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# An equilateral triangle that has an area of 9 is inscribed

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Manager
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An equilateral triangle that has an area of 9 is inscribed [#permalink]  23 Dec 2005, 01:32
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An equilateral triangle that has an area of 9 is inscribed in a circle. What is the area of the circle?

A. 6 pie
B. 9 pie
C. 12 pie
D. 9 pie (root 3)
E. 18 pie (root 3)
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JAI HIND!

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How come I'm getting r^2 =4 sqrt 3

Area = 4 sqrt 3 pie

???????
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Re: PS: Area of a circle - good one! [#permalink]  23 Dec 2005, 04:36
JAI HIND wrote:
An equilateral triangle that has an area of 9 is inscribed in a circle. What is the area of the circle?

A. 6 pie
B. 9 pie
C. 12 pie
D. 9 pie (root 3)
E. 18 pie (root 3)

It is D. Here is my working.

Area of equivalateral trngle = s^2 (SQRT(3))/4
9 = s^2*SQRT(3)/4
s^2 = 36/Sqrt(3)
This is the square of the diamter so the radius is
9/Sqrt(3)
Area of cirlce = PI r^2 = 9 pie (root 3)
Director
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@krisni

have you deduced the area formula by yourself, I can't catch it.

I tried to express one site of the equilateral as the height so that, (s*h)/2=9.
s should therefore be 4h/sqrt3. But in the end I've exactly half of the area, namely 4,5*sqrt3*PI.

Maybe I understand my mistake if you can clear up your formula.
Manager
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D is not the OA! I will wait for a couple more tries before i post the OA and the explanation!
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JAI HIND!

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OK will give it another try...

Director
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Everything I get is 4,5 PI.

The relation of the sides of an isosceles triangle is 1:1:sqrt2.

So that we know that (x*(x*sqrt2))/2=6, which yields x=3/sqrt2. The y, namely one side of of the two equal sides is the radius, so that we yield finally 4,5*PI.

On test day I would choose 9*PI, since it is a multiple of my result, but for sure that's not recommendable.
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Manager
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I am getting 4 pie sqrt(3).

But the option is not in the choices.
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JAI HIND, are you sure there is no typo in the question?
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Re: PS: Area of a circle - good one! [#permalink]  23 Dec 2005, 19:50
area of equi. triangle = [sqrt(3) (s^2)] / (4)
9 = [sqrt(3) (s^2) / (4)
a side of equi. triangle, s = 6/3^(1/4)

to find the length of the equi. triangle,

(1/2) (L x s) = 9
(1/2) (L) 6/3^(1/4) = 9
L = (3) 3^(1/4)

now divide the equi triangle into 3 equal triangles.
the area of each of small triangles = 3
1/2 (l x s) = 3
1/2 (l) 6/3^(1/4) = 3
l = 3^(1/4). this is the height of each of the small triangles.

r = L - l =
r = (3) 3^(1/4) - 3^(1/4) = (2) 3^(1/4)

area of the circle = pi r^2 = (pi) (2) [(2) 3^(1/4)]^2 = 4 pi sqrt(3)

did i miss anything?
Manager
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The area of Equilateral Triangle = sqrt (3)/4*a^2 (where a= side of triangle)

sqrt (3)/4*a^2 = 9 ==> that a = 6/Fourthroot(3)

The radius of the circle = a/sqrt (3).........[(Height of triangle = sqrt (3)/2*a) and radius of circle is 2/3*Height of Triangle]

Therefore Area of circle = 4pie*sqrt (3)
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There must be a typo in the question, I got exactly what others do:
4*pi*sqrt3
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The area of an equilateral triangle is s^2*root(3)/2. We can find the value of side of equilateral traingle from the area given.
Once we find the side length we can use one of the properties of equilateral triangle to find the median of the triangle (which happens to be the radius of the circle).
The median length is s*root(3)/2
So the answer should be (E)
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--gregspirited

Manager
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gregspirited wrote:
The area of an equilateral triangle is s^2*root(3)/2. We can find the value of side of equilateral traingle from the area given.
Once we find the side length we can use one of the properties of equilateral triangle to find the median of the triangle (which happens to be the radius of the circle).
The median length is s*root(3)/2
So the answer should be (E)

gregspirited, you are few steps closer to the answer.
The radius of the circle is 2/3 of the Median length and the area of the circle results to 4*pi*sqrt3
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Re: PS: Area of a circle - good one! [#permalink]  24 Dec 2005, 12:41
I am confused with the relation between height of triangle and radius of the circle?

can some one explain it please!

Last edited by SunShine on 24 Dec 2005, 14:27, edited 1 time in total.
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I am not alone with the 4 pie sqrt(3)

JaiHind, Are you sure there is no typo in the question ?
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Re: PS: Area of a circle - good one! [#permalink]  24 Dec 2005, 13:44
krisrini,

can you please explain how did you get the bold red part?

thanks

It is D. Here is my working.

Area of equivalateral trngle = s^2 (SQRT(3))/4
9 = s^2*SQRT(3)/4
s^2 = 36/Sqrt(3)
This is the square of the diamter so the radius is
9/Sqrt(3)
Area of cirlce = PI r^2 = 9 pie (root 3)[/quote]
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Re: PS: Area of a circle - good one! [#permalink]  24 Dec 2005, 16:19
SunShine wrote:
krisrini, can you please explain how did you get the bold red part?
This is the square of the diamter so the radius is
9/Sqrt(3)
Area of cirlce = PI r^2 = 9 pie (root 3)

in equilateral triangle inscribed in a circle, the radius of the circle = s/sqrt(3)
so diameter = 2[s/sqrt(3)]
Re: PS: Area of a circle - good one!   [#permalink] 24 Dec 2005, 16:19
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# An equilateral triangle that has an area of 9 is inscribed

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