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# If points A , B , and C lie on a circle of radius 1, what is

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If points A , B , and C lie on a circle of radius 1, what is [#permalink]  27 Sep 2009, 09:00
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If points A , B , and C lie on a circle of radius 1, what is the area of triangle ABC ?

1. AB^2 = BC^2 + AC^2
2. \angle CAB equals 30 degrees
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Re: Triangle inscribed in circle [#permalink]  27 Sep 2009, 09:15
1, ABC is right triangle, Angle C=90, cant figure out the 3 sides ->insuff
2, Ang A=30 degrees, insuff

both 1&2, ABC is right triangle, with 1 angle=30degrees, ABC is half of a equa triangle, then AB=2, AC=2BC... suff
AC^2+BC^2=AB^2=4 -> AC, BC -> area

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Re: Triangle inscribed in circle [#permalink]  27 Sep 2009, 09:32
Yep, answer is C as per the explanation from Mikko.
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Re: Triangle inscribed in circle [#permalink]  27 Sep 2009, 12:01
Question: Is it possible to have a right triangle in a circle without having hypotenuse as the diameter? If no, then stmt 1 should be sufficient as the size and shape of the triangle is always the same !!
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Re: Triangle inscribed in circle [#permalink]  27 Sep 2009, 13:28
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Economist wrote:
Question: Is it possible to have a right triangle in a circle without having hypotenuse as the diameter? If no, then stmt 1 should be sufficient as the size and shape of the triangle is always the same !!

I disagree. The height of such triangle depends on where the 3rd point lies on the circle, and so is its area. Consider following figure:

Attachment:

Triangles.jpg [ 10.69 KiB | Viewed 1668 times ]

So Area of Triangle 1 < Area of Triangle 2
So statement 1 is insufficient.
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Re: Triangle inscribed in circle [#permalink]  27 Sep 2009, 21:30
ahhh okay...thanks hgp2k ..+1 to you. This clears my misconception that there is only one way to draw a right triangle inscribed in a circle
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Re: Triangle inscribed in circle [#permalink]  27 Sep 2009, 21:35
Economist wrote:
ahhh okay...thanks hgp2k ..+1 to you. This clears my misconception that there is only one way to draw a right triangle inscribed in a circle

Thanks mate
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Re: Triangle inscribed in circle [#permalink]  27 Sep 2009, 21:58
Economist wrote:
If points A , B , and C lie on a circle of radius 1, what is the area of triangle ABC ?

1. AB^2 = BC^2 + AC^2
2. \angle CAB equals 30 degrees

What is the purpose of radius = 1?We haven't used it anywhere in the problem.
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Re: Triangle inscribed in circle [#permalink]  27 Sep 2009, 22:08
deepakraam wrote:
Economist wrote:
If points A , B , and C lie on a circle of radius 1, what is the area of triangle ABC ?

1. AB^2 = BC^2 + AC^2
2. \angle CAB equals 30 degrees

What is the purpose of radius = 1?We haven't used it anywhere in the problem.

That gives us the length of base of triangle. Remember, the diameter forms a right angle triangle in the circle? So diameter = base of the triangle, which will be used to calculate the area of it.
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Re: Triangle inscribed in circle [#permalink]  28 Sep 2009, 06:26
Economist wrote:
If points A , B , and C lie on a circle of radius 1, what is the area of triangle ABC ?

1. AB^2 = BC^2 + AC^2
2. \angle CAB equals 30 degrees

1. AB^2 = BC^2 + AC^2
2. \angle CAB equals 30 degrees
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Re: Triangle inscribed in circle [#permalink]  28 Sep 2009, 08:14
hgp2k wrote:
maratikus wrote:
hgp2k wrote:

Try again.

Whats the catch?

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Re: Triangle inscribed in circle [#permalink]  28 Sep 2009, 08:36
maratikus wrote:

1. AB^2 = BC^2 + AC^2
2. \angle CAB equals 30 degrees

It is nowhere indicated in question that it is a right angle triangle or one of the sides of triangle is diameter.

maratikus wrote:

State 1: From the first statement we just come to know that line AB = diameter of the circle. But we still don't know anything about line BC. It is not possible to find the length of this line using given information. So insufficient.
State 2: We just know that the angle opposite to line BC = 30. But we do not have any additional information to find the length of line BC. So insufficient.

Together we can derive that the \angle ACB equals 90 degrees, and \angle CAB equals 30 degrees.
So we can derive that BC = AB/2.

Answer could have been B, if the question were like this: A, B, C lie on a circle of radius 1, where points A and B are two ends of the diameter. What is the length of BC?

Please correct me if I am wrong, or missing something.
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Re: Triangle inscribed in circle [#permalink]  28 Sep 2009, 12:44
You are missing something. B is the correct answer.
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Re: Triangle inscribed in circle [#permalink]  28 Sep 2009, 12:46
maratikus wrote:
You are missing something. B is the correct answer.

Could you please explain how B is correct? I am not able to understand
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Re: Triangle inscribed in circle [#permalink]  28 Sep 2009, 13:00
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hgp2k wrote:
maratikus wrote:
You are missing something. B is the correct answer.

Could you please explain how B is correct? I am not able to understand

Based on the sine theorem, BC/sin(BAC)=2*R -> BC = sin(30 degrees)*2*R = (1/2)*2*1=1
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Re: Triangle inscribed in circle [#permalink]  28 Sep 2009, 14:38
maratikus wrote:
hgp2k wrote:
maratikus wrote:
You are missing something. B is the correct answer.

Could you please explain how B is correct? I am not able to understand

Based on the sine theorem, BC/sin(BAC)=2*R -> BC = sin(30 degrees)*2*R = (1/2)*2*1=1

You are assuming that this is a right angle triangle. Question does not indicate so. Please recheck.
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Re: Triangle inscribed in circle [#permalink]  29 Sep 2009, 04:35
hgp2k wrote:
You are assuming that this is a right angle triangle. Question does not indicate so. Please recheck.

I'm not making that assumption.
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Re: Triangle inscribed in circle [#permalink]  29 Sep 2009, 08:16
maratikus wrote:
hgp2k wrote:
You are assuming that this is a right angle triangle. Question does not indicate so. Please recheck.

I'm not making that assumption.

OK, GOT IT!!!! I am such a DUMB person I did not first understand that you are using the sine rule. I am in total agreement now that Statement 2 is sufficient. Thanks a lot for reminding me of that rule.
+1 to you maratikus
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Re: Triangle inscribed in circle [#permalink]  29 Sep 2009, 09:22
I hope that complicated formulas are not tested in GMAT.
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Re: Triangle inscribed in circle [#permalink]  02 Oct 2009, 15:51
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maratikus wrote:
hgp2k wrote:
You are assuming that this is a right angle triangle. Question does not indicate so. Please recheck.

I'm not making that assumption.

Sure, no such assumption was made. But to make clear that the answer to maratikus q is B, no sine theorem is needed:

Assume that O is the center of circle, so if BAC=30 degrees --> BOC=60 degrees, BO=OC=r and triangle BOC is equilateral, BOC=OBC=OCB=60 degrees, BC=r=1
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Re: Triangle inscribed in circle   [#permalink] 02 Oct 2009, 15:51
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