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

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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, 23:44
Jaisri wrote:
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.


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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New post 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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New post 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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New post 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]

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New post 07 Oct 2016, 17:38
Hello from the GMAT Club BumpBot!

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

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New post 14 Jul 2017, 07:36
Can this formula for the radius be applied to any kind of inscribed polygon??

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

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New post 20 Aug 2017, 08:25
meshtrap wrote:
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,
meshtrap


Does the formula highlighted and bold works for all types of triangle or only for an equilateral triangle?

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

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New post 26 Aug 2017, 11:55
Use the Formula

Area of Triangle = (S x S x S) / 4R

In this case Area of Triangle = 9 sqrt(3) ---- Given
Side (s) = 6------( using area of Equi triangle formula)

So just put in the values and-- R---- can be found------9sqrt(3)= 6*6*6 / 4R
R= 216/ 4 x 9sqrt(3)


R==. 6/ sqrt(3)

therefore the area of the circle ==== [6/sqrt(3)] ^2 * pie. ========12Pie

hope it helps

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Re: An equilateral triangle that has an area of 9*root(3)   [#permalink] 26 Aug 2017, 11:55

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