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# The points R, T, and U lie on a circle that has radius 4. If

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Manager
Joined: 05 Jul 2009
Posts: 181

Kudos [?]: 52 [1], given: 5

The points R, T, and U lie on a circle that has radius 4. If [#permalink]

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15 Oct 2009, 15:52
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Difficulty:

35% (medium)

Question Stats:

78% (01:09) correct 22% (02:28) wrong based on 113 sessions

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The points R, T, and U lie on a circle that has radius 4. If the length of arc RTU is $$\frac{4*\pi}{3}$$, what is the length of line segment RU?

A. 4/3
B. 8/3
C. 3
D. 4
E. 6

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-points-r-t-and-u-lie-on-a-circle-that-has-radius-86376.html
[Reveal] Spoiler: OA

Kudos [?]: 52 [1], given: 5

Senior Manager
Joined: 31 Aug 2009
Posts: 415

Kudos [?]: 349 [1], given: 20

Location: Sydney, Australia
Re: Geometry Arc? [#permalink]

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15 Oct 2009, 16:36
1
KUDOS
Diameter = 2(Pi)r = 8Pi
Let O be the centre of the circle. Angle of ROU is [(4Pi/3)/8Pi] * 360 = 60
OR = OU =4 (radii) so the triangle ROU is isosceles.
Since angle ROU is 60 this makes the triangle equilateral.
RU = OR = OU = 4
ANS = D

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Manager
Joined: 04 Aug 2009
Posts: 53

Kudos [?]: 3 [0], given: 2

Location: pitt
Re: Geometry Arc? [#permalink]

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15 Oct 2009, 16:43
2 ( pi) r = x + 4(pi) / 3 .

why will this not work ..
as perimeter of circle should be = length of 2 arc ....

Kudos [?]: 3 [0], given: 2

Senior Manager
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Re: Geometry Arc? [#permalink]

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15 Oct 2009, 16:47
pittgreek wrote:
2 ( pi) r = x + 4(pi) / 3 .

why will this not work ..
as perimeter of circle should be = length of 2 arc ....

Not quite sure what X is meant to be in your equation.
4(pi)/3 is a part of the overall diameter which is 2(pi)r = 8(pi).

Kudos [?]: 349 [0], given: 20

Senior Manager
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Kudos [?]: 349 [0], given: 20

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Re: Geometry Arc? [#permalink]

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15 Oct 2009, 17:02
pittgreek wrote:
x= length of RU ...

sorry not absolutely sure where you got the original equation from. The arc (4Pi/3) is defined by a portion of the overall diameter.
Arc + Line Segment will not give you the perimeter of the circle, only of the area bound by the line segment.

Unless I've misunderstood the question what is it asking for is the straight line segment. which will be 4.

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Manager
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Re: Geometry Arc? [#permalink]

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17 Oct 2009, 16:15
@ yangsta8
You understood the question right and I think your approach to the solution is also correct. Thanks a lot.

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Director
Joined: 23 Apr 2010
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Re: Geometry Arc? [#permalink]

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02 Nov 2010, 05:00
Can someone please look at this question? I got $$4sqrt{2}$$

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Posts: 41892

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Re: Geometry Arc? [#permalink]

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02 Nov 2010, 05:03
Expert's post
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nonameee wrote:
Can someone please look at this question? I got $$4sqrt{2}$$

The points R, T, and U lie on a circle that has radius 4. If the length of arc RTU is 4*PI/3, what is the length of line segment RU?
A. 4/3
B. 8/3
C. 3
D. 4
E. 6

The circumference of a circle=$$2*\pi*r=8*\pi$$, $$\frac{RTU}{8*\pi}= \frac{(\frac{4*\pi}{3})}{8\pi}=\frac{1}{6}$$. --> Angle $$\angle{RCU}=\frac{360}{6}=60$$ degrees (C center of the circle).

RCU is isosceles triangle as $$RC=CU=r$$ and $$RCU=CRU=CUR=60$$ degrees. Hence $$RU=r=4$$.

Also discussed here: circle-arc-86376.html?hilit=length%20radius%20segment
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Director
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Re: Geometry Arc? [#permalink]

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02 Nov 2010, 05:13
Oh, yes, thank you, Bunuel. They are referring to the length of arc and not to the measure of a corresponding angle (which is 120 degrees).

Thanks a lot again.

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Re: Geometry Arc? [#permalink]

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14 Nov 2013, 12:57
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Re: The points R, T, and U lie on a circle that has radius 4. If [#permalink]

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14 Nov 2013, 13:11
OPEN DISCUSSION OF THIS QUESTION IS HERE: the-points-r-t-and-u-lie-on-a-circle-that-has-radius-86376.html
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Re: The points R, T, and U lie on a circle that has radius 4. If   [#permalink] 14 Nov 2013, 13:11
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# The points R, T, and U lie on a circle that has radius 4. If

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