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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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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) \(9z^2-\pi y=0\)

B) \(3z^2-\pi y=0\)

C) \(\pi z^2-3y=0\)

D) \(\pi z^2-9y=0\)

E) \(\pi z^2-27y=0\)
[Reveal] Spoiler: OA

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

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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: \(\pi{z^2}-27y=9a^2\pi-9a^2\pi=0\).

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

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New post 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
Cheers
J :)

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

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Attachment:
File comment: circle and equilateral
gmatcircle.jpg
gmatcircle.jpg [ 17.44 KiB | Viewed 5210 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

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

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New post 30 Nov 2017, 10:53
Can I have better explanation of this question

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Re: An equilateral triangle is inscribed in a circle. If the   [#permalink] 30 Nov 2017, 10:53
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