GMATPrepNow wrote:
The two lines are tangent to the circle. If AC = 10 and AB = 10√3, what is the area of the circle?
A) 100π
B) 150π
C) 200π
D) 250π
E) 300π
*kudos for all correct solutions
If AC = 10, then BC = 10
Since ABC is an isosceles triangle, the following gray line will create two right triangles...
Now focus on the following blue triangle. Its measurements have a lot in common with the BASE 30-60-90 special triangle
In fact, if we take the BASE 30-60-90 special triangle and multiply all sides by 5 we see that the sides are the same as the sides of the blue triangle.
So, we can now add in the 30-degree and 60-degree angles
Now add a point for the circle's center and draw a line to the point of tangency. The two lines will create a right triangle (circle property)
We can see that the missing angle is 60 degrees
Now create the following right triangle
We already know that one side has length 5√3
Since we have a 30-60-90 special triangle, we know that the hypotenuse is twice as long as the side opposite the 30-degree angle.
So, the hypotenuse must have length 10√3
In other words, the radius has length 10√3
What is the area of the circle? Area = πr²
= π(10√3)²
= π(10√3)(10√3)
= 300π
Answer: E
Cheers,
Brent
Is there any property that states that a perpendicular bisector from C to AB will also pass through the center O? Else we will not be able to confirm the length 5*(root 3) is the same in the two triangles. I guess I am missing a trick here..Thanks!