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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*3^(1/2) [#permalink]

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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: OA

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Re: An equilateral triangle that has an area of 9*3^(1/2) [#permalink]

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New post 07 Jul 2010, 11:14
From the properties of equilateral triangles, we can determine that one side of the triangle is 6:

1 / 2 * [ 6 * 3√3 ] = 9√3

We need to find the radius - we must subtract something from 3√3 (the height of the triangle). The way to do this is to divide the triangle by 3 using corresponding radii. The area of these triangles would be 3√3. We can use the segment we need to subtract from 3√3 and 6 to determine the area of the newly formed triangles:

3√3 = 1/2 * 6 * x
3√3 = 3x
x = √3

Subtracting √3 from 3 would give us the radius:

3√3 - √3 = √3 ( 3 - 1 ) = 2√3

As such the circle's area is:

(2√3)^2 * π = 12 π

Hence C.

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Re: An equilateral triangle that has an area of 9*3^(1/2) [#permalink]

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New post 07 Jul 2010, 11:24
RGM wrote:
From the properties of equilateral triangles, we can determine that one side of the triangle is 6:

1 / 2 * [ 6 * 3√3 ] = 9√3

We need to find the radius - we must subtract something from 3√3 (the height of the triangle). The way to do this is to divide the triangle by 3 using corresponding radii. The area of these triangles would be 3√3. We can use the segment we need to subtract from 3√3 and 6 to determine the area of the newly formed triangles:

3√3 = 1/2 * 6 * x
3√3 = 3x
x = √3

Subtracting √3 from 3 would give us the radius:

3√3 - √3 = √3 ( 3 - 1 ) = 2√3

As such the circle's area is:

(2√3)^2 * π = 12 π

Hence C.


Couldnt get the red highlighted part!
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Re: An equilateral triangle that has an area of 9*3^(1/2) [#permalink]

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New post 07 Jul 2010, 11:36
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Given that the area of the equilateral triangle is \(9\sqrt{3}\).

Assuming the side of the triangle is a, we get:

\(\frac{1}{2}*\frac{\sqrt{3}}{2}*a = 9\sqrt{3}\)

Solving this we get \(a = 6\)

Now, in the figure, consider the shaded triangle. Clearly we can see that the area of this triangle is a third of the area of the whole triangle.
Attachment:
EqTriangle.JPG
EqTriangle.JPG [ 41.88 KiB | Viewed 11039 times ]


So we have \(\frac{1}{2}*h*\frac{a}{2} = \frac{1}{3}*9\sqrt{3}\) so from here we get \(h=\sqrt{3}\)

This means that the radius, as we can see from the picture = \(3\sqrt{3}-\sqrt{3}=2\sqrt{3}\)

Hence area of circle = \(\pi*(2\sqrt{3})^2 = 12\pi\)

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Re: An equilateral triangle that has an area of 9*3^(1/2) [#permalink]

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New post 07 Jul 2010, 11:39
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Hussain15 wrote:
RGM wrote:
From the properties of equilateral triangles, we can determine that one side of the triangle is 6:

1 / 2 * [ 6 * 3√3 ] = 9√3

We need to find the radius - we must subtract something from 3√3 (the height of the triangle). The way to do this is to divide the triangle by 3 using corresponding radii. The area of these triangles would be 3√3. We can use the segment we need to subtract from 3√3 and 6 to determine the area of the newly formed triangles:

3√3 = 1/2 * 6 * x
3√3 = 3x
x = √3

Subtracting √3 from 3 would give us the radius:

3√3 - √3 = √3 ( 3 - 1 ) = 2√3

As such the circle's area is:

(2√3)^2 * π = 12 π

Hence C.


Couldnt get the red highlighted part!


I wish I could attach a diagram! It's hard to explain!

I'll try to illustrate with words.

An equilateral triangle can be bisected into portions, leaving us with two identical right triangles. Let's set one side of the triangle as 2X, another as X and using triangle properties, the length of the bisector is X√3.

We computed this X√3 as 3√3 from the equations earlier. This segment goes through the center, and contains the radius and a length that goes straight from the center up to one side of the triangle. By subtracting this length from 3√3, we would get the radius.

How to get this? What I did was to divide the equilateral triangle by three radii. This leaves us with three identical triangles, with angles 30-30-120. Dividing the 9√3 area of the equilateral triangle gives us 3√3. I did this because this is the area of the new triangles and the area could be computed by multiplying one side of the triangle (6) and the segment from the center up to the side of the triangle (the one we're interested in subtracting). Backsolving would give us the length of this mystery segment and enable us to solve the problem.

Hope this clarifies. :) Just ask me for clarifications - geometry is hard to describe with just words.

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Re: An equilateral triangle that has an area of 9*3^(1/2) [#permalink]

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New post 07 Jul 2010, 11:40
Et voila, a diagram. Thanks whiplash.

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Re: An equilateral triangle that has an area of 9*3^(1/2) [#permalink]

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New post 07 Jul 2010, 11:48
No problem, folks.

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Re: An equilateral triangle that has an area of 9*3^(1/2) [#permalink]

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New post 08 Jul 2010, 10:14
Thanks to both of you for giving your valuable feedback.

After spending 15 minutes, now I got the issue. Actually both of you mentioned \(3\sqrt{3}\) as the area of the smaller triangle, & thats correct, but when you have subtracted \(\sqrt{3}\) from \(3\sqrt{3}\), you didnt mention that this \(3\sqrt{3}\) is not the area of smaller triangle, rather its the height of the larger triangle. Then you have subtracted the height of smaller triangle from the height of larger triangle, ultimately giving us the radius of the circle. Now its making sense for me. :)

Thanks again guys!!,
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Re: An equilateral triangle that has an area of 9*3^(1/2) [#permalink]

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Hussain15 wrote:
Thanks to both of you for giving your valuable feedback.

After spending 15 minutes, now I got the issue. Actually both of you mentioned \(3\sqrt{3}\) as the area of the smaller triangle, & thats correct, but when you have subtracted \(\sqrt{3}\) from \(3\sqrt{3}\), you didnt mention that this \(3\sqrt{3}\) is not the area of smaller triangle, rather its the height of the larger triangle. Then you have subtracted the height of smaller triangle from the height of larger triangle, ultimately giving us the radius of the circle. Now its making sense for me. :)

Thanks again guys!!,


Just wanted to contribute how I solved the problem...I used the centroid concept

The side of the equilateral triangle can be calculated to 6 units.
s=6, therefore height =\({(sqrt(3)/2)s}\) = \({3sqrt(3)}\)

When an equilateral triangle is inscribed in a circle, the centroid(point where the three medians meet) coincides with the centre of the circle. So all we need to calculate the radius is to calculate the length from the centroid to one of the vertices of the triangle.

The length of the segment from the centroid to the vertex(i.e. radius of the circle) is (2/3)* height of the triangle

Therefore, radius = \({(2/3)3sqrt(3)}\) = \({2sqrt(3)}\)

Thus area of the circle = \(12pi\)

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

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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 ....

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

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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 \(r=a*\frac{\sqrt{3}}{6}\)) --> \(area_{circle}=\pi{R^2}=12\pi\).

Answer: C.

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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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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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

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

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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

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

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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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New post 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
r - radius.

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

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New post 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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New post 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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New post 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.


The correct answer cannot be D...the answer must be C...
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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New post 28 Jul 2013, 23:37
rite2deepti wrote:
An equilateral triangle that has an area of \(9\sqrt{3}\) is inscribed in a circle.


I am trying to relearn about the term "inscribed". Please help:
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] 28 Jul 2013, 23:37

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