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Goemetry - Circles 2!

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varun7nurav: Senior | Next Rank: 100 Posts; Posts: 33; Joined: Fri Nov 19, 2010 8:07 am

by varun7nurav » Sun Sep 18, 2011 8:53 am

Dear guys, this is another problem. I am able to determine the answer but the method is pretty lengthy. Please do suggest any shorter way to get the solution.
Please find attached a snapshop of the problem.

Thank you in advance!

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Source: Beat The GMAT — Problem Solving | Sun Sep 18, 2011 8:53 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Sep 18, 2011 9:26 am

I posted a solution to a very similar problem here:

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knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Sun Sep 18, 2011 9:49 am

For equilateral triangle with side S inscribed in circle the radius r is given by

r=âˆš3*Side/3
or
Side=3r/âˆš3=âˆš3*r=4âˆš3

Perimeter=3*Side=3*4âˆš3=12âˆš3 Hence D

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cbaum: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Wed Jul 06, 2011 8:24 am

by cbaum » Mon Sep 19, 2011 2:15 pm

The way I would have done it is create an isosceles triangle within the equilateral triangle using the radius (the sides of this new triangle will go from the center of the circle to two different points of the equilateral triangle). Then I would have used the isosceles relationship 1:1:2^(1/2) to determine that the length of one side of the equilateral triangle is 4*2^(1/2). Multiply that by 3 (since we're looking for the perimeter) and you get the answer 12*2^(1/2), or C.

What's the OA?

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varun7nurav: Senior | Next Rank: 100 Posts; Posts: 33; Joined: Fri Nov 19, 2010 8:07 am

by varun7nurav » Mon Sep 19, 2011 8:27 pm

cbaum,
your approach of using the radius to get the side is right.Although, you do not get a 1-1-2^1/2 here as its NOT a 45-45-90 triangle. Instead if u split that isos triangle into 2 triangles by dropping a perpendicular, u get a 30-60-90 i.e. 1-3^(1/2)-3 triangle. From this we get that (side of the equilateral triangle/2) = 4root3. Answer is therefore 12root3.

Cheers!

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cbaum: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Wed Jul 06, 2011 8:24 am

by cbaum » Tue Sep 20, 2011 8:24 am

varun7nurav wrote:cbaum,
your approach of using the radius to get the side is right.Although, you do not get a 1-1-2^1/2 here as its NOT a 45-45-90 triangle. Instead if u split that isos triangle into 2 triangles by dropping a perpendicular, u get a 30-60-90 i.e. 1-3^(1/2)-3 triangle. From this we get that (side of the equilateral triangle/2) = 4root3. Answer is therefore 12root3.

Cheers!

Thanks! Realized shortly after I posted that with my way the circle would only equal 270 degrees

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sl750: Master | Next Rank: 500 Posts; Posts: 496; Joined: Tue Jun 07, 2011 5:34 amFollowed by: 1 member; Thanked: 38 times

by sl750 » Tue Sep 20, 2011 9:53 am

You should get a 30-30-120 triangle r:r:r*sqrt(3)

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