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16 Sep 2014, 01:07
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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
(1) \(AB^2 = BC^2 + AC^2\)
(2) \(\angle CAB\) equals 30 degrees
16 Sep 2014, 01:07
Kudos
Bookmarks
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
(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
28 Jan 2015, 18:24
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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
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
