fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 7)
In the figure shown, AB = AC = 12. Which of the following is closest to the area of the circle with center O that is tangent, at points M, N, and P, to the circular sector ABPC with center A and central angle of 60 degrees?
(A) 40.1
(B) 43.7
(C) 46.8
(D) 50.2
(E) 53.7
AOP passes through the center of the inscribed circle and thus must BISECT sector ACB.
AB, AP and AC are all radii of sector ACB and thus are equal.
The prompt indicates AB=AC=12.
Thus:
AP=12.
As shown in the figure, AOM is a 30-60-90 triangle
Since OM is a radius of the circle, let OM = r.
In a 30-60-90 triangle, the hypotenuse is twice the shorter leg.
OM is opposite the 30-degree angle in triangle AOM and thus is the shorter leg.
Since OM=r, hypotenuse AO=2r, with the result that AP = 2r+r = 3r.
Since AP=12, we get:
3r = 12
r = 4.
Thus, the area of the inscribed circle = πr² = π(4²) ≈ 50.
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