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# Points A, B, and C lie on a circle of radius 1. What is the

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Director
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Points A, B, and C lie on a circle of radius 1. What is the [#permalink]  27 Sep 2009, 08:00
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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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Jul 2013, 23:12, edited 1 time in total.
Manager
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Re: Triangle inscribed in circle [#permalink]  27 Sep 2009, 08:15
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

Director
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Re: Triangle inscribed in circle [#permalink]  27 Sep 2009, 11:01
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, 12:28
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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 3556 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, 20:58
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, 21:08
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, 05:26
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

[Reveal] Spoiler:
B
. Why?
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Re: Triangle inscribed in circle [#permalink]  28 Sep 2009, 07:36
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, 11:46
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, 12:00
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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, 13:38
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, 03:35
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, 07:16
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, 08:22
I hope that complicated formulas are not tested in GMAT.
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Re: Triangle inscribed in circle [#permalink]  02 Oct 2009, 14:51
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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]  03 Oct 2009, 05:12
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

Does GMAT testers expect the test takers to know SIN, COS and TAN formulas ?
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Re: If points A , B , and C lie on a circle of radius 1, what is [#permalink]  30 Jul 2013, 18:12
If AB^2=AC^2+BC^2 then AB is hypotenuse=any chord which "can" also be diameter
Also, AC=r=1 and BC=r=1 and area is 1*1/2

Hence, 1 alone sufficient.
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Re: If points A , B , and C lie on a circle of radius 1, what is [#permalink]  30 Jul 2013, 18:16
DO NOT step into trignometry to try to arrive at answers bcos GMAT questions are not built around it.
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Re: Points A, B, and C lie on a circle of radius 1. What is the [#permalink]  30 Jul 2013, 23:12
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Expert's post
Points A, B, and C lie on a circle of radius 1. What is the area of triangle ABC?

(1) $$AB^2 = BC^2 + AC^2$$ --> triangle ABC is a right triangle with AB as hypotenuse --> $$area=\frac{BC*AC}{2}$$. Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle (the reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle). So, hypotenuse AB=diameter=2*radius=2, but just knowing the length of the hypotenuse is not enough to calculate the legs of a right triangle thus we can not get the area. Not sufficient.

(2) $$\angle CAB$$ equals 30 degrees. Clearly insufficient.

(1)+(2) From (1) ABC is a right triangle and from (2) $$\angle CAB=30$$ --> we have 30°-60°-90° right triangle and as AB=hypotenuse=2 then the legs equal to 1 and $$\sqrt{3}$$ --> $$area=\frac{BC*AC}{2}=\frac{\sqrt{3}}{2}$$. Sufficient.

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Re: Points A, B, and C lie on a circle of radius 1. What is the [#permalink]  06 Oct 2014, 04:31
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