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A circle is circumscribed about an equilateral triangle of

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qhoc0010
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A circle is circumscribed about an equilateral triangle of [#permalink] New post   07 Jan 2005, 07:05
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A
B
C
D
E

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A circle is circumscribed about an equilateral triangle of side 'a'. The diameter of the circle is

(A) a v3 (where v3 denotes 'root-3')

(B) a / v3

(C) 2a/ v3

(D) 2 v(3a)

(E) 3 v(a)

HIGHLIGHT BELOW FOR OA:
Actual Answer : C

Explanation
Let R be the circumradius, then,
Area of the triangle = Product of sides of a triangle/4R.
Here, side of equilateral triangle = a. Area of equilateral triangle = 3^½/4 a². Therefore a * a * a/4R = (3^½/4)a². R = a/3^½. Diameter = 2R = 2a/3^½.

Hence (C) is the correct answer






This question is from Crack GMAT. I think the OA is incorrect. I approach this problem using trig as below

tan(30) = R/(a/2)
1/(3^1/2) = R/(a/2)
R = a/(2*(3^1/2))
=> Diameter = 2R = a/(3^1/2)
So the answer should be (B)

Tell me what you think.

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riteshe
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[#permalink] New post   07 Jan 2005, 07:51
Answer is C. Here is the trigonometry solution.
Let R be the radius of the circumscribed circle.

RCos30 = a/2 => R = a/sqrt(3)

Therefore diameter = 2a/sqrt(3)
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[#permalink] New post   07 Jan 2005, 08:16
would go with C. (2a/sqrt(3))

if the circle would have been inscribed then Radius would have been
a/2 * sqrt(3)
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[#permalink] New post   07 Jan 2005, 08:17
the following should throw some ligh on how they are calculated.

https://mathforum.org/dr.math/faq/formul ... angle.html
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qhoc0010
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[#permalink] New post   10 Jan 2005, 09:06
vprabhala wrote:
would go with C. (2a/sqrt(3))

if the circle would have been inscribed then Radius would have been
a/2 * sqrt(3)


If R = a/2 * sqrt(3)
Then Diameter must be 2R = a/sqrt(3) which is (B)



Quote:
Answer is C. Here is the trigonometry solution.
Let R be the radius of the circumscribed circle.

RCos30 = a/2 => R = a/sqrt(3)

Therefore diameter = 2a/sqrt(3)

This RCos30 = a/2 cannot be true
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gmatclubot
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10 Jan 2005, 09:06
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