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An equilateral triangle is inscribed in a circle. If the

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BN1989
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An equilateral triangle is inscribed in a circle. If the [#permalink] New post   11 Apr 2012, 07:58
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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) \(9z^2-\pi y=0\)

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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E
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Re: An equilateral triangle is inscribed in a circle. If the [#permalink] New post   11 Apr 2012, 08:20
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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: \(\pi{z^2}-27y=9a^2\pi-9a^2\pi=0\).

Answer: E.
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Re: An equilateral triangle is inscribed in a circle. If the [#permalink] New post   11 May 2014, 06:44
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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] New post   21 May 2014, 06:01
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Attachment:
File comment: circle and equilateral
gmatcircle.jpg
gmatcircle.jpg [ 17.44 KiB | Viewed 15333 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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An equilateral triangle is inscribed in a circle. If the [#permalink] New post   23 Apr 2018, 09:55
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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) \(9z^2-\pi y=0\)

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 = \(\pi {z^2}\)
\(\pi z^2-27y=0\)


Answer E
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An equilateral triangle is inscribed in a circle. If the [#permalink] New post   25 Jun 2020, 10:25
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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) \(9z^2-\pi y=0\)

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

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

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

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




We're dealing with 30-60-90 triangles, or rather 1:sqrt(3):2 triangles

STRATEGY: find "a" in function of "z" and plug that value into the answers.

Equilateral triangle side is z/3
Working with ratios btw similar 30-60-90 triangles we find that the radius "R" of the circle is z/(3*sqrt(3))

Therefore,
a=R^2*pi=((z^2)/27)*pi

Quickly plug this into each answer choice.
Only one equation turns true: CORRECT ANSWER is E
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An equilateral triangle is inscribed in a circle. If the [#permalink] New post   30 Nov 2021, 06:48
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Attachment:
temp-2.JPG
temp-2.JPG [ 30.92 KiB | Viewed 3942 times ]


Side of an equilateral triangle (a) inscribed in a circle with radius r is given by a = r√𝟑
[ Watch this video to know how ]

=> Perimeter of triangle = 3a = 3 * √𝟑 r = z (given) => r = \(\frac{z }{ 3 √𝟑}\)

Area of circle with radius r = \(\pi r^2\) = y (given)
=> \(\pi\) * \((\frac{z }{ 3 √𝟑})^2\) = y (substituting value of r from (1) )
=> \(\pi\) * \(\frac{z^2}{9*3}\) = y
=> \(\pi\) * \(\frac{z^2}{27}\) = y
=> \(\pi\) * \(z^2\) = 27 * y
=> \(\pi z^2-27y=0\)

So, Answer will be E
Hope it helps!

Watch the following video to Learn Properties of Equilateral Triangle inscribed in a Circle

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Re: An equilateral triangle is inscribed in a circle. If the [#permalink] New post   28 Jun 2022, 16:10
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The relation of the side of an equilateral triangle inscribed in a circumference with the radius of a circumference is given by s = r \sqrt{3}, where s is the side of the triangle.
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Re: An equilateral triangle is inscribed in a circle. If the [#permalink] New post   03 Jul 2023, 10:43
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