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In the figure shown, AB = AC = 12. Which of the following is

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In the figure shown, AB = AC = 12. Which of the following is

by fskilnik@GMATH » Thu Feb 21, 2019 1:55 pm

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GMATH practice exercise (Quant Class 7)

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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

Answer: [spoiler]____(D)__[/spoiler]
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circle in a sector

by GMATGuruNY » Thu Feb 21, 2019 4:11 pm
fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 7)

Image

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


Image

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.

The correct answer is D.
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In the figure shown, AB = AC = 12. Which of the following is

by fskilnik@GMATH » Fri Feb 22, 2019 6:28 am
fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 7)

Image

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
Image

$$?\,\, \cong \,\,\pi {r^2}$$
Point O is equidistant from the rays AM and AN, hence AO must be contained in the angle bisector of angle MAN = angle BAC.
$$\left\{ \matrix{
\,OM = r\,\,\,\left( {{{30}^ \circ }} \right) \hfill \cr
\Delta AOM\,\,\,\left( {{{30}^ \circ },{{60}^ \circ },{{90}^ \circ }} \right) \hfill \cr} \right.\,\,\,\, \Rightarrow \,\,\,\,AO = 2r$$
Point P also belongs to the ray AO (by symmetry in the arc BPC) and AP=AB=12. Therefore:
$$12 = AP = AO + OP = 3r\,\,\,\,\, \Rightarrow \,\,\,\,\,r = 4$$
$$?\,\, = \,\,16\pi \,\, \cong \,\,\left( {14 + 2} \right) \cdot {{22} \over 7}\,\, = \,\,44 + {{42 + 2} \over 7} = 50{2 \over 7}$$

The correct answer is (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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