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kuristar
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A circle is circumscribed about an equilateral traingle of 04 May 2006, 09:27
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A circle is circumscribed about an equilateral traingle of side 'a'. The diameter of the circle is:

A. a v3 (where v3 denotes 'root-3'
B. a/v3
C. 2a/v3
D. 2v(3a)
E. 3v(a)

Please explain your solution.



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willget800
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A circle is circumscribed about an equilateral traingle of 04 May 2006, 10:04
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A...

some thing that is greater than the height of the equilateral triangle.. which is r > sqrt(3)*a/2
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kuristar
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A circle is circumscribed about an equilateral traingle of 04 May 2006, 10:51
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If we simply need a number that is larger than the height of the triangle, couldn't AC E also be the answer?
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mattflow
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A circle is circumscribed about an equilateral traingle of 04 May 2006, 11:35
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Area of triangle = 1/2*a*sqrt(3)/2*a = sqrt(3)/4*a^2
Area/3 = a^2/(4*sqrt(3)

we use the third of the area to find the height of the base of the triangle to the center of the triangle

a^2/(4*sqrt3) = 1/2*a*h

h = a/(2*sqrt(3))

since we know the height from the base to the center, and that one half of the base is a/2, then we can find the radius of the circle

(a/2*sqrt(3))^2 + (1/2*a)^2 = r^2

r = a/sqrt(3)
diameter = 2*r = 2a/sqrt(3)
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Loner
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A circle is circumscribed about an equilateral traingle of 05 May 2006, 03:24
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I think it is C as well. If you use 30 60 90 triangles after dropping a perpendicual from the vertex of any one side to the circumference.
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sm176811
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A circle is circumscribed about an equilateral traingle of 07 May 2006, 15:03
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Yeah its C..

I used Trig to solve


cos 30 = a/2*1/r


or sqrt(3)/2 =a/2*1/r

so r = a/sqrt(3)

Thus, diameter = 2a/sqrt(3)
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shevy
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A circle is circumscribed about an equilateral traingle of 07 May 2006, 15:03
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I agree with C :wink:
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trivikram
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A circle is circumscribed about an equilateral traingle of 07 May 2006, 16:36
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Can we do this way?

Theory of Apollonius

2*Median^2 + 2 (1/2*third side)^2 = Sum of squares of other two sides

Since the givevn triangle is equilateral lets suppose each side is 'a'.

2*Median^2 + 2(a/2)^2 = a^2 + a^2

Median^2 + (a/2)^2 = a^2

Median = sqrt(3)a/2

Since centroid divides median in 2:1 ratio circumcentre from each vertice is 2*sart(3)*a/3*2 which is the circum radius

So diameter = 2a/sqrt(3)



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