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

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

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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.

Originally posted by Economist on 27 Sep 2009, 08:00.
Last edited by Bunuel on 30 Jul 2013, 23:12, edited 1 time in total.
Added the OA.
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Re: Triangle inscribed in circle  [#permalink]

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New post 27 Sep 2009, 08:15
2
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

answer is C
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Re: Triangle inscribed in circle  [#permalink]

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New post 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]

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New post 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
Triangles.jpg [ 10.69 KiB | Viewed 9223 times ]


AD1<AD2
So Area of Triangle 1 < Area of Triangle 2
So statement 1 is insufficient.
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Re: Triangle inscribed in circle  [#permalink]

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New post 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]

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New post 27 Sep 2009, 21:08
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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]

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New post 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


How about this problem: A, B, C lie on a circle of radius 1, what is the length of BC.

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

The answer is . Why?
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Re: Triangle inscribed in circle  [#permalink]

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New post 28 Sep 2009, 07:36
maratikus wrote:
How about this problem: A, B, C lie on a circle of radius 1, what is the length of BC.

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:
The answer is B. Why?


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 is C.

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]

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New post 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 :oops:
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Re: Triangle inscribed in circle  [#permalink]

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New post 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 :oops:


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]

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New post 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 :oops:


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]

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New post 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]

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New post 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 :cry: 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 :beer
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Re: Triangle inscribed in circle  [#permalink]

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New post 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]

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New post 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]

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New post 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 :oops:


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]

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New post 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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New post 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]

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New post 30 Jul 2013, 23:12
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Points A, B, and C lie on a circle of radius 1. What is the area of triangle ABC?

(1) \(AB^2 = BC^2 + AC^2\) --> 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.

Answer: C.
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Re: Points A, B, and C lie on a circle of radius 1. What is the  [#permalink]

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New post 22 Feb 2018, 17:46
Economist wrote:
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.


Imo C

Excellent question
From 1 We only know that the triangle is a right triangle but to find area we have to know the two legs of the right triangle .
If one of the acute angle is not know the area can very as when the triangle is isosceles it will have different area than that of triangle with 3 different sides.

Clearly 2 is insufficient as we do know anything about the other two angle and sides.

Together we know the triangle is 30-60-90 and we can calculate the area
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Re: Points A, B, and C lie on a circle of radius 1. What is the &nbs [#permalink] 22 Feb 2018, 17:46
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