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??
equilateral triangle pS
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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)
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)
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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
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
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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
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
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
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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
using the property of the 30-60-90 with base as 3, we get radius as 2*3^1/2
- logitech
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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 )
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 )
LGTCH
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