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The circle with center O has a circumference of 6π

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The circle with center O has a circumference of 6π  [#permalink]

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New post 16 Jun 2015, 03:28
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The circle with center O has a circumference of \(6\pi{\sqrt{3}}\). If AC is a diameter of the circle, what is the length of line segment BC?

(A) \(\frac{3}{\sqrt{2}}\)

(B) 3

(C) \(3\sqrt{3}\)

(D) 9

(E) \(9\sqrt{3}\)


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Attachment:
2015-06-16_1428.png
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Re: The circle with center O has a circumference of 6π  [#permalink]

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New post 16 Jun 2015, 05:22
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Bunuel wrote:
Image
The circle with center O has a circumference of \(6\pi{\sqrt{3}}\). If AC is a diameter of the circle, what is the length of line segment BC?

(A) \(\frac{3}{\sqrt{2}}\)

(B) 3

(C) \(3\sqrt{3}\)

(D) 9

(E) \(9\sqrt{3}\)


Kudos for a correct solution.
Attachment:
2015-06-16_1428.png


Circumference of Circle = \(6\pi{\sqrt{3}}\) = \(2\pi{r}\) where r is the radius of circle

i.e. \(r = 3{\sqrt{3}}\) = AC/2

Since, Side AC of Triangle is equal to the Diameter therefore Angle ABC must be 90 degree

CONCEPT: Angle drawn in a semicircle on the circumference is always 90 degrees

i.e. Triangle ABC is 30-60-90 Traingle with ratio of the sides \(x:x{\sqrt{3}}:2x\)

and AC = \(2x = 2r = 6{\sqrt{3}}\)

therefore, BC = \(x{\sqrt{3}} = 3{\sqrt{3}}*{\sqrt{3}}\)

i.e. BC = 9

Answer: Option
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Re: The circle with center O has a circumference of 6π  [#permalink]

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New post 16 Jun 2015, 06:48
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Circumference of a circle = 2\(\pi\)r = 6\(\pi\sqrt{3}\)
This means r(radius)= 3\(\sqrt{3}\)
AC= diameter= 2r = 6\(\sqrt{3}\)

In triangle ABC,
cos 30 = \(\frac{BC}{AC}\)
\(\sqrt{3}\)/2 = BC/6\(\sqrt{3}\)
BC= 9

Option D
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The circle with center O has a circumference of 6π  [#permalink]

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New post Updated on: 16 Jun 2015, 14:01
Bunuel wrote:
Image
The circle with center O has a circumference of \(6\pi{\sqrt{3}}\). If AC is a diameter of the circle, what is the length of line segment BC?

(A) \(\frac{3}{\sqrt{2}}\)

(B) 3

(C) \(3\sqrt{3}\)

(D) 9

(E) \(9\sqrt{3}\)


Kudos for a correct solution.
Attachment:
2015-06-16_1428.png


This is a 30-60-90 triangle and we are told that the circumference is \(6\pi{\sqrt{3}}\). Set \(6\pi{\sqrt{3}}\) = \(2(\pi)r\) giving you r = \(3{\sqrt{3}}\)

You know the diameter, line AC equal 2r or \(6{\sqrt{3}}\)

If you remember your 30-60-90 angles, you can calculate the 60 degree angle, \(3{\sqrt{3}}\)*\({\sqrt{3}}\) BC = 9

Originally posted by BuggerinOn on 16 Jun 2015, 08:14.
Last edited by BuggerinOn on 16 Jun 2015, 14:01, edited 2 times in total.
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Re: The circle with center O has a circumference of 6π  [#permalink]

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New post 16 Jun 2015, 10:36
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The circle with center O has a circumference of \(6\pi{\sqrt{3}}\). If AC is a diameter of the circle, what is the length of line segment BC?

Solution -

Angle B is always 90 degrees. So C=30 and A=60.

Triangle ABC is 30-60-90 Triangle with ratio of the sides x:x√3:2x.

2\(\pi\)r=\(6\pi{\sqrt{3}}\) ->Diameter(2r) = AC = 2x = 6√3 ->x=3√3.

BC = x√3 = 9. ANS D.

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Re: The circle with center O has a circumference of 6π  [#permalink]

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New post 16 Jun 2015, 11:20
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Bunuel wrote:
Image
The circle with center O has a circumference of \(6\pi{\sqrt{3}}\). If AC is a diameter of the circle, what is the length of line segment BC?

(A) \(\frac{3}{\sqrt{2}}\)

(B) 3

(C) \(3\sqrt{3}\)

(D) 9

(E) \(9\sqrt{3}\)


Kudos for a correct solution.

Attachment:
2015-06-16_1428.png


Inscribed triangle ABC is a 30-60-90 triangle. Angle ABC = 90 because the opposite line is the diameter of the Circle. Angle CAB is 60 to add up to a Total of 180 Degrees.

AC = Diameter of the Circle = \(6\sqrt{3}\) which equals 2x in the special right triangle. Solve for x by dividing by 2: \(3\sqrt{3}\) = x. Finally BC = \(x\sqrt{3}\) = \(3\sqrt{3}*\sqrt{3}\)=9

Answer Choice D
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Re: The circle with center O has a circumference of 6π  [#permalink]

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New post 22 Jun 2015, 06:49
Bunuel wrote:
Image
The circle with center O has a circumference of \(6\pi{\sqrt{3}}\). If AC is a diameter of the circle, what is the length of line segment BC?

(A) \(\frac{3}{\sqrt{2}}\)

(B) 3

(C) \(3\sqrt{3}\)

(D) 9

(E) \(9\sqrt{3}\)


Kudos for a correct solution.
Attachment:
The attachment 2015-06-16_1428.png is no longer available


MANHATTAN GMAT OFFICIAL SOLUTION:

Some intuitive recollection of geometry rules and a picture drawn to scale can help us determine reasonable answer choices. If AC is a diameter of the circle, then triangle ABC is a right triangle, with angle ABC = 90 degrees. The shortest side of a triangle is across from its smallest angle, and the longest side of a triangle is across from its largest angle. Therefore, AC > BC > AB.

The circumference of the circle = \(\pi{d}=6\pi{\sqrt{3}}\), so \(d=6\sqrt{3}\approx{6*1.7}=10.2\) Thus, AC ≈ 10.2 and BC < 10.2. But we can clearly see from our picture drawn to scale that BC is longer than half the diameter, so we conservatively determine that BC > 5.1.
Image

Attachment:
2015-06-22_1748.png
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Re: The circle with center O has a circumference of 6π  [#permalink]

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Re: The circle with center O has a circumference of 6π   [#permalink] 04 Aug 2019, 21:51
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