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# M20-07

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:07
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65% (hard)

Question Stats:

45% (00:55) correct 55% (01:02) wrong based on 121 sessions

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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) $$AB^2 = BC^2 + AC^2$$

(2) $$\angle CAB$$ equals 30 degrees

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16 Sep 2014, 01:07
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Official Solution:

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

(1) $$AB^2 = BC^2 + AC^2$$. This means that triangle $$ABC$$ is a right triangle with $$AB$$ as hypotenuse, so $$area=\frac{BC*AC}{2}$$. 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 side, 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) $$\angle CAB=30$$. Hence we have 30°-60°-90° right triangle and as $$AB=hypotenuse=2$$ then the legs equal to 1 and $$\sqrt{3}$$ (in 30°-60°-90° right triangle the sides are always in the ratio $$1:\sqrt{3}:2$$). Therefore $$area=\frac{BC*AC}{2}=\frac{\sqrt{3}}{2}$$. Sufficient.

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28 Jan 2015, 18:24
Hi Bunuel,

I have a doubt here. Area of triangle is (base * Height)/2 . here Since AB is the hypotenuse it is 2 and I can use it as base. Now the height has to be the radius of the circle which is 1. The area will be 1/2*2*1 = 1. I marked A. Please explain why I am wrong?

Regards,
Arun

Bunuel wrote:
Official Solution:

(1) $$AB^2 = BC^2 + AC^2$$. This means that triangle $$ABC$$ is a right triangle with $$AB$$ as hypotenuse, so $$area=\frac{BC*AC}{2}$$. 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 side, 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) $$\angle CAB=30$$. Hence we have 30°-60°-90° right triangle and as $$AB=hypotenuse=2$$ then the legs equal to 1 and $$\sqrt{3}$$ (in 30°-60°-90° right triangle the sides are always in the ratio $$1:\sqrt{3}:2$$). Therefore $$area=\frac{BC*AC}{2}=\frac{\sqrt{3}}{2}$$. Sufficient.

Math Expert
Joined: 02 Sep 2009
Posts: 47025

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29 Jan 2015, 05:28
amariappan wrote:
Hi Bunuel,

I have a doubt here. Area of triangle is (base * Height)/2 . here Since AB is the hypotenuse it is 2 and I can use it as base. Now the height has to be the radius of the circle which is 1. The area will be 1/2*2*1 = 1. I marked A. Please explain why I am wrong?

Regards,
Arun

Bunuel wrote:
Official Solution:

(1) $$AB^2 = BC^2 + AC^2$$. This means that triangle $$ABC$$ is a right triangle with $$AB$$ as hypotenuse, so $$area=\frac{BC*AC}{2}$$. 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 side, 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) $$\angle CAB=30$$. Hence we have 30°-60°-90° right triangle and as $$AB=hypotenuse=2$$ then the legs equal to 1 and $$\sqrt{3}$$ (in 30°-60°-90° right triangle the sides are always in the ratio $$1:\sqrt{3}:2$$). Therefore $$area=\frac{BC*AC}{2}=\frac{\sqrt{3}}{2}$$. Sufficient.

The point is that the height from C is not necessarily the radius of the circle:
Attachment:
Untitled.png
You'd be right if it were right isosceles triangle.
>> !!!

You do not have the required permissions to view the files attached to this post.

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18 Jan 2017, 08:33
I think this is a high-quality question. Area of triangle-1/2*b*h. I understand that the 2 legs vale needed to solve the area, however- to calculate area - cant we use 1/2*(hypotenuse as base)*radius(height from point c to centre of circle )

For ex- if AB is hypotenuse and altitude is CO(O is centre )
in that case option 1 is sufficient.
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03 Aug 2017, 11:46
I marked it wrong because statement 2 said angle CAB = 30 which conflict with the statement 1
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18 Feb 2018, 09:34
Dear Bunuel,

does a triangle inscribed in a circle not always have a 90 degree angle? If this is true, isn't statement 2 sufficient, telling us that we are dealing with a 30-60-90 triangle?

Would appreciate guidance highly!

best
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Posts: 47025

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18 Feb 2018, 09:40
gmatc2018 wrote:
Dear Bunuel,

does a triangle inscribed in a circle not always have a 90 degree angle? If this is true, isn't statement 2 sufficient, telling us that we are dealing with a 30-60-90 triangle?

Would appreciate guidance highly!

best

A triangle inscribed in a circle will be a right angled only if one of its side coincides with the diameter of the circle.

If the diameter of a circle is also the triangle’s side, then that triangle is a right triangle (the reverse is also true: a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle.)
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Re: M20-07   [#permalink] 18 Feb 2018, 09:40
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# M20-07

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