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ps geometry [#permalink] New post 31 May 2005, 14:14
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A
B
C
D
E

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hello
this is from ETS can anyone helps
ty

If A is the center of the circle shown above and AB=BC=CD, what is the value of x?
(A) 15
(B) 30
(C) 45
(D) 60
(E) 75

Thanks
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Senior Manager
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 [#permalink] New post 31 May 2005, 14:55
Mandy, please attach JPEGs for everyone convenience.

2*PI*R extends 360 degrees

R extends 360/2PI degrees.

x = 180 - 90 - 360/2PI = approx 30 degrees
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 [#permalink] New post 31 May 2005, 17:15
is there some data missing in this figure like if AD is perpedicular then X will be 30 degrees as ABC and BCD are equilateral trianglels
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 [#permalink] New post 31 May 2005, 17:31
rxs0005 wrote:
is there some data missing in this figure like if AD is perpedicular then X will be 30 degrees as ABC and BCD are equilateral trianglels


The secant is bisected because the arc itself is bisected. Therefore the line will be a perpendicular from center to secant. This has to be deduced.

Data provided is sufficient.
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 [#permalink] New post 02 Jun 2005, 07:51
B.

In triangle ABM ( M - point where AC intersects BD)
AB = 1/2 AM (coz ABD and BCD are congruent triangles)
so L BAM = 60
Hence X = 90 - 60 = 30 ( right angle triangle)

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 [#permalink] New post 02 Jun 2005, 14:23
the figure ABCD is a rhombus with equals sides and equal diagonals. so ABC is an equilateral triangle with all angles are 60. the intersection of AC is the midpoint and bisects ABC. so ABD is 30.
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 [#permalink] New post 02 Jun 2005, 19:09
HMTG and Christoph,

I think you cannot assume that BC is a straight line and call the figure a rhombus.

Although the answers come the same in both cases ( I got 30 too )
I think your approach is a little flawed.
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 [#permalink] New post 02 Jun 2005, 21:13
My understanding is AB=BC=CD are talking about line BC and line CD, not the arcs. Therefore we can get the equalateral trangle and the 30 degree.
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 [#permalink] New post 02 Jun 2005, 22:56
ashkg wrote:
HMTG and Christoph,

I think you cannot assume that BC is a straight line and call the figure a rhombus.

Although the answers come the same in both cases ( I got 30 too )
I think your approach is a little flawed.


but all sides are equal. the diagonals are perpendicular. AB=AC=AC=radii. it must be a rhombus or a equilateral parallelogram. otherwise BC wouldnt equal CD.
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 [#permalink] New post 03 Jun 2005, 00:26
HongHu wrote:
My understanding is AB=BC=CD are talking about line BC and line CD, not the arcs. Therefore we can get the equalateral trangle and the 30 degree.


How did I miss that ! ( equilateral triangle :oops: )

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 [#permalink] New post 03 Jun 2005, 07:33
christoph wrote:
ashkg wrote:
HMTG and Christoph,

I think you cannot assume that BC is a straight line and call the figure a rhombus.

Although the answers come the same in both cases ( I got 30 too )
I think your approach is a little flawed.


but all sides are equal. the diagonals are perpendicular. AB=AC=AC=radii. it must be a rhombus or a equilateral parallelogram. otherwise BC wouldnt equal CD.


Chris, I didnt understand, can you explain why BC cannot equal CD ?

All,
The confusion is whether BC and CD are arcs or straight lines. The problem can be solved assuming either.

I solved it assuming BC and CD are arcs because they are shown as arcs and not as straight lines.

What happens if we get somethign like this in the exam ? What do you assume then ? :roll:
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 [#permalink] New post 03 Jun 2005, 09:12
ashkg wrote:
rxs0005 wrote:
is there some data missing in this figure like if AD is perpedicular then X will be 30 degrees as ABC and BCD are equilateral trianglels


The secant is bisected because the arc itself is bisected. Therefore the line will be a perpendicular from center to secant. This has to be deduced.

Data provided is sufficient.


I'm not sure I understand how it can be assumed that BD is perpendicular to AC
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 [#permalink] New post 03 Jun 2005, 16:34
cloudz9 wrote:
ashkg wrote:
rxs0005 wrote:
is there some data missing in this figure like if AD is perpedicular then X will be 30 degrees as ABC and BCD are equilateral trianglels


The secant is bisected because the arc itself is bisected. Therefore the line will be a perpendicular from center to secant. This has to be deduced.

Data provided is sufficient.


I'm not sure I understand how it can be assumed that BD is perpendicular to AC


OK.. this is a rule, any line that bisects a chord is perpendicular to it.

Try deducing it yourself. The line bisects the chord and creates two congruent triangles. The angles that hit the chord will therefore be equal to each other.

Angle on the chord = 180/2 = 90
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 [#permalink] New post 03 Jun 2005, 20:41
ashkg wrote:
All,
The confusion is whether BC and CD are arcs or straight lines. The problem can be solved assuming either.

I solved it assuming BC and CD are arcs because they are shown as arcs and not as straight lines.


You can't assume the BC and CD in AB=BC=CD (given in the stem) are arcs ashkg. AB should be less then arc BC and arc CD if AB is going to be equal to segment BC and segment CD.
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 [#permalink] New post 03 Jun 2005, 21:44
BAC = 60
CAD = 60

That's a total of 120 in triangle BAD, where AB = AD

ABD = BDA = (180-120)/2 =30.
  [#permalink] 03 Jun 2005, 21:44
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