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Goemetry - Circles 2!

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Goemetry - Circles 2!

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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by GMATGuruNY » Sun Sep 18, 2011 9:26 am
I posted a solution to a very similar problem here:

https://www.beatthegmat.com/triangle-ins ... 90961.html
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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=√3*Side/3
or
Side=3r/√3=√3*r=4√3

Perimeter=3*Side=3*4√3=12√3 Hence D
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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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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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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 :D
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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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