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

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Manager
Joined: 12 Oct 2011
Posts: 112
GMAT 1: 700 Q48 V37
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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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Difficulty:

85% (hard)

Question Stats:

57% (02:52) correct 43% (02:48) wrong based on 292 sessions

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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) $$\pi z^2-27y=0$$
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Joined: 02 Sep 2009
Posts: 51218
Re: An equilateral triangle is inscribed in a circle. If the  [#permalink]

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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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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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Joined: 14 May 2014
Posts: 41
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 7492 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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An equilateral triangle is inscribed in a circle. If the  [#permalink]

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23 Apr 2018, 08:55
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) $$9z^2-\pi y=0$$

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

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

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

E) $$\pi z^2-27y=0$$

Perimeter of equilateral triangle = z
So, side of equilateral triangle = z/3
Median of equilateral triangle = $$\frac{\sqrt{3}}{2} * (z/3)$$

Radius of circle = 2/3 * Median of equilateral triangle = 2/3 * $$\frac{\sqrt{3}}{2} * (z/3)$$ = $$\frac{z}{3 \sqrt{3}}$$
Area of circle = y = $$\pi {\frac{z}{3 \sqrt{3}}}^2$$ = $$\pi \frac{z^2}{27}$$
27y = $$\pi {z^2}$$
$$\pi z^2-27y=0$$

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An equilateral triangle is inscribed in a circle. If the &nbs [#permalink] 23 Apr 2018, 08:55
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