. AC passes through the center of the circle thus it's the diameter.

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B.

I understand the rule, but in other words, how do we know that the drawing isn't misleading and that AC passes above the center?

OG13, page 150:
Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

OG13, page 272:
A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2).
Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight.
You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees.
All figures lie in a plane unless otherwise indicated.
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Re: In the figure above, showing circle with center O and points [#permalink]

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28 Jul 2014, 21:27
Bunuel wrote:

In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?

A. 12
B. 18
C. 24
D. 36
E. 72

Good question. It tests three must know properties for the GMAT:

1. A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B.

2. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

According to the above, central angle AOB is twice inscribed angle ACB (notice that both subtend the same arc AB). Hence since AOB=120 degrees (third of 360 degrees), then ACB=60 degrees. So, we have that triangle ABC is 30°-60°-90° right triangle.

3. In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$.

Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

Thus, $$AB:AC=\sqrt{3}: 2$$ --> $$\frac{12\sqrt{3}}{AC}=\frac{\sqrt{3}}{2}$$ --> AC=diameter=24 --> radius=12.

For more on triangles check: math-triangles-87197.html
For more on circles check: math-circles-87957.html

Hope it helps.

Hi Bunuel

Particularly in this question with the given options , do we even need this much detail.

We know that the angle AOB = 120 degrees as it is 1/3 the circumference. AO=OB as they are the radius. So the two angles OAB and OBA will be equal in the triangle AOB. Thus Angle OBA=OBA=30 ( 180-120)/2. In a triangle the greatest side is opposite the largest angle . Here AB is opposite the largest angle and is 12sqrt2 so the other sides ( the radii) will have to be less that this value and only A satisfies the condition over here
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Re: In the figure above, showing circle with center O and points [#permalink]

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23 Feb 2016, 03:38

12$$\sqrt{3}$$ = $$\frac{1}{3}$$ * 2 * pie * R
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In the figure above, showing circle with center O and points [#permalink]

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23 Feb 2016, 03:49
emmak wrote:
Attachment:
The attachment 1.jpg is no longer available
given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ ,

does the wording simply states that the length of the arc AB is one third WITHOUT INDICATING THE EXACT VALUE, and the value of 12 root 3 is given for the side AB (rather than the arc AB)? i think the probles is poorly worded. look how it looks in the original. veritas should not have placed that bar above AB if they meant it was not the arc but the length
Attachments

2016-02-23 16-47-01 gmat.veritasprep.com gmat exams 1105058 responses 18191158 - Google Chrome.png [ 92.84 KiB | Viewed 2969 times ]

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Re: In the figure above, showing circle with center O and points [#permalink]

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31 Jul 2016, 00:24
suk1234 wrote:
emmak wrote:
Attachment:
1.jpg
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?

A. 12
B. 18
C. 24
D. 36
E. 72

Guys I did this question with this approach, can you kindly evaluate the approach
$$AB= \frac{1}{3} Circumference$$
So,$$AB = \frac{2}{3}Pi* R$$
Also $$AB=12\sqrt{3}$$
Therefore, $$12\sqrt{3} = \frac{2}{3}Pi* R$$
Manipulating$$R = 12$$

AB Arc and not AB line segment is equal to 2/3 of circumference. Moreover AB arc is not 12root3, AB Line segment is.
Re: In the figure above, showing circle with center O and points   [#permalink] 31 Jul 2016, 00:24
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