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For any triangle inscribed in a circle
K=abc/4R where K=area of triangle, abc are the three sides and R is the radius of the circle---------------(1)
Since this is an equilateral triangle abc=a^3
area of equilateral triangle= 9*(3^1/2)
thus a=6
putting in equation(1)
R=6/srqt 3
Thus area of triangle =pi (r^2)=12pi

puma
An equilatteral 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) 12pi
d) 9pi 3^1/2
e) 18pi 3^1/2
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ritula
For any triangle inscribed in a circle
K=abc/4R where K=area of triangle, abc are the three sides and R is the radius of the circle---------------(1)

Thanks This was new to me.
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Me too. I found the formula to compute the radius of a circle circumscribed around a triangle, but didn't find the other formula. I like the forumula to find the radius as it is simpler, even though it has more steps to answer the question once the formula is solved. I must also consider my own accuracy when answering these questions and sometimes, a very complicated formula (or one with many parts) leads to incorrect use of that formula.
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why is the radius of the cricle r= a/sqrt(3) where a is the side of the equilateral triangle?

why cant it be r=a/sqrt(2) ?
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puma
An equilatteral 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) 12pi
d) 9pi 3^1/2
e) 18pi 3^1/2


We have 18sqrt3 as B*H.

Now the small side of the triangle =x. Thus the height = xsqrt3. A large side = 2x.

thus 18sqrt3= 2x^2sqrt3 ---> 18=2x^2 --> x=3

Now the radius of the circle is a line that can be drawn from the point of the trangle to the center of the triangle.

This line is equal to 2*3/sqrt3 --> 6sqrt3/ --> 2sqrt3.

the area is now just pir^2

(2sqrt3)^2*pi --> 12pi

C
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puma
An equilatteral 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) 12pi
d) 9pi 3^1/2
e) 18pi 3^1/2

This problem is very similar to one that I posted a while ago and that was on my actual GMAT test.
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GMATBLACKBELT

We have 18sqrt3 as B*H.

Now the small side of the triangle =x. Thus the height = xsqrt3. A large side = 2x.

thus 18sqrt3= 2x^2sqrt3 ---> 18=2x^2 --> x=3

Now the radius of the circle is a line that can be drawn from the point of the trangle to the center of the triangle.

This line is equal to 2*3/sqrt3 --> 6sqrt3/ --> 2sqrt3.

the area is now just pir^2

(2sqrt3)^2*pi --> 12pi

C

hey i have a question..if we know the side a of the equilateral triangle..why cant we draw an issoceles triangle from the center of the cirle..now if we do that then r the radius of the cricle becomes a/sqrt(2).

i am drawing this..and i am begining to question why r=a/sqrt(3)
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fresinha12
GMATBLACKBELT

We have 18sqrt3 as B*H.

Now the small side of the triangle =x. Thus the height = xsqrt3. A large side = 2x.

thus 18sqrt3= 2x^2sqrt3 ---> 18=2x^2 --> x=3

Now the radius of the circle is a line that can be drawn from the point of the trangle to the center of the triangle.

This line is equal to 2*3/sqrt3 --> 6sqrt3/ --> 2sqrt3.

the area is now just pir^2

(2sqrt3)^2*pi --> 12pi

C

hey i have a question..if we know the side a of the equilateral triangle..why cant we draw an issoceles triangle from the center of the cirle..now if we do that then r the radius of the cricle becomes a/sqrt(2).

i am drawing this..and i am begining to question why r=a/sqrt(3)

B/c you assuming that iscoles triangle is a 45,45, 90.
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fresinha12
GMATBLACKBELT

We have 18sqrt3 as B*H.

Now the small side of the triangle =x. Thus the height = xsqrt3. A large side = 2x.

thus 18sqrt3= 2x^2sqrt3 ---> 18=2x^2 --> x=3

Now the radius of the circle is a line that can be drawn from the point of the trangle to the center of the triangle.

This line is equal to 2*3/sqrt3 --> 6sqrt3/ --> 2sqrt3.

the area is now just pir^2

(2sqrt3)^2*pi --> 12pi

C

hey i have a question..if we know the side a of the equilateral triangle..why cant we draw an issoceles triangle from the center of the cirle..now if we do that then r the radius of the cricle becomes a/sqrt(2).

i am drawing this..and i am begining to question why r=a/sqrt(3)

Actually, the formula for the Radius(R) is this:

\(R = a\frac{\sqrt{3}}{3}\) for an equilateral traingle inscribed inside of a circle.
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fresinha12
GMATBLACKBELT

We have 18sqrt3 as B*H.

Now the small side of the triangle =x. Thus the height = xsqrt3. A large side = 2x.

thus 18sqrt3= 2x^2sqrt3 ---> 18=2x^2 --> x=3

Now the radius of the circle is a line that can be drawn from the point of the trangle to the center of the triangle.

This line is equal to 2*3/sqrt3 --> 6sqrt3/ --> 2sqrt3.

the area is now just pir^2

(2sqrt3)^2*pi --> 12pi

C

hey i have a question..if we know the side a of the equilateral triangle..why cant we draw an issoceles triangle from the center of the cirle..now if we do that then r the radius of the cricle becomes a/sqrt(2).

i am drawing this..and i am begining to question why r=a/sqrt(3)

Actually, the formula for the Radius(R) is this:

\(R = a\frac{\sqrt{3}}{3}\) for an equilateral traingle inscribed inside of a circle.


yes..but its the same as a/sqrt(3)
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How is \(a\frac{\sqrt{3}}{3}\) the same as \(\frac{a}{\sqrt{3}}\)?
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How is \(a\frac{\sqrt{3}}{3}\) the same as \(\frac{a}{\sqrt{3}}\)?

good lord.. a^2*3/9=a^2/3 sqrt it now..a/sqrt(3)
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This board doesn't need people like you that are condescending. This is supposed to be a place we can all ask questions and get answers without judgment. Please edit your post so you don't continue to look like a jerk.

fresinha12
jallenmorris
How is \(a\frac{\sqrt{3}}{3}\) the same as \(\frac{a}{\sqrt{3}}\)?

good lord.. a^2*3/9=a^2/3 sqrt it now..a/sqrt(3)
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jallenmorris
This board doesn't need people like you that are condescending. This is supposed to be a place we can all ask questions and get answers without judgment. Please edit your post so you don't continue to look like a jerk.

fresinha12
jallenmorris
How is \(a\frac{\sqrt{3}}{3}\) the same as \(\frac{a}{\sqrt{3}}\)?

good lord.. a^2*3/9=a^2/3 sqrt it now..a/sqrt(3)

sorry didnt mean to sound condescending..
you have to realize not all of us are native speakers so what we type is not usually what intend to say ..
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I believe you knew exactly what you intended to say by "good lord". That's not somethign non-native english speakers pick up on accident. Don't hide behind ESL. If you're sorry, say it.

fresinha12
jallenmorris
How is \(a\frac{\sqrt{3}}{3}\) the same as \(\frac{a}{\sqrt{3}}\)?

good lord.. a^2*3/9=a^2/3 sqrt it now..a/sqrt(3)

sorry didnt mean to sound condescending..
you have to realize not all of us are native speakers so what we type is not usually what intend to say ..[/quote]
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i am not a native speaker..i am from india..just watch a lot of hollywood movies..

jallenmorris
I believe you knew exactly what you intended to say by "good lord". That's not somethign non-native english speakers pick up on accident. Don't hide behind ESL. If you're sorry, say it.

fresinha12
jallenmorris
How is \(a\frac{\sqrt{3}}{3}\) the same as \(\frac{a}{\sqrt{3}}\)?

good lord.. a^2*3/9=a^2/3 sqrt it now..a/sqrt(3)

sorry didnt mean to sound condescending..
you have to realize not all of us are native speakers so what we type is not usually what intend to say ..
[/quote]
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If you are in NYC as your information suggests, you know now that Hollywood is not at all reality here. Hope you enjoy NYC. I've never been, but want to go someday. Maybe to attend Columbia? Who knows.

J Allen Morris



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