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Circle O is inscribed in equilateral triangle ABC. If the

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Circle O is inscribed in equilateral triangle ABC. If the [#permalink] New post 18 Nov 2012, 14:16
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Circle O is inscribed in equilateral triangle ABC. If the area of ABC is \(24 \sqrt{3}\), what is area of circle O?

A. 2\(\pi\)\sqrt{3}
B. 4\(\pi\)
C. 4\(\pi\)\sqrt{3}
D. 8\(\pi\)
E. 12\(\pi\)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 19 Nov 2012, 02:27, edited 2 times in total.
Renamed the topic and edited the question.
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Re: In the figure above... [#permalink] New post 18 Nov 2012, 16:13
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Skientist wrote:
Circle O is inscribed in equilateral triangle ABC. If the area of ABC is \(24 \sqrt{3}\), what is area of circle O?
2\(\pi\)\sqrt{3}
4\(\pi\)
4\(\pi\)\sqrt{3}
8\(\pi\)
12\(\pi\)

See the attached figure.
Area of triangle is given as \(24 \sqrt{3}\). using area for equilateral trianlge, side of triangle is \(4 \sqrt{6}\).
Also using pythagorean theorem height of triangle is \(6 \sqrt{2}\)

Now note in figure, height = r + p => r+p =\(6 \sqrt{2}\)

Also, \(p^2 = r^2 + (2 \sqrt{6})^2\)

=> \((6 \sqrt{2} -r)^2 =r^2 + 24\)
=> \(r = 2 \sqrt{2}\)

Thus area of circle = \(\pi*r^2\) = \(8*\pi\)

Ans D it is!
Attachments

circle_triangle.jpg
circle_triangle.jpg [ 11.71 KiB | Viewed 2088 times ]


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Re: In the figure above... [#permalink] New post 19 Nov 2012, 01:36
Quote:
Area of triangle is given as . using area for equilateral trianlge, side of triangle is .
Also using pythagorean theorem height of triangle is


Hi Vips0000

Can you please expand out this element? I can't follow what you've done here (in either case).

Thanks

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Re: In the figure above... [#permalink] New post 19 Nov 2012, 02:05
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Skientist wrote:
Quote:
Area of triangle is given as . using area for equilateral trianlge, side of triangle is .
Also using pythagorean theorem height of triangle is


Hi Vips0000
Can you please expand out this element? I can't follow what you've done here (in either case).
Thanks
Skientist

1: "using area for equilateral trianlge, side of triangle is"
area for a equilateral triangle is given by \((\sqrt{3}/4)a^2\) , where a is the side of equilateral triangle.
thus \((\sqrt{3}/4)a^2\) =\(24\sqrt{3}\)
=> side of triangle = a =\(4\sqrt{6}\)

2: "using pythagorean theorem height of triangle is"

From picture you can see,
\(height^2 + (base/2)^2 = Side^2\)
base and side are same and equal to a, as defined earlier.
=> \(height^2 = a^2 - (a/2)^2\)
using value of a obtained earlier.
=>height = \(6\sqrt{2}\)

Hope it is clear. You better give me 10 kudos now :-D
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Re: Circle O is inscribed in equilateral triangle ABC. If the [#permalink] New post 19 Nov 2012, 06:03
Skientist wrote:
Circle O is inscribed in equilateral triangle ABC. If the area of ABC is \(24 \sqrt{3}\), what is area of circle O?

A. 2\(\pi\)\sqrt{3}
B. 4\(\pi\)
C. 4\(\pi\)\sqrt{3}
D. 8\(\pi\)
E. 12\(\pi\)


Simple one....
Remember the incircle and circumcircle relation always
Area of Triangle = abc/(4*circumradius) = inradius * semi perimeter
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Re: Circle O is inscribed in equilateral triangle ABC. If the [#permalink] New post 07 Oct 2013, 05:48
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This one turns out to be a pretty straight forward question if you relay on a couple of formulas.

Basically, we are seeking to estimate the area of a circle and to do that we need a radius. There must be a relationship between the radius of the incircle and the area of the equilateral triangle.
Let's start off recalling the formula of the equilateral triangle's area.

A= s^2 (√3/4) (where s is a side of the equilateral triangle)

Now from the above expression we ensue that s= 4 √6

Now we know that the radius of a circle inscribed in an equilateral triangle is equal to the length of the side multiplied by √3/6

From here we just need to plug the values in and solve the equation accordingly which will yield 8(22/7) and the correct answer is D.
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Re: Circle O is inscribed in equilateral triangle ABC. If the [#permalink] New post 09 Nov 2014, 07:47
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Re: Circle O is inscribed in equilateral triangle ABC. If the   [#permalink] 09 Nov 2014, 07:47
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