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stonecold
An equilateral triangle is inscribed in a circle of diameter D. If the length of the arc bounded by the adjacent corners of the triangle is between 4π and 6π, then which of the following could be the value of D?

A)6.5
B)9
C)11.9
D)15
E)23.5
\(?\,\,\,:\,\,\,D = 2r\,\,\underline {{\rm{could}}} \,\,{\rm{be}}\)

\(4\pi \,\,\, < \,\,\,{\rm{arc}}\,\,{\rm{length}}\,\,{\rm{ = }}\,\,{1 \over 3}\left( {2\pi r} \right)\,\,\, < \,\,\,6\pi \,\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,\pi \,\,\,{\rm{and}}\,\,\, \cdot \,\,3} \,\,\,\,\,4 \cdot 3\,\,\, < \,\,2r\,\, < \,\,6 \cdot 3\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left( D \right)\)


This solution follows the notations and rationale taught in the GMATH method.

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Hi All,

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:

GMAT assassins aren't born, they're made,
Rich
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stonecold
An equilateral triangle is inscribed in a circle of diameter D. If the length of the arc bounded by the adjacent corners of the triangle is between 4π and 6π, then which of the following could be the value of D?

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

step 1:
Sum of the 2 angles formed by a chord at the circumference in the two opposite segments of a circle is equal to 180 degrees.
60+x=180
ie., x=120
step 2:
length of an arc=(x/360)*2πr
Take length of arc=5π( as it is between 4π and 5π)
5π=(120/360)*2π*r
15π=2πr
15=2r=D

That's it...
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