An equilateral triangle is inscribed in a circle. If the

Question

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

Bunuel · Accepted Answer

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.

jlgdr · Answer

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

kundankshrivastava · Answer

gmatcircle.jpg 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

shashankism · Answer

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

mdanieletto · Answer

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

BrushMyQuant · Answer

temp-2.JPG Side of an equilateral triangle (a) inscribed in a circle with radius r is given by a = r√𝟑 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 https://www.youtube.com/watch?v=PwhV06rECYI

MBAHOUSE · Answer

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.

bumpbot · Answer

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An equilateral triangle is inscribed in a circle. If the

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