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I am looking for a quicker method ?? If equilateral triangle

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I am looking for a quicker method ?? If equilateral triangle [#permalink] New post 27 Mar 2005, 07:18
I am looking for a quicker method ??

If equilateral triangle of area 9Sqrt(3) is inscribed with in a circle what is the area of the circle?
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 [#permalink] New post 27 Mar 2005, 08:33
length of side of equilateral triangle divided by 4 should be equal to the radius of the circle which would then give you the area.
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 [#permalink] New post 27 Mar 2005, 08:46
apologies confused on this one...please ignore the answer posted.

:?
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Re: quicker method [#permalink] New post 27 Mar 2005, 11:34
praveen_rao7 wrote:
I am looking for a quicker method ?? If equilateral triangle of area 9Sqrt(3) is inscribed with in a circle what is the area of the circle?

i do not have any quicker methos than this.........
[sqrt(3) L^2]/4= 9sqrt(3), which gives you L=6

draw Prependicular lines from each vertex of the triangle to the other respective sides(base) of the triangle. now we have different triangle. take one triangle and find the H (Hypoteneous) of that triangle and the H is the redius of that triangle.

H of smal triangle = r = 2sqrt(3). then, area of the circle= formula.......

Praveen, if you have any short cut method, pls do share with us.
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Re: quicker method [#permalink] New post 27 Mar 2005, 12:11
praveen_rao7 wrote:
I am looking for a quicker method ??

If equilateral triangle of area 9Sqrt(3) is inscribed with in a circle what is the area of the circle?

Is it an equilateral triangle or a right triangle.
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 [#permalink] New post 27 Mar 2005, 14:29
sqrt(3)/4*s^2 is the area of an equi triangle.

Since the area of the triangle = 9*sqrt(3)

we get the side as 6 for the triangle

each side is a chord for the circle

So find the length of the arc for the 3 semicircles

Since side =6 ; the length of the arc would be 3PI

Also we have 3 semicircles => length of 3 semicirles = 3*3PI=9PI

the 3 semicircles would give the cirsumfrence of the circle

2PI*r = 9PI

r=4.5

Area = PI*(4.5)^2
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 [#permalink] New post 27 Mar 2005, 14:33
gmat2me2 wrote:
sqrt(3)/4*s^2 is the area of an equi triangle.



gmat2me2, can you be more brief as to how you deduced
"sqrt(3)/4*s^2 is the area of an equi triangle."
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 [#permalink] New post 27 Mar 2005, 14:35
Area of an equi triangle whose side is a

is sqrt(3)/4*a^2 Thats a formula....
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 [#permalink] New post 27 Mar 2005, 14:44
Actually this is how we can proceed.

A triangle if inscribed in a circle is a circumcircle. So lets suppose S is the centre of the circle and A.B and C be the vertices.
Then SA=SB=SC=circumradius


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

a=6 side of triangle

Now angle ASB = 2*Angle C = 2* 60=120(Where S is the Circumcenter . Angle at vertex is half at the circumcenter ,S)

Since AB=6 ===>> AS = 6*Sin120 = 3

So circumradius = 3 and the area of circumcircle = PI*9

PLease ignore the previous posting of mine...

Last edited by gmat2me2 on 27 Mar 2005, 15:48, edited 1 time in total.
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 [#permalink] New post 27 Mar 2005, 15:48
gmat2me2 wrote:
sqrt(3)/4*s^2 is the area of an equi triangle.

Since the area of the triangle = 9*sqrt(3)

we get the side as 6 for the triangle

each side is a chord for the circle

So find the length of the arc for the 3 semicircles

Since side =6 ; the length of the arc would be 3PI

Also we have 3 semicircles => length of 3 semicirles = 3*3PI=9PI

the 3 semicircles would give the cirsumfrence of the circle

2PI*r = 9PI

r=4.5

Area = PI*(4.5)^2


What is the formula to find the lenght of an arc? I thought it was n/360 * 2.pi.r. Is there another twist to finding the lenght of an arc?
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 [#permalink] New post 27 Mar 2005, 15:49
Folaa3 wrote:
gmat2me2 wrote:
sqrt(3)/4*s^2 is the area of an equi triangle.

Since the area of the triangle = 9*sqrt(3)

we get the side as 6 for the triangle

each side is a chord for the circle

So find the length of the arc for the 3 semicircles

Since side =6 ; the length of the arc would be 3PI

Also we have 3 semicircles => length of 3 semicirles = 3*3PI=9PI

the 3 semicircles would give the cirsumfrence of the circle

2PI*r = 9PI

r=4.5

Area = PI*(4.5)^2


What is the formula to find the lenght of an arc? I thought it was n/360 * 2.pi.r. Is there another twist to finding the lenght of an arc?


Please ignore this message ...The one previous to your message is the valid one
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 [#permalink] New post 27 Mar 2005, 17:10
I don't have OA

solution I am getting is Area of Circle = 4*pi*Sqrt(3)
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 [#permalink] New post 27 Mar 2005, 17:13
Well Given (sqrt3)/4 a^2 = 9sqrt3 => a = 6 length of equilateral triangle

Now The centroid of an equilateral triangle is where the 3 medians cut each other and it should be the centre of the circle and the radius then becomes (2/3)*a*cos 60 deg = 2/3*6*1/2 = 2
Therefore area = pi84 = 4pi
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 [#permalink] New post 27 Mar 2005, 19:44
What I would do is to memorize the special right triangle whose three angels are 90, 60, 30. The ratio of its three sides are 1:sqrt(3):2.

The unit area of this special triangle is sqrt(3)/2. Since the area of our triangle (Call it ABC) here is 9sqrt(3), half of the triangle (ABD) is 9sqrt(3)/2. We know that the triangle's three sides very quickly: 3,3sqrt(3),6.

AOD is again this same kind of special triangle and AD=3. We know that AO:AD=2:sqrt(3). So AO=r=6/sqrt(3)=2sqrt(3).

Check: AD=AO+OD
OD=3sqrt(3)-2sqrt(3)=sqrt(3)
AD^2+OD^2=9+3=12
AO^2=12
matches

From here it's easy to get the area of the circle. S=pi*r^2=12pi.

Verification: The area of the triangle is 9sqrt(3)=9*1.7=15
Area of the circle =12pi=12*3.14=37

If you draw the graph, you'll see that the area that is formed by AB and the arc AB is a little bit larger than the area of the triangle ABO. It is true for the other two triangles. In other words, the total area of the triangle ABC is a little smaller than the area within the circle and outside of the triangle. In other words it should be a little less than half of the area of the circle. Our results checked out.

If it is a PS question and you have different choices to choose from. You may not even need to calculate if you remember the area of the circle is a little bit larger than twice of the area of the triangle.
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 [#permalink] New post 27 Mar 2005, 20:23
The simplest solution is here

Area of any triangle is (a*b*c)/4R

Where a,b,c are sides of the triangle

R is the circumradius

Area is given as 9*sqrt(3)

=======>>> 9*sqrt(3) = (6*6*6)/4*R (you get 6 by solving sqrt(3)/4*s^2)

here since it is equi trangle a=b=c=s

You get R as 2sqrt(3)

So area of the circum circle is PI*R^2 = 12PI
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 [#permalink] New post 27 Mar 2005, 20:25
anirban16 wrote:
Well Given (sqrt3)/4 a^2 = 9sqrt3 => a = 6 length of equilateral triangle

Now The centroid of an equilateral triangle is where the 3 medians cut each other and it should be the centre of the circle and the radius then becomes (2/3)*a*cos 60 deg = 2/3*6*1/2 = 2
Therefore area = pi84 = 4pi


Sorry Anirban you cannot take centroid in all cases... But it is an eq trangle you may probably but it fails for othercases......
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 [#permalink] New post 27 Mar 2005, 22:24
praveen_rao7 wrote:
I don't have OA
solution I am getting is Area of Circle = 4*pi*Sqrt(3)

praveen, it should be = pi*[2Sqrt(3)]^2=12 pi
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 [#permalink] New post 28 Mar 2005, 14:13
12pi

Same explanation as Gmat2me2..
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Re: quicker method [#permalink] New post 28 Mar 2005, 23:44
For a eq triangle, [sqrt(3)/4] a^2 = Area, where a is a side.

Also, for the arrangement of eq triangle inscribed in a circle, the radius of the circle is a/sqrt(3).

Therefore in this case, a = 6 and radius of the circle = 2 sqrt(3).

Thus area = 12 pi

Short enough?
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Re: quicker method [#permalink] New post 29 Mar 2005, 20:50
kapslock wrote:
For a eq triangle, [sqrt(3)/4] a^2 = Area, where a is a side.

Also, for the arrangement of eq triangle inscribed in a circle, the radius of the circle is a/sqrt(3).

Therefore in this case, a = 6 and radius of the circle = 2 sqrt(3).

Thus area = 12 pi

Short enough?


How did you get the radius of the circle?
Re: quicker method   [#permalink] 29 Mar 2005, 20:50
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