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A triangle with three equal sides is inscribed inside a

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A triangle with three equal sides is inscribed inside a [#permalink]

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A triangle with three equal sides is inscribed inside a circle. A point is selected at random inside the circle. What is the probability that the point selected is inside the triangle?

A) \(\frac{3}{4pi}\)
B) \(3\sqrt{2}\) / 5pi
C) \(3\sqrt{3}\) / 4pi
D) \(5\sqrt{3}\) / 4pi
E) \(3\sqrt{3}\) / 2pi

P.S.: Apologies for the poor formatting of the answer options. I am not able to format the options properly with the available GMATCLUB tags :(
[Reveal] Spoiler: OA

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New post 02 Oct 2013, 01:22
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TirthankarP wrote:
A triangle with three equal sides is inscribed inside a circle. A point is selected at random inside the circle. What is the probability that the point selected is inside the triangle?

A) \(\frac{3}{4pi}\)
B) \(3\sqrt{2}\) / 5pi
C) \(3\sqrt{3}\) / 4pi
D) \(5\sqrt{3}\) / 4pi
E) \(3\sqrt{3}\) / 2pi

P.S.: Apologies for the poor formatting of the answer options. I am not able to format the options properly with the available GMATCLUB tags :(


The area of equilateral triangle is \(area=side^2*\frac{\sqrt{3}}{4}\).

For equilateral triangle the radius of the circumscribed circle is \(R=side*\frac{\sqrt{3}}{3}\), thus the area of that circle is \(\pi{R^2}=\frac{\pi*side^2}{3}\).

P = (the area of triangle)/(area of circle) = \(\frac{3\sqrt{3}}{4\pi}\).

Answer: C.
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Re: A triangle with three equal sides is inscribed inside a [#permalink]

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New post 23 Feb 2015, 11:30
area of triangle in circumcsribed circle=abc/4r,
Eq triangle, so a=b=c.
Area of triangle will be = a^3/4r
If we draw the eq. triangle inside a circle, then a=2rcos30=sqrt3 r
Probability= Area of Triangle/ Area of Circle
= (3(sqrt3)(r^3)/4r)/(pi*r^2)
= 3(sqrt3)/4pi
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Re: A triangle with three equal sides is inscribed inside a [#permalink]

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New post 25 Jan 2016, 17:20
Can someone please tell me where I went wrong? I created this illustration to show my working.

The selected triangle has an area of 1, and you can make 6 such triangles in the large triangle

Image

Then the circle has an area of pi.r^2, which is simply 5.pi

The probability is therefore 6/(5.pi)

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Re: A triangle with three equal sides is inscribed inside a [#permalink]

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New post 22 Oct 2016, 09:41
Create your own diagram and assign a value you choose to serve as the side for the equilateral triangle.

Then, from the center of the circle, draw a line to the edge of the triangle and you will form a 30-60-90 triangle.

I chose 4 to serve as the value of a side of the triangle. The base of my triangle, opposite 60 degrees was 2, making the hypotenuse (i.e. the radius of the circle) [4sqrt(3)]/3

So... Area triangle = 4sqrt(3)
Area circle = 16/3

Answer = [4sqrt(3)]/(16/3) = 3sqrt(3)/4

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A triangle with three equal sides is inscribed inside a [#permalink]

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New post 25 Oct 2016, 00:09
Bunuel , how do you get \(\)"For equilateral triangle the radius of the circumscribed circle is R =side*\sqrt{3}/3, thus the area of that circle is πR2=π∗side23πR2=π∗side23.[/m]

you mind elaborating the steps in between? thank you !

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A triangle with three equal sides is inscribed inside a [#permalink]

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xnthic wrote:
Bunuel , how do you get \(\)"For equilateral triangle the radius of the circumscribed circle is R =side*\sqrt{3}/3, thus the area of that circle is πR2=π∗side23πR2=π∗side23.[/m]

you mind elaborating the steps in between? thank you !



Its an equilateral triangle, we can draw median/altitude from all the vertices which will meet at the center of the circle. That point is called centroid and it divides the length into 2:1 ratio.

So we know the length of median in an equilateral triangle is = \(\frac{Side}{2}\)*\sqrt{3}

Radius of the circle if you refer to the figure is the length OA which is 2 parts of the median:

= \(\frac{2}{3}\)*\(\frac{Side}{2}\)*\sqrt{3}

= \sqrt{3}*\(\frac{Side}{3}\)

HTH
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Re: A triangle with three equal sides is inscribed inside a   [#permalink] 06 Nov 2017, 07:59
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