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A triangle with three equal sides is inscribed inside a : Problem Solving (PS)
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Re: A triangle with three equal sides is inscribed inside a [#permalink]
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



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]
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]
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]
fast & clean!
As always Bunuel surpassing :thumbup:
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Re: A triangle with three equal sides is inscribed inside a [#permalink]
Bunuel wrote:
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.


I get the above, but what I got caught up with here is whether we need to consider that the equilateral triangle's vertices touch a point on the circumference? Or does the above apply even if the equilateral triangle is placed at any random location.
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Re: A triangle with three equal sides is inscribed inside a [#permalink]
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CEdward wrote:
Bunuel wrote:
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.


I get the above, but what I got caught up with here is whether we need to consider that the equilateral triangle's vertices touch a point on the circumference? Or does the above apply even if the equilateral triangle is placed at any random location.


A polygon is inscribed in a circle means that all of its vertices are on the circumference.
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Re: A triangle with three equal sides is inscribed inside a [#permalink]
For Equilateral triangle the radius of the circumscribed circle is R= side * root3/3, thus the area of the circle is pie r^2 = pie * side^2 /3
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AD990A79-8C60-40B8-8C98-C962C1DCE56C.jpeg
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Re: A triangle with three equal sides is inscribed inside a [#permalink]
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adityasuresh wrote:
For Equilateral triangle the radius of the circumscribed circle is R= side * root3/3, thus the area of the circle is pie r^2 = pie * side^2 /3


The system limits the number of kudos that can be given to a user to five per day. However, I would like to take this opportunity to express my appreciation for the effort you are putting into these Geometry questions. Your sketches are very neat and helpful. Thank you!
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Re: A triangle with three equal sides is inscribed inside a [#permalink]
Bunuel wrote:
adityasuresh wrote:
For Equilateral triangle the radius of the circumscribed circle is R= side * root3/3, thus the area of the circle is pie r^2 = pie * side^2 /3


The system limits the number of kudos that can be given to a user to five per day. However, I would like to take this opportunity to express my appreciation for the effort you are putting into these Geometry questions. Your sketches are very neat and helpful. Thank you!


Sir, you made my day!, geometry has been a huge pain point for me ever since high school!, have devoured the quant mega thread, that has been my go to source, along with quarter wit quarter wisdom blog.
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Re: A triangle with three equal sides is inscribed inside a [#permalink]
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