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# An equilateral triangle that has an area of 9*root(3)

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An equilateral triangle that has an area of 9*root(3) [#permalink]

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25 Nov 2010, 02:11
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An equilateral triangle that has an area of $$9\sqrt{3}$$ is inscribed in a circle. What is the area of the circle?

A. $$6\pi$$
B. $$9\pi$$
C. $$12\pi$$
D. $$9\pi \sqrt{3}$$
E. $$18\pi \sqrt{3}$$

[Reveal] Spoiler:
This is how I solved it area of equilateral triangle =Square root 3/4*a^2=9 square root 3
we get a =6
know to calculate radius of an equlateral triangle in an inscribed circle we can use formulae

r=a*square root 3/6
with this I get r=square root 3 and then area =3 pi
but its not in the answer choices
OA is something else ....
[Reveal] Spoiler: OA
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An equilateral triangle that has an area of 9*root(3) [#permalink]

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25 Nov 2010, 02:30
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rite2deepti wrote:
An equilateral triangle that has an area of $$9\sqrt{3}$$ is inscribed in a circle. What is the area of the circle?

A. $$6\pi$$
B. $$9\pi$$
C. $$12\pi$$
D. $$9\pi \sqrt{3}$$
E. $$18\pi \sqrt{3}$$

This is how I solved it area of equilateral triangle =Square root 3/4*a^2=9 square root 3
we get a =6
know to calculate radius of an equlateral triangle in an inscribed circle we can use formulae

r=a*square root 3/6
with this I get r=square root 3 and then area =3 pi
but its not in the answer choices
OA is something else ....

I'd recommend to know the ways some basic formulas can be derived in geometry rather than memorizing them.

Anyway, $$area_{equilateral}=a^2*\frac{\sqrt{3}}{4}$$, where $$a$$ is the length of a side --> as given that $$area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=9\sqrt{3}$$ then $$a=6$$;

Now, given that this triangle is inscribed in circle (not circle is inscribed in triangle). The radius of the circumscribed circle is $$R=a*\frac{\sqrt{3}}{3}=2\sqrt{3}$$ (you used the formula for the radius of the inscribed circle $$r=a*\frac{\sqrt{3}}{6}$$) --> $$area_{circle}=\pi{R^2}=12\pi$$.

Check Triangles chapter of Math Book for more: math-triangles-87197.html

Hope it helps.
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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27 Nov 2010, 18:50
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Use the formula to calculate area of the inscribed circle to the equliateral triangle. Bunuel's guide on Triangles has all the important formulae in it..
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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15 May 2012, 21:07
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monir6000 wrote:
An equilateral triangle that has an area of 9 root 3 inscribed in a circle. What is the area of the circle?

1. 6Pi
2. 9 Pi
3. 12Pi
4. 9Pi*3^1/2
5. 18Pi*3^1/2

Hi Moneer

First determine the side of Eq triangle by formula area of eq T= (Sqrt3/4)* (side)^2

This gives side of eq T= 6

Also The radius of the circumscribed circle is R= (Sqrt3/3)*(Side)

So R= 2 Sqrt 3

Area of circle= Pie*R^2 = 12 Pie

Hope this clrifies

Best
Vaibhav
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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15 May 2012, 21:15
1
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monir6000 wrote:
An equilateral triangle that has an area of 9 root 3 inscribed in a circle. What is the area of the circle?

1. 6Pi
2. 9 Pi
3. 12Pi
4. 9Pi*3^1/2
5. 18Pi*3^1/2

Hi Monir

First determine the side of Eq triangle by formula area of eq T= (Sqrt3/4)* (side)^2

This gives side of eq T= 6

Also The radius of the circumscribed circle is R= (Sqrt3/3)*(Side)

So R= 2 Sqrt 3

Area of circle= Pie*R^2 = 12 Pie

Hope this clrifies

Best
Vaibhav
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Best
Vaibhav

If you found my contribution helpful, please click the +1 Kudos button on the left, Thanks

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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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15 May 2012, 21:28
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Area of the triangle = a^3/4R where a is a side of the triangle and R is the radius of the circle

sqrt(3)(side^2)/4 = 9*sqrt(3)
=> side = a = 6

9*sqrt(3) = 216/4R
=> R = 6/sqrt(3)

So area of the circle = pi*36/3 = 12pi

Option 3
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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18 Dec 2012, 04:32
Formula
Triangle inscribed in a circle.
area of the triangle inscribed in a circle with radius "r" = (abc)/4r
a,b,c - are the sides of a triangle inscribed
r - is the radius of the circle.

Circle inscribed in a Triangle
are of the triangle = (a+b+c)r/2
a,b,c - sides of a triangle
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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13 May 2013, 00:03
rite2deepti wrote:
An equilateral triangle that has an area of $$9\sqrt{3}$$ 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 is how I solved it area of equilateral triangle =Square root 3/4*a^2=9 square root 3
we get a =6
know to calculate radius of an equlateral triangle in an inscribed circle we can use formulae

r=a*square root 3/6
with this I get r=square root 3 and then area =3 pi
but its not in the answer choices
OA is something else ....

the above calculation is correct.
since 9pi3^1/2= 9pi/3=3pi
so the correct answer is D.
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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13 May 2013, 01:27
rite2deepti wrote:
An equilateral triangle that has an area of $$9\sqrt{3}$$ 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

using the area of the triangle we can get the side of the equilateral triangle using area = (\sqrt{3}/4) x a^2
We can get the side as 6.

now we have the sine formula for the triangle. ie a/sinA = 2R. where R the circumradius of the triangle. a is the side and A is the corresponding angle.

We have A = 60 a = 6 we can get R.which will be 2\sqrt{3}. So the area of the circle will be 12 pi
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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13 May 2013, 01:28
khosru wrote:
rite2deepti wrote:
An equilateral triangle that has an area of $$9\sqrt{3}$$ 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 is how I solved it area of equilateral triangle =Square root 3/4*a^2=9 square root 3
we get a =6
know to calculate radius of an equlateral triangle in an inscribed circle we can use formulae

r=a*square root 3/6
with this I get r=square root 3 and then area =3 pi
but its not in the answer choices
OA is something else ....

the above calculation is correct.
since 9pi3^1/2= 9pi/3=3pi
so the correct answer is D.

from the formula of area of triangle you can calculate the side of triangle ie 6.
Now the radius of circumcircle (for equilateral triangle) is side/root(3)
So substitute to get the final answer ie C

Consider kudos if my post helps!!!!!!!!!!!!!1

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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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28 Jul 2013, 23:37
rite2deepti wrote:
An equilateral triangle that has an area of $$9\sqrt{3}$$ is inscribed in a circle.

Does "inscribed" mean all the edges of the triangle are just touching the circle or can it mean that the trianlge lies completely inside the circle (meaning the edges need not touch the circle and can range from very small size to size that exactly fits in the circle).

In some questions I have seen that it is necessary to assume that inscribed means the triangle completely lies inside (and not necessarily have the edges touch the circle) while for this question is means a perfectly inscribed triangle.
Same doubt applies for circumscribed as well.
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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28 Jul 2013, 23:44
Jaisri wrote:
rite2deepti wrote:
An equilateral triangle that has an area of $$9\sqrt{3}$$ is inscribed in a circle.

Does "inscribed" mean all the edges of the triangle are just touching the circle or can it mean that the trianlge lies completely inside the circle (meaning the edges need not touch the circle and can range from very small size to size that exactly fits in the circle).

In some questions I have seen that it is necessary to assume that inscribed means the triangle completely lies inside (and not necessarily have the edges touch the circle) while for this question is means a perfectly inscribed triangle.
Same doubt applies for circumscribed as well.

A triangle inscribed in a circle does NOT mean that it is simply "inside" the circle, it means that the triangle's vertices are on the circumference of the circle.
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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29 Jul 2013, 08:21
Bunuel wrote:
A triangle inscribed in a circle does NOT mean that it is simply "inside" the circle, it means that the triangle's vertices are on the circumference of the circle.

Thanks for the clarification Bunuel! Seeing the explanation " The lower limit of the perimeter of an inscribed triangle in a circle of ANY radius is 0: P>0." in http://gmatclub.com/forum/which-of-the-following-can-be-a-perimeter-of-a-triangle-68310.html, I mistakenly assumed the triangle has to just lie inside the circle. I now understand that the triangle's edges should touch the circle and have perimeter >0.
Thanks again!
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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12 Jul 2014, 12:00
Hi Bunnel,

My logic-- side of equi triangle --6

now, the formula for area of a inscribed trianle is (p.r)/2=9sqrrt(3)

p=6+6+6=18

therefore r=sqrt(3)

there area(circle)=pi.3

is this incorrect?
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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12 Jul 2014, 12:26
manish2014 wrote:
Hi Bunnel,

My logic-- side of equi triangle --6

now, the formula for area of a inscribed trianle is (p.r)/2=9sqrrt(3)

p=6+6+6=18

therefore r=sqrt(3)

there area(circle)=pi.3

is this incorrect?

Since your answer does not match any of the options, then obviously you are doing something wrong there. Also, I don't know what kind of formula you are using in your solution...
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Re: An equilateral triangle that has an area of 9*root(3) [#permalink]

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Re: An equilateral triangle that has an area of 9*root(3)   [#permalink] 26 Sep 2015, 01:28
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