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M20-07

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M20-07 [#permalink]

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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
[Reveal] Spoiler: OA

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M20-07 [#permalink]

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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.


Answer: C
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Re: M20-07 [#permalink]

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New post 28 Jan 2015, 17:24
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?

Thanks in advance.

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.


Answer: C
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Re: M20-07 [#permalink]

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New post 29 Jan 2015, 04:28
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?

Thanks in advance.

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.


Answer: C



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

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Re M20-07 [#permalink]

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New post 18 Jan 2017, 07:33
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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Re: M20-07 [#permalink]

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New post 03 Aug 2017, 10:46
I marked it wrong because statement 2 said angle CAB = 30 which conflict with the statement 1
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Re: M20-07 [#permalink]

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New post 18 Feb 2018, 08:34
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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Re: M20-07 [#permalink]

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New post 18 Feb 2018, 08:40
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M20-07   [#permalink] 18 Feb 2018, 08:40
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