An equilateral triangle that has an area of 9*3^1/2 : GMAT Problem Solving (PS)
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# An equilateral triangle that has an area of 9*3^1/2

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

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05 May 2006, 23:28
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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. $$18\pi \sqrt{3}$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: an-equilateral-triangle-that-has-an-area-of-9-root-105356.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Nov 2013, 00:46, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: The area of the circle [#permalink]

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16 May 2006, 11:11
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.
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16 May 2006, 12:15
I got 24pi, but my approach is slightly different.

What is the OA?
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Re: The area of the circle [#permalink]

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16 May 2006, 12:23
Professor wrote:
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.

Prof, what is 'a'? I have never seen these equations for equilateral triangles before.
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Re: The area of the circle [#permalink]

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16 May 2006, 12:49
shampoo wrote:
Professor wrote:
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.

Prof, what is 'a'? I have never seen these equations for equilateral triangles before.

a = a side of an equilateral triangle..
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Re: The area of the circle [#permalink]

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16 May 2006, 15:59
Professor wrote:
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.

Prof is correct

Also you may take a longer route

After you find a=6(whihc is the side of eq triangle)

You may get the ht = sqrt(3)/2*a = sqrt(3)*3

now circum radius = 2/3*ht = 2*sqrt(3)

Area = 12pi
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Re: The area of the circle [#permalink]

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17 May 2006, 05:50
Professor wrote:
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.

Is there any way to solve these without knowingn these?
just using the standard properties of triangles & circles?
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Re: The area of the circle [#permalink]

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17 May 2006, 06:35
kook44 wrote:
Is there any way to solve these without knowingn these?
just using the standard properties of triangles & circles?

Yes, I used another approach

Assume the triangle is ABC and O is the center of the circle.
AO, BO and CO divide the triangle ABC into 3 equal isosceles triangles,
when each has an area of 9sqrt(3) / 3 = 3sqrt(3) .

Extending the AO up to BC , we get a height AD , so OD is the height of triangle BOC.

Now assume OD = x , hence DC = x*sqrt(3) and OC=2 *x = R ( ODC is a right triangle 30-60-90 )

the area of ODC = 3sqrt(3) / 2

x*x*sqrt(3) / 2 = 3sqrt(3)/2
x = sqrt(3)
OC = R = 2*sqrt(3) , hence the area of the circle is 12pi

Last edited by deowl on 17 May 2006, 06:55, edited 1 time in total.
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Re: The area of the circle [#permalink]

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17 May 2006, 06:38
kook44 wrote:
Professor wrote:
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.

Is there any way to solve these without knowingn these?
just using the standard properties of triangles & circles?

Yes there are ways. One of them is here

Side of triangle = a
Height of triangle using pythagoras = SQRT(a^2 - (a/2)^2 ) = a * SQRT(3)/2

Area of triangle = 1/2 * Base * Height = 1/2 * a * a * SQRT(3)/2 = a^2 * SQRT(3)/4 = 9 * SQRT(3)

So a = 6
Height = 6 * SQRT(3)/2 = 3 * SQRT(3)

Now if you drop lines from all three vertices of triangle to the centre of sides then these three lines will meet at the center of the circle. The ratio of the lines from centre to the vertex and centre to the side is 2:1 (This you have to remember or you may waste time to prove this using similar triangles). The line from centre to the vertex is the radius of circle.
So r = 2/3 * Height = 2/3 * 3 * SQRT(3) = 2 * SQRT3

Area of circle = PI * r^2 = PI * (2 * SQRT(3)) ^2 = 12 PI

Hope this helps.
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Re: An equilateral triangle is inscribed in a circle and has the [#permalink]

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20 Nov 2013, 17:48
$$\sqrt{3}$$/4*$$a^2$$=9$$\sqrt{3}$$. a=6. Now shortcut formula for radius of circle circumsubcribed over an equilateral is $$\sqrt{3}$$a/3= 2$$\sqrt{3}$$. $$area=pi*r^2$$=12pi.
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Re: An equilateral triangle that has an area of 9*3^1/2 [#permalink]

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21 Nov 2013, 00:49
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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. $$18\pi \sqrt{3}$$

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

We are given that this triangle is inscribed in circle. The radius of the circumscribed circle is $$R=a*\frac{\sqrt{3}}{3}=2\sqrt{3}$$ (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

OPEN DISCUSSION OF THIS QUESTION IS HERE: an-equilateral-triangle-that-has-an-area-of-9-root-105356.html
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Re: An equilateral triangle that has an area of 9*3^1/2   [#permalink] 21 Nov 2013, 00:49
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