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equilateral triangle inscribed in circle.

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by santa_dem » Wed Aug 13, 2008 7:44 am
the area of the triangle is

a^2*sqrt(3)/4=9*sqrt(3) => a =6, where a is the side of the triangle.

the radius of the circle is 2/3*a*sqrt(3)/2=a*sqrt(3)/3

square of the circle = a^2*(pi)/3=36*(pi)/3=12*pi
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by pepeprepa » Wed Aug 13, 2008 8:06 am
For "a" we can use Pythagore theorem and (b*H)/2
Do you use it ?

Could you explain more clearly how you find the radius?
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by sudhir3127 » Wed Aug 13, 2008 9:03 am
Hi pepeprepa.. this is how i do it.. i know its a bit difficult to remember the formulae but we have no options...

area of equilateral triangle is sqrt3/4*a^2

sqrt3/4*a^2 =9 rt 3

hence a^2 = 36, a =6

now we need to find the height

height of equilateral triangle is sqroot 3/2 * a

which is rt3/2*6 = 3sqrt3

now u shud know a rule ...

The centroid (i.e the centre of circumcircle divides this) height in 2 : 1 ratio

thus radius would be

2/3* 3 sqrt 3 = 2sqrt 3.

now area of a circle is pi* r^2

thus its pi*(2sqrt3)^2 = 12pi.

i hope its clear now..

do let me know if u have any doubts...
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by pepeprepa » Wed Aug 13, 2008 9:40 am
Thanks a lot for the knowledge sudhir.

I think you could have skipped the calculation of height.
Because the radius of a circumcircle with equilateral triangle inscribed is also
(one side)/sqrt(3)
So we have two formulas to find this radius with equilateral triangles and this one is more direct.
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by arorag » Wed Aug 13, 2008 3:28 pm
Here is another way to do this if you don't know centeriod formula

Once you side of triangle as 6. Draw a perpendicular from cebter of circle to one side of triangle, this will bisect the side of triangle.
Make triangle 90-60-30
get radius as 6/squat3
Area as 12 pie
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by rahul.s » Fri Jan 22, 2010 5:03 am
sudhir3127 wrote:Hi pepeprepa.. this is how i do it.. i know its a bit difficult to remember the formulae but we have no options...

area of equilateral triangle is sqrt3/4*a^2

sqrt3/4*a^2 =9 rt 3

hence a^2 = 36, a =6

now we need to find the height

height of equilateral triangle is sqroot 3/2 * a

which is rt3/2*6 = 3sqrt3

now u shud know a rule ...

The centroid (i.e the centre of circumcircle divides this) height in 2 : 1 ratio

thus radius would be

2/3* 3 sqrt 3 = 2sqrt 3.

now area of a circle is pi* r^2

thus its pi*(2sqrt3)^2 = 12pi.

i hope its clear now..

do let me know if u have any doubts...
Awesome stuff! Thank you :D
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by zander21 » Mon Sep 12, 2011 11:01 am
arorag wrote:Here is another way to do this if you don't know centeriod formula

Once you side of triangle as 6. Draw a perpendicular from cebter of circle to one side of triangle, this will bisect the side of triangle.
Make triangle 90-60-30
get radius as 6/squat3
Area as 12 pie
This is obviously the easiest way to find. The only formula you have to know is area of equiateral triangle (squareroot3/4 *(side^2)) which gives 6 as side of triangle. Then use 30-60-90 rules (x, x*squareroot3, 2x) to find radius which is 6/squareroot3. Anyway, ya I like this way.
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by MartyMurray » Sun Jul 20, 2014 2:56 pm
I always do it zander21's way. Just use basics and ideas rather than memorizing formulas.
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by GMATinsight » Mon Jul 21, 2014 6:49 am
Truly Basic methods are the best... However for those who are Quant Freaks...

There is an expression to calculate the radius of the circle encircling any Triangle

This is called Circumradious, R = (a x b x c)/ (4 x Area of Triangle)

where a, b and c are sides of Triangle

Area od Equilateral Triangle = (√3/4)Side^2 = 9√3 ==> Side^2 = 36 ==> Side = 6

Here R = (6x6x6)/(4x9√3) = 6/√3 = 2√3

Area of Circle = � R^2 = � (2√3)^2 = 12�

Answer: Option C
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by lutp44 » Thu Nov 05, 2015 5:08 am
Can someone please explain to me how we get the radius? In a 30-60-90 triangle, the side corresponding to 60 degree angle is 3. I don't understand how the radius is found to be 6sqrt3.
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by GMATGuruNY » Thu Nov 05, 2015 6:33 am
lutp44 wrote:Can someone please explain to me how we get the radius? In a 30-60-90 triangle, the side corresponding to 60 degree angle is 3. I don't understand how the radius is found to be 6sqrt3.
I posted an explanation here:

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