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by timothy.shaha » Thu Apr 11, 2013 3:05 pm
How can I prove that the radius of a circle inside an equilateral triangle is one third the altitude of the triangle? Please Help Me.....
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by Brent@GMATPrepNow » Thu Apr 11, 2013 3:54 pm
A proof? What are the 5 answer choices :-) (GMAT questions always have 5 answer choices).
Sorry, but proofs aren't part of the GMAT.

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by Lifetron » Thu Apr 11, 2013 9:22 pm
The radius of a circle inside an equilateral triangle is one third the altitude of the triangle
I think, it has got to do with in-radius and circum-radius. They will divide the altitude in the ratio 1:2. Not sure, which is what though.

Is the statement right ?

Atleast, we can use that for problems !
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by timothy.shaha » Sun Apr 14, 2013 2:15 am
At least thank you for your concern. I honestly did't know that you can't
that my question is irrelevant.
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by veenu08 » Thu Apr 18, 2013 4:54 am
Property of equilateral triangle-1.All centroid, circum centre, in center meet are concurrent
2. Centroid divide the median in 2:1 ratio

In equilateral triangle length of median(altitude)- (sqrt3)/2

Using point 2, the radius of incircle will be 1/3(altitude)

Hope this clears your doubt
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by Brent@GMATPrepNow » Thu Apr 18, 2013 8:37 am
timothy.shaha wrote:How can I prove that the radius of a circle inside an equilateral triangle is one third the altitude of the triangle? Please Help Me.....
In case any of this comes in handy, here's the proof:

Start with this.
Image

Draw line that bisects 60-degree angle to create 30-degree angle
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Draw line from center to point of tangency to create 90-degree angle
Image

IMPORTANT: This creates a special 30-60-90 triangle. So. let's let the radius of the circle have length 1
Image

This means that the hypotenuse of the 30-60-90 triangle will have length 2
Image

Finally, we can add another radius of length 1 to show that the equilateral triangle's altitude is 3 times the radius of the inscribed triangle.
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Cheers,
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