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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 =AC^2+BC^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.

If points A, B, and C lie on a circle of radius 1, what is the area of triangle ABC?

1. AB^2 =AC^2+BC^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.

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.

Attachment:

Ques2.jpg [ 7.77 KiB | Viewed 4586 times ]

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. _________________

Except for the rt angled triangle the geometry cannot be defined with two parameters. First assume S1 right angled -one parameter (hyp is known). The other parameter is S2 (one angle is known)

If points A, B, and C lie on a circle of radius 1, what is the area of triangle ABC?

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

my take is A as by having 1 AB^2 =AC^2+BC^2 we are clear that it will be a right angle triangle with right angle at C and then AB will be the diameter = 2 and then we know all the angles hence can find out the area

while with statement 2. Angle CAB equals 30 degrees there are various possibilities of triangles with different hieghts and diffferent bases

please clarify _________________

WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html do not hesitate me giving kudos if you like my post.

Re: If points A, B, and C lie on a circle of radius 1, what is [#permalink]

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27 Dec 2013, 16:02

zaur2010 wrote:

Quote:

If points A, B, and C lie on a circle of radius 1, what is the area of triangle ABC?

1. AB^2 =AC^2+BC^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.

I think I may have made a mistake

See, from Statement 1 we have that ABC is a right triangle, since it is inscribed in the circle then of course hypothenuse = diameter = 2

So then how can we find the area. Well can't we extend a height perpendicular to the diameter which will in fact be the radius = 1 to find it. With base and height we could have the area

Would anybody be so kind to explain why this reasoning is wrong?

Re: If points A, B, and C lie on a circle of radius 1, what is [#permalink]

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27 Dec 2013, 19:47

Jlgdr,Warlock, I think you are considering the triangle as isosceles triangle with diameter AB. But the point C can be very near to say Point A or B and still be a right angle and statement 1 would be true. But the height ( perpendicular on AB from C ) would change and thus the area.

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.

Attachment:

Triangle 2.jpg

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.

(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.

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.

Attachment:

Triangle 2.jpg

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. _________________

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

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01 May 2014, 22:21

Bunuel wrote:

qlx wrote:

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.

Attachment:

Triangle 2.jpg

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.

(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.

Hi Bunuel,

In the figure it is given that BD is the diameter. So if from statement 1, if AB is also a diameter, then BD and AB should be intersecting in the middle. However in the figure it is not shown in this way.

Can we still go ahead and think AB as diameter? Pls clarify. _________________

_________________________________ Consider Kudos if helpful

Last edited by 12Sums on 01 Apr 2015, 04:37, edited 1 time in total.

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.

Attachment:

Triangle 2.jpg

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.

(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.

Hi Bunuel,

In the figure it is given that BD is the diameter. So if from statement 1, if AB is also a diameter, then BD and AB should be intersecting in the middle. However in the figure it is not shown in this way.

Can we still go ahead and think AB as diameter? Pls clarify.

Regards, Thoufique

Figure in the original post is not a part of the question, so don't refer to it while solving. _________________

Re: If points A, B, and C lie on a circle of radius 1, what is [#permalink]

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01 Apr 2015, 00:00

VeritasPrepKarishma wrote:

qlx wrote:

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.

Attachment:

Triangle 2.jpg

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?

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.

Attachment:

Triangle 2.jpg

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

You can consider AB as base but is it necessary that CO will be the height? The height has to be perpendicular to the base. Draw a right triangle ABC in a circle. You will see that CO needn't be perpendicular to the diameter/hypotenuse. Hence we cannot say that it will be the height. _________________

Re: If points A, B, and C lie on a circle of radius 1, what is [#permalink]

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01 Apr 2015, 05:10

Attachment:

Triangle 2.jpg

[/quote]

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.[/quote]

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[/quote]

You can consider AB as base but is it necessary that CO will be the height? The height has to be perpendicular to the base. Draw a right triangle ABC in a circle. You will see that CO needn't be perpendicular to the diameter/hypotenuse. Hence we cannot say that it will be the height.[/quote]

Re: If points A, B, and C lie on a circle of radius 1, what is [#permalink]

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27 Jun 2016, 02:05

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