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

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

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 !!

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?

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



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

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.

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

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.


The previous answers in this forum tended for C as the correct answer. I've marked B not C and let me explain why

statement (1) suggests that there's a right triangle, BUT the angle sides might be different and the area of triangle might vary with these angle mesaures. E.g. when angles follow 45-45-90 the area of triangle would be 1, while with 30-60-90 the area of triangle is Sqrt(3)/2 Not Sufficient;

statement (2) Very interesting statement offering the inscribed angle measurement. If we find the angle CAB intercepted at the center, we get (30`)*2 OR 60`. Additionally, with the centrally intercepted angle we have the isosceles triangle with the base angles 60` which convert into the equilateral triangle, since all angles are 60` (BC=OC=OB). SO, side BC is equal to radius 1.

If we continue the line BO from the point O up-to the point D we receive height DC for the side BC of triangle ABC. Now we need to calculate the height which is easy by knowing triangle BCD is a right triangle and angle CBD=60`. So, DC is Sqrt(3). The area of triangle ABC using all these properties ---> base (BC)*height (CD)/2 = 1*Sqrt(3)/2, Sufficient as we can answer the questions area of triangle ABC=Sqrt(3)/2 therefore answer B.[/spoiler]

The answer is C.

Take a minute and think why the height of triangle ABC is CD.

The height of triangle ABC should AE where E is the point extended the line BC from C.

Ans is C.

Possible inscribed triangles with \(30^{\circ}\) angle.

All these triangles have different areas.

First of all, I think it's a great effort. It is always refreshing when people try to analyze from different perspectives. There was one error though... Look at the diagram below and figure out which of the following colorful altitudes could help you find the area of the triangle? They are all perpendicular to their respective bases.


I think you will agree that the purple line cannot be used as an altitude to find the area of this triangle... I hope this helps you in identifying your mistake.

if we draw a right triangle in a circle , does the triangle have to have Hypotenuse as the Diameter of the Circle?


From this question it does seem so.

Can we not have something as shown in the figure attached.

Yes, 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)

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.

Think about it logically - in your figure, say the hypotenuse is AC. Now arc AC subtends an inscribed angle of 90 degrees. So the central angle it subtends must be 180 degrees (since it is twice the inscribed angle). Angle of 180 degrees at the center means it is a straight angle and AC is the diameter. So no matter how you draw the figure. If you have made a right triangle in a circle, its hypotenuse will be the diameter.

Hi Karishma

1 doubt - maybe conceptual understanding

In Stmnt 1

I understand that AB is the Diameter and Angle C formed is Right angle.
Now Area of Triangle ABC = 1/2 base * height
If I consider base as AB (which is 2) and height as CO (O is centre) which makes CO = Radius = 1
This is sufficient. Isnt it?

Pls clarify . Thanks