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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 !!

Re: Triangle inscribed in circle [#permalink]
27 Sep 2009, 12:28

3

This post received KUDOS

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 2751 times ]

AD1<AD2 So Area of Triangle 1 < Area of Triangle 2 So statement 1 is insufficient.

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.

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. _____________________ Consider KUDOS for good post

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

Re: Triangle inscribed in circle [#permalink]
02 Oct 2009, 14:51

1

This post received KUDOS

Expert's post

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 _________________

Re: Points A, B, and C lie on a circle of radius 1. What is the [#permalink]
30 Jul 2013, 23:12

2

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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.

(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.