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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]  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]  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 [ 11.71 KiB | Viewed 2088 times ]

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Manager
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Re: In the figure above... [#permalink]  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

Skientist
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Re: In the figure above... [#permalink]  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
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Re: Circle O is inscribed in equilateral triangle ABC. If the [#permalink]  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
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Re: Circle O is inscribed in equilateral triangle ABC. If the [#permalink]  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]  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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