myfish wrote:There are two concentric circles such that the area of the outer circle is four times the area of the inner circle. Let A, B, and C be three distinct points on the perimeter of the outer circle such that AB and AC are tangents to the inner circle. If the outer circle is 12 square centimeters, then the area of the triangle ABC is
A) PI Root12
B) 9/PI
C) 9Root3/PI
D) 6Root3/PI
E) 3Root3/PI
I get it to the point of the radius of the smaller circle. But from there to the area is incredible. Anyone?
P.S. I can get to the area of the circle with given lengts of the triangle but somehow not the other way around. THanks so much.
Outer circle:
Since �R² = 12, R = √(12/�) = 2√(3/�).
Inner circle:
Since the area is 1/4 the area of the outer circle, �r² = 3 and r = √(3/�).
The result is the following figure:
AO is the Radius of the outer circle.
OD is the radius of the inner circle.
A line tangent to a circle and a radius drawn to the point of tangency form a right angle.
Thus, ∆AOD is a right triangle.
Hypotenuse AO is twice the length of leg OD.
A right triangle whose hypotenuse is twice the length of a leg is a 30-60-90 triangle.
Thus, ∆AOD is a 30-60-90 triangle.
In a 30-60-90 triangle, the sides are proportioned x : x√3 : 2x.
Thus, AD = √3 * √(3/�) * = 3/√�.
Using the same reasoning for ∆OCD, we can deduce that CD = 3/√�.
Thus, AC = 2 * 3/√� = 6/√�.
If we continue with this reasoning, we'll get the following figure:
The figure indicates that ∆ABC is an equilateral triangle.
The formula for the area of an equilateral triangle = (s²√3)/4.
Since AC = 6/√�, the area of ∆ABC = (6/√�)² * √3/4 = 36/� * √3/4 = (9√3)/�.
The correct answer is
C.
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