It is currently 18 Nov 2017, 18:37

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# ps geometry

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Senior Manager
Joined: 30 May 2005
Posts: 274

Kudos [?]: 44 [0], given: 0

### Show Tags

31 May 2005, 15:14
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Attachments

File comment: the circle
Doc2.doc [20.5 KiB]

Kudos [?]: 44 [0], given: 0

Senior Manager
Joined: 21 Mar 2004
Posts: 445

Kudos [?]: 76 [0], given: 0

Location: Cary,NC

### Show Tags

31 May 2005, 15:55
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
_________________

ash
________________________
I'm crossing the bridge.........

Kudos [?]: 76 [0], given: 0

Director
Joined: 07 Jun 2004
Posts: 610

Kudos [?]: 948 [0], given: 22

Location: PA

### Show Tags

31 May 2005, 18:15
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

Kudos [?]: 948 [0], given: 22

Senior Manager
Joined: 21 Mar 2004
Posts: 445

Kudos [?]: 76 [0], given: 0

Location: Cary,NC

### Show Tags

31 May 2005, 18:31
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.
_________________

ash
________________________
I'm crossing the bridge.........

Kudos [?]: 76 [0], given: 0

Senior Manager
Joined: 17 Apr 2005
Posts: 372

Kudos [?]: 30 [0], given: 0

Location: India

### Show Tags

02 Jun 2005, 08:51
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)

HMTG.

Kudos [?]: 30 [0], given: 0

VP
Joined: 30 Sep 2004
Posts: 1480

Kudos [?]: 425 [0], given: 0

Location: Germany

### Show Tags

02 Jun 2005, 15:23
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.
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Kudos [?]: 425 [0], given: 0

Senior Manager
Joined: 21 Mar 2004
Posts: 445

Kudos [?]: 76 [0], given: 0

Location: Cary,NC

### Show Tags

02 Jun 2005, 20:09
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.
_________________

ash
________________________
I'm crossing the bridge.........

Kudos [?]: 76 [0], given: 0

SVP
Joined: 03 Jan 2005
Posts: 2228

Kudos [?]: 385 [0], given: 0

### Show Tags

02 Jun 2005, 22:13
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.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Kudos [?]: 385 [0], given: 0

VP
Joined: 30 Sep 2004
Posts: 1480

Kudos [?]: 425 [0], given: 0

Location: Germany

### Show Tags

02 Jun 2005, 23:56
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.
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Kudos [?]: 425 [0], given: 0

Senior Manager
Joined: 17 Apr 2005
Posts: 372

Kudos [?]: 30 [0], given: 0

Location: India

### Show Tags

03 Jun 2005, 01:26
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 )

HMTG.

Kudos [?]: 30 [0], given: 0

Senior Manager
Joined: 21 Mar 2004
Posts: 445

Kudos [?]: 76 [0], given: 0

Location: Cary,NC

### Show Tags

03 Jun 2005, 08:33
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 ?
_________________

ash
________________________
I'm crossing the bridge.........

Kudos [?]: 76 [0], given: 0

Senior Manager
Joined: 17 May 2005
Posts: 270

Kudos [?]: 19 [0], given: 0

Location: Auckland, New Zealand

### Show Tags

03 Jun 2005, 10:12
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

Kudos [?]: 19 [0], given: 0

Senior Manager
Joined: 21 Mar 2004
Posts: 445

Kudos [?]: 76 [0], given: 0

Location: Cary,NC

### Show Tags

03 Jun 2005, 17:34
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
_________________

ash
________________________
I'm crossing the bridge.........

Kudos [?]: 76 [0], given: 0

SVP
Joined: 03 Jan 2005
Posts: 2228

Kudos [?]: 385 [0], given: 0

### Show Tags

03 Jun 2005, 21:41
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.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Kudos [?]: 385 [0], given: 0

Manager
Joined: 28 Aug 2004
Posts: 205

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

### Show Tags

03 Jun 2005, 22:44
BAC = 60

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

ABD = BDA = (180-120)/2 =30.

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

03 Jun 2005, 22:44
Display posts from previous: Sort by

# ps geometry

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.