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In the figure above, an equilateral triangle is inscribed

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In the figure above, an equilateral triangle is inscribed

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In the figure above, an equilateral triangle is inscribed in a circle. If the arc bounded by adjacent corners of the triangle is between 4π and 6π long, which of the following could be the diameter of the circle?

(A) 6.5
(B) 9
(C) 11.9
(D) 15
(E) 23.5

OA=D

Source: Manhattan GMAT

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Hi VJesus12,

We're told that in the figure above, an EQUILATERAL triangle is inscribed in a circle and the arc bounded by adjacent corners of the triangle is between 4π and 6π long. We're asked which of the following COULD be the diameter of the circle. As scary as this question might look, it's based on a couple of standard Geometry rules, so you can answer it with just a little work.

To start, an equilateral triangle has 3 equal angles - and since the triangle is inscribed in the circle, each of the three 'arc pieces' is equal in length. Thus, the total of those three arcs (re: the circumference of the circle) is between (3)(4π) and (3)(6π). If the total circumference is between 12π and 18π, then we can 'work backwards' to find the possible diameter....

12π = 2π(R) = πD..... Diameter = 12
18π = 2π(R) = πD..... Diameter = 18

There's only one answer that's between 12 and 18...

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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VJesus12 wrote:


In the figure above, an equilateral triangle is inscribed in a circle. If the arc bounded by adjacent corners of the triangle is between 4π and 6π long, which of the following could be the diameter of the circle?

(A) 6.5
(B) 9
(C) 11.9
(D) 15
(E) 23.5

OA=D

Source: Manhattan GMAT
Since each arc is bounded by adjacent corners of the triangle representing 1/3 of the circumference, the range of values of the circumference is:

Minimum:

1/3(C) = 4π

C = 12π, so the diameter would be 12.

Maximum:

1/3(C) = 6π

C = 18π, so the diameter would be 18.

The diameter is between 12 and 18, so a possible diameter is 15.

Answer: D

_________________
Scott Woodbury-Stewart Founder and CEO

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