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Area of a triangle in a circle

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by infiniti007 » Sun Oct 25, 2015 8:22 pm
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If the circle has radius 6, what is the area of the triangle?

1.) AC = AB
2.) BC = 12

For Statement 2, is it possible to draw a perpendicular from the 90 degree angle to the center of the circle (midpoint of BC) and deduce that we have two similar right isosceles triangles?
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Source: — Data Sufficiency |

by MartyMurray » Sun Oct 25, 2015 8:49 pm
From Statement 1 you can obviously deduce that triangle ABC is isosceles, but although segment BC seems to be the diameter of the circle, you can't assume that and so you can't determine the area of the triangle from Statement 1.

From Statement 2 you can determine that BC is a diameter of the circle because the length of BC is twice the radius of the circle.

Regarding your question, you can draw a line segment that goes from A to BC and is perpendicular to BC and thus create two similar right triangles, but the segment will not necessarily intersect BC at the center of the circle, and the two triangles created will not necessarily be isosceles right triangles. The only way that the segment will intersect BC at the center of the circle and they will be isosceles is if A is at the midpoint of arc BC, which case would be the case if AB were equal to BC.
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by Matt@VeritasPrep » Sun Oct 25, 2015 10:23 pm
You can draw a line from <BAC to the center of the circle and create two isosceles triangles in the process, but you wouldn't know that line AO is a perpendicular UNLESS you knew that ∆ABC is itself isosceles. For instance, if ∆ABC isn't isosceles, your perpendicular could look like this:

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