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# An equilateral triangle is inscribed in a circle. If the

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An equilateral triangle is inscribed in a circle. If the [#permalink]

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11 Apr 2012, 06:58
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An equilateral triangle is inscribed in a circle. If the perimeter of the triangle is z inches and the area of the circle is y square inches, which of the following equations must be true?

A) 9z²-pi*y=0
B) 3z²-pi*y=0
C) pi*z²-3y=0
D) pi*z²-9y=0
E) pi*z²-27y=0
[Reveal] Spoiler: OA
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11 Apr 2012, 07:20
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BN1989 wrote:
An equilateral triangle is inscribed in a circle. If the perimeter of the triangle is z inches and the area of the circle is y square inches, which of the following equations must be true?

A) 9z²-pi*y=0
B) 3z²-pi*y=0
C) pi*z²-3y=0
D) pi*z²-9y=0
E) pi*z²-27y=0

We need to establish a relationship between the perimeter of the triangle and the area of the circle.

The radius of the circumscribed circle is $$R=a\frac{\sqrt{3}}{3}$$, where $$a$$ is the side of the inscribed equilateral triangle (check this for more: math-triangles-87197.html).

Now, the area of the circle is $$\pi{r^2}=\pi{\frac{a^2}{3}}=y$$ and the perimeter of the triangle is $$3a=z$$. Now, you can plug these values in answer choices to see which is correct.

Option E fits: $$\pi{z^2}-27y=9a^2\pi-9a^2\pi=0$$.

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Re: An equilateral triangle is inscribed in a circle. If the [#permalink]

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16 Jan 2014, 21:40
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Re: An equilateral triangle is inscribed in a circle. If the [#permalink]

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11 May 2014, 05:44
Perimeter of triangle is 3s = z
Area of circle is pi (r^2) = y

Now then, r = s (sqrt 3) / 3

Let us say that s=1.
Therefore z = 3

Now let's replace on area

pi ((s)(sqrt (3)) /3)^2 = y
Given s=1 then
y=pi / 3

Replacing in answer choices only E works

Hope it helps
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Re: An equilateral triangle is inscribed in a circle. If the [#permalink]

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21 May 2014, 05:01
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Attachment:
File comment: circle and equilateral

gmatcircle.jpg [ 17.44 KiB | Viewed 3664 times ]

Above image shows a pictorial view of the problem.

Perimeter of equilateral triangle = z
side of the equilateral triangle = z/3

A line from center to the Vertex A of the triangle will make 30 degree angle from base.
A perpendicular from center will bisect the base. Hence AP = z/6

Now , Cos 30 = sqrt(3)/2 = AP/OA

or , OA = 2* AP/sqrt(3)
Radius of circle = OA = 2*(z/6)/sqrt(3) = z/3 *sqrt(3)
Area of circly = y = pi* radius^2

y = pi (z/3*sqrt(3))^2

y = pi*z^2/27

27y = pi* z^2

pi*z^2 - 27y = 0

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Re: An equilateral triangle is inscribed in a circle. If the [#permalink]

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16 Feb 2016, 10:22
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Re: An equilateral triangle is inscribed in a circle. If the   [#permalink] 16 Feb 2016, 10:22
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