circle and triangle

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nidhis.1408: Master | Next Rank: 500 Posts; Posts: 135; Joined: Mon Oct 03, 2011 6:54 am; Followed by:4 members

circle and triangle

by nidhis.1408 » Wed Nov 28, 2012 2:05 pm

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?

1. 3/4pi
2. 3âˆš2/5pi
3. 3âˆš3/4pi
4. 5âˆš3/4pi
5. 3âˆš3/2pi

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Source: Beat The GMAT — Problem Solving | Wed Nov 28, 2012 2:05 pm

The Iceman: Master | Next Rank: 500 Posts; Posts: 194; Joined: Mon Oct 15, 2012 7:14 pm; Location: India; Thanked: 47 times; Followed by:6 members

by The Iceman » Wed Nov 28, 2012 11:33 pm

nidhis.1408 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?

1. 3/4pi
2. 3âˆš2/5pi
3. 3âˆš3/4pi
4. 5âˆš3/4pi
5. 3âˆš3/2pi

if r is the radius of the circle, then length of each side of the triangle is âˆš3r.

So the required probability = area of triangle / area of the circle = 3âˆš3/4pi

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Bill@VeritasPrep: GMAT Instructor; Posts: 1248; Joined: Thu Mar 29, 2012 2:57 pm; Location: Everywhere; Thanked: 503 times; Followed by:192 members; GMAT Score:780

by Bill@VeritasPrep » Wed Nov 28, 2012 11:41 pm

I assigned values to the sides of the triangles, then determined the area of the circle.

If the triangle's side length is 6, its area is 9sqrt3.

The circle must have a radius of 6/(sqrt3), and we can then find the area of 12pi.

P(point in triangle) = 9sqrt3/12pi = 3sqrt3/4pi.

I'll attach a diagram in a second.

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Bill@VeritasPrep: GMAT Instructor; Posts: 1248; Joined: Thu Mar 29, 2012 2:57 pm; Location: Everywhere; Thanked: 503 times; Followed by:192 members; GMAT Score:780