equilateral triangle pS

[vishubn]

An equilateral triangle that has an area of 9 3^1/2 is inscribed in a circle. What is the
area of the circle?
A . 6pi
B. 9pi
C. 12 pi
D. 9pi 3^1/2
E. 18pi 3^1/2

i was able to dril down to side of eaxch triangle beign 6 !! further/? i guess i am blacked out !

Any comments please??

[vishubn]

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)

[vishubn]

no take on this ??

[amitansu]

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

[vishubn]

hey

nice !!

Wasnt aware of that formula

it is the correct answer !

Vishui

[vishubn]

thanks amitansu

[mental]

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

[giatch]

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?

[vishubn]

No thats the Formula !!

Vishu

[giatch]

what formula is that?

[DeepakR]

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

[69aero69]

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

[tritrantran]

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

[scoobydooby]

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

[logitech]

Guys I have seen this same problem so many times over and over again..I carried the whole calculation and found out:

Area of Triangle / Area of Circle = 3xSQRT(3)/4Pi

In our example:

3xSQRT(3)/4Pi = 9SQRT(3)/Area of Circle

Area of Circle : 12 pi

In your face GMAT!!



Put it in a flash card ( if you have time and patience, you can drive the formula )