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A B -------------- E --- C ------- D (Since, I cannot attach

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gayathri
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A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 08:47
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-------A

B--------------E

---C-------D

(Since, I cannot attach a jpg, connect the pts ABCDE to form a pentagon.)

In the figure, What is the measure of angle ACB?

1) Triangle ACD is an equilateral triangle.
2) AB=BC=AE=DE



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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 08:58
I would have to go with C

As I would be the inner triangle is 60-60-60 but we cannot make out about ACB

from II nothign can be said

from I and II all sidesof the traingle become equal so it will be 60 deg
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:03
rxs0005 wrote:

from I and II all sidesof the traingle become equal so it will be 60 deg


How do you get all sides are equal?
1) gives you AC=AD=CD
2) gives you AB=BC=AE=DE

How do you combine the two?
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:04
I pick C as statement 2 is not enough.
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:06
I would go with E. how is it C? guys please explain..
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:06
twixt wrote:
I pick C as statement 2 is not enough.


Twixt can you explain your approach?
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:08
I am getting it as E. I may not be able to explain at length here because of the diagrams. However, both the stems either individually or together are not giving us any conclusice stuff.

so, E..I am eager to know the answer 8-)

gayathri wrote:
-------A

B--------------E

---C-------D

(Since, I cannot attach a jpg, connect the pts ABCDE to form a pentagon.)

In the figure, What is the measure of angle ACB?

1) Triangle ACD is an equilateral triangle.
2) AB=BC=AE=DE
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:09
banerjeea_98 wrote:
"C" for me.

can you explain how you are getting C
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:10
The stem is asking ACB angle but not ACD. Only ACD can be 60 degrees.

rxs0005 wrote:
I would have to go with C

As I would be the inner triangle is 60-60-60 but we cannot make out about ACB

from II nothign can be said

from I and II all sidesof the traingle become equal so it will be 60 deg
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:11
All sides are not equal actually...
If they were, pentagon would become an hexagon...
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twixt
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:29
There is no conclusion from statement 1 neither from statement 2.

Both together : considering that 0 is the circle center then OA=OC=OB
then ACB=ACO

As ACD is equilateral O must be the barycenter (?) of the triangle and in this specific case line CO divides equally the angle C
So ACO=ACB = 30 degrees to me.
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twixt
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:29
There is no conclusion from statement 1 neither from statement 2.

Both together : considering that 0 is the circle center then OA=OC=OB
then ACB=ACO

As ACD is equilateral O must be the barycenter (?) of the triangle and in this specific case line CO divides equally the angle C
So ACO=ACB = 30 degrees to me.
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gayathri
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:48
OA is E

Twixt, from your explanation it looks like you are assuming that the pentagon is a regular pentagon (OA=OC=OB). But we do not know that.
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rxs0005
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:53
I think i know what mistake i made

i did not read the Q properly i assumed

CD was = too in the II statement since CD is common to ACD and CA common to CBA
i concluded that both ABC and ACD are equilateral Triangle

good Q G!
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twixt
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 09:54
My mistake gayathri, I do not know why I was thinking all your points were standing on a circle... :oops:
If not E is clear
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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink] New post   13 Jan 2005, 10:10
Hmmm...i made the assumption of regular pentagon....my bad, in that case "E" is correct.



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Re: A B -------------- E --- C ------- D (Since, I cannot attach [#permalink]
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