i also found this

Radius is equal to:
area of triangle/(1/2(triangle side 1 + triangle side 2 + triangle side 3)

with this i calculated radius =root 3

and the area was 3pi (which was not even an option)

no take on this ??

There are formuae somewhere i read :

Area of equilateral triangle=3^1/2/4 x (side)^2
from here, side=6 (putting all given values)

Also Height=3^1/2/2 X Side=>Height=3* 3^1/2
Finally , R (Circum radius)=2/3 * Height=2/3*3*3^1/2=2*3^1/2

Area of circle =Pi* R^2=Pi * (2*3^1/2)^2=12Pi

Is it the right ans..... else i am assuming formulae somewhere wrong.


Amit

hey

nice !!

Wasnt aware of that formula

it is the correct answer !

Vishui

thanks amitansu

Amit all the formulas you have written are correct

they can be derived using properties of equilateral triangle

also may find in std 10th/ 11th geometry section

You've lost me here..

"Finally , R (Circum radius)=2/3 * Height=2/3*3*3^1/2=2*3^1/2"

How did you get 2/3 x h? I've figured out the height and the length of each side of the triangle...but now, how do i find the radius? what is the relationship? how do u know this?

No thats the Formula !!

Vishu

what formula is that?

Relationship between Inradius (r), Circumradius(R) and area of triangle is as follows:

Area of triangle = Sqrt of (S(S-a)(S-b)(S-c)) = abc/4R = S*r

In the above case we know A=9 sqrt(3) and a=b=c=6 and hence R=abc/4A
= 6*6*6/(4*9sqrt(3)) = 6/sqrt(3). Hence Area of circle=pi * 36/3= 12 * pi

- Deepak

The correct anwser is 12Pi.

Regarding the formula R=2/3 x H

first, we know that H=sqrt(3)/2 x L
second, consider the triangle of the attached drawing (two blue dots and the center of the circle)

tan(30º)=Z / (L/2) <=> Z= sqrt(3) / 6 x L
but L= 2 x H /sqrt(3)

than, Z=1/3 x H

Also, H = 1/3 x H + R
therefore, R=2/3 x H

Hope it helps.
Rgs

How would we solve this problem if we did not know the equation Radius = (2/3)*height?

altitude from a vertex of a equilateral triangle bisects the base. if we join the sides of the triangle to centre, we a triangle OAB say, angle AOB will be 120 (2*60, as the angle at the center is twice that at the arc). the altitude from the centre the the base makes two 30-60-90 triangles, the base of each is 3 (half of 6 as altitude bisects the base)

using the property of the 30-60-90 with base as 3, we get radius as 2*3^1/2