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Old 2007 October 17th, 05:47 PM   #1 (permalink)
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650 + questions geometry

The area of an equilateral triangle is 9.what is the area of it circumcircle.
A.10
π
B.12π
C.14π
D.16π
E.18π



Could anybody give me a step by step solution?

Thanks
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Old 2007 October 18th, 12:46 AM   #2 (permalink)
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is it 10 pi
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Old 2007 October 18th, 12:50 AM   #3 (permalink)
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"a" = length of a side of equilateral triangle

sqrt(3) / 4 * a^2 = 9 as per the question stem

=> a^2 = 12 * sqrt(3)

Radius of circumcricle of an equilateral triangle ( this is special case ) = a / sqrt(3)

=> R = a / sqrt(3)

=> R^2 = a^2 / 3 = 4 * sqrt(3)

Area of circumcircle = pi * R^2 = pi * 4 * sqrt(3)


I am not getting any of the above answer choices..Not sure where I am going wrong..

Last edited by tublai : 2007 October 18th at 03:17 AM.
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Old 2007 October 18th, 02:26 AM   #4 (permalink)
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Ken English can you explain your answer?

Tublai I don't get it at all. What does "a" mean? If "a" is area of the triangle we're given that.

I'm sorry but both of these are totally unhelpful.
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Old 2007 October 18th, 03:16 AM   #5 (permalink)
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Quote:
Originally Posted by notsofast View Post
Ken English can you explain your answer?

Tublai I don't get it at all. What does "a" mean? If "a" is area of the triangle we're given that.

I'm sorry but both of these are totally unhelpful.
I'm sorry notsofast..I forgot to mention what is "a"

a = side of equilateral triangle, then area of that equilateral triangle is sqrt(3) / 4 * a^2

I have edited my earlier post..
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Old 2007 November 3rd, 09:50 PM   #6 (permalink)
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My answer is 4*rt3*pi
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Old 2007 November 3rd, 09:56 PM   #7 (permalink)
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Here are some Equilateral Triangle properties :

For an
Equilateral Triangle with side 'a':

Circumscribed Circle Radius

Inscribed Circle Radius

Area


Perimeter


Altitude


Median
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Old 2007 November 4th, 05:32 PM   #8 (permalink)
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answer is 4 pi sqrt 3
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Old 2007 November 4th, 05:35 PM   #9 (permalink)
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you can solve it even if you dont know the formulaes

lets assume radius is r; side is a; ht is h

we can form the equations to arrive at the solution...

let me see if I can draw something and attach it here
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Old 2007 November 4th, 05:59 PM   #10 (permalink)
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yes you are right, all the above formulae can be derived easily
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