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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] New post 25 Feb 2013, 01:21
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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, 00:43, edited 1 time in total.
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Re: In the figure above, showing circle with center O and points [#permalink] New post 25 Feb 2013, 03:17
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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

Answer is A
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Re: In the figure above, showing circle with center O and points [#permalink] New post 26 Feb 2013, 01:09
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Image
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.
Image
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.

Answer: A.

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] New post 22 May 2013, 17:38
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] New post 23 May 2013, 01:35
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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


Can you please show your work?
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Re: In the figure above, showing circle with center O and points [#permalink] New post 24 Nov 2013, 04:44
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

Answer is A


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

:banana
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Re: In the figure above, showing circle with center O and points [#permalink] New post 24 Nov 2013, 10:28
Bunuel wrote:
Image
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.
Image
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.

Answer: A.

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] New post 24 Nov 2013, 15:55
AccipiterQ wrote:
Bunuel wrote:
Image
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.
Image
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.

Answer: A.

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


Thnaks bunel for your posts
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Re: In the figure above, showing circle with center O and points [#permalink] New post 26 Nov 2013, 07:26
Expert's post
AccipiterQ wrote:
Bunuel wrote:
Image
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.
Image
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.

Answer: A.

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


Please, follow this link: math-circles-87957.html
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: In the figure above, showing circle with center O and points [#permalink] New post 08 Dec 2013, 22:26
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
ManipulatingR = 12
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Re: In the figure above, showing circle with center O and points [#permalink] New post 11 Dec 2013, 13:37
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] New post 11 Dec 2013, 14:06
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] New post 12 Dec 2013, 01:20
Expert's post
WholeLottaLove wrote:
Do we just assume that AC is the diameter?


Please read here: in-the-figure-above-showing-circle-with-center-o-and-points-147885.html#p1187477

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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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: In the figure above, showing circle with center O and points [#permalink] New post 12 Dec 2013, 04:21
Bunuel wrote:
WholeLottaLove wrote:
Do we just assume that AC is the diameter?


Please read here: in-the-figure-above-showing-circle-with-center-o-and-points-147885.html#p1187477

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] New post 12 Dec 2013, 04:31
Expert's post
WholeLottaLove wrote:
Bunuel wrote:
WholeLottaLove wrote:
Do we just assume that AC is the diameter?


Please read here: in-the-figure-above-showing-circle-with-center-o-and-points-147885.html#p1187477

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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In the figure above, showing circle with center O and points [#permalink] New post 28 Jul 2014, 20:27
Bunuel wrote:
Image
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.
Image
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.

Answer: A.

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
Re: In the figure above, showing circle with center O and points   [#permalink] 28 Jul 2014, 20:27
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