A triangle with three equal sides is inscribed inside a

Question

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

Bunuel · Accepted Answer

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.

veergmat · Answer

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

lamode · Answer

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

https://i1248.photobucket.com/albums/hh489/top_photographer/circ.jpg

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

The probability is therefore 6/(5.pi)

law258 · Answer

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)  /3

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

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

xnthic · Answer

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.

you mind elaborating the steps in between? thank you !

gauravk · Answer

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

CEdward · Answer

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.

Bunuel · Answer

A polygon is inscribed in a circle means that all of its vertices are on the circumference.

adityasuresh · Answer

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

Bunuel · Answer

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!

adityasuresh · Answer

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.

Mirko92 · Answer

Are questions of this style still in the GMAT?

Bunuel · Answer

_____________________
No. Geometry no longer tested on the GMAT.

A triangle with three equal sides is inscribed inside a

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