equilateral triangle inscribed in circle.

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

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?

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...

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.

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

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

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.

I always do it zander21's way. Just use basics and ideas rather than memorizing formulas.

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

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.

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:

https://www.beatthegmat.com/gmat-set-7-q8-t282169.html