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An equilatteral triangle that has an area of 9*(3^1/2) is

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An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   05 Jun 2008, 12:37
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   05 Jun 2008, 13:14
Come on now, give the explanation. Stating what your answer is doesn't help much.

OK. Figured it out now.

If you have an equilateral triangle inscribed inside a circle, the formula to find the Radius of the circle is:

Radius = \(a\frac{\sqrt{3}}{3}\) where is a is the length of 1 side of the equilateral triangle.

An equilateral triangle is made up of two 30-60-90 triangles, so the sides are in proportion of Base = 1, Height = \(\sqrt{3}\), Hypotnuse = 2.

So the equation is:
\(x * x\sqrt{3} = 9\sqrt{3}\)
\(x^2\sqrt{3} = 9\sqrt{3}\)
\(x^2 = 9\)
\(x = 3\)

Don't forget that this is only the base of 1 of the 30-60-90 triangles, so double it to get one complete side of the equilateral triangle. 1 side = 6.

Back to the equation to find the Radius of the circumscribed circle:

a = a side, so

\(6 * \frac{\sqrt{3}}{3}\) = \(\frac{6\sqrt{3}}{3}\) = \(\2\sqrt{3}\)

This is the radius so the area of the circle = \((2\sqrt{3})^2\pi = 4 * 3 = 12\pi\)


prasannar wrote:
C ->12 Pi
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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   05 Jun 2008, 13:57
Area of equilateral triangle = (3^1/2)/4 r^2 = 9(3^1/2)

=> r = 6........1
Also, 1/2* base*height = 9(3^1/2)
base = 6 from step 1

so height = 3(3^1/2)

for a circle with an inscribed equilateral traingle, the center of the circle divides the height in the ratio 2:1( the numerical value of the first part is the radius).
Property of circumcirle and circumradius.

so r = 2/3*3(3^1/2) = 2(3^1/2)

Area = pi(r^2)= pi.4*3 = 12.pi


C
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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 00:13
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 wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 00:35
ritula wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 05:57
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 08:16
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 08:28
puma wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 08:28
puma wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 08:35
GMATBLACKBELT wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 08:36
fresinha12 wrote:
GMATBLACKBELT wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 08:39
fresinha12 wrote:
GMATBLACKBELT wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 08:47
jallenmorris wrote:
fresinha12 wrote:
GMATBLACKBELT wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 09:03
How is \(a\frac{\sqrt{3}}{3}\) the same as \(\frac{a}{\sqrt{3}}\)?
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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 09:23
jallenmorris wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 10:02
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 wrote:
jallenmorris wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 10:29
jallenmorris wrote:
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 wrote:
jallenmorris wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 10:44
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 wrote:
jallenmorris wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 11:01
i am not a native speaker..i am from india..just watch a lot of hollywood movies..

jallenmorris wrote:
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 wrote:
jallenmorris wrote:
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink] New post   06 Jun 2008, 11:12
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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Re: An equilatteral triangle that has an area of 9*(3^1/2) is [#permalink]
06 Jun 2008, 11:12
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