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Lets say the triangle is ABC. There for AB = BC = AC = 10 The circle has the center O and touches BC at point D. Since ABC is equilateral, (1) BD = DE = 5, (2) CO bisects the angle BCA, and (3) OD is perpendicular to DC.

From the right triangle DCO, we get sin (angle DCO) = sin 30 = 1/2. Also, sine (angle DCO) = OD/OC. This means OD is 5, which is the radius of the circle. Therefore the area of the circle is 25π.

Lets say the triangle is ABC. There for AB = BC = AC = 10 The circle has the center O and touches BC at point D. Since ABC is equilateral, (1) BD = DE = 5, (2) CO bisects the angle BCA, and (3) OD is perpendicular to DC.

From the right triangle DCO, we get sin (angle DCO) = sin 30 = 1/2. Also, sine (angle DCO) = OD/OC. This means OD is 5, which is the radius of the circle. Therefore the area of the circle is 25π.

Is that the right answer?

Hi, it seems you were on the right track but lost track of some of the information.

You equate 1/2 to OD/OC, which is unsolvable because OD and OC are unknown.

The length we do know is DC which is 5. so using the law of tangent,

\(\frac{1}{\sqrt{3}}=\frac{OD}{DC}=\frac{OD}{5}\)

Therefore \(OD=\frac{5}{\sqrt{3}}\)

Since OD = Radius of the circle,

Area of circle = \(pi*(\frac{5}{\sqrt{3}})^2=\frac{25*pi}{3}\)

Is this right? Its late at night and I can't focus well at the moment. haha

Last edited by chaoswithin on 07 Nov 2010, 01:18, edited 1 time in total.

@Chaoswithin - ooops.. yeah.. its difficult without the figure ... yeah we don't know OC but we do know DC and therefore we could use tan 30 and get the values ... and reach the area. I believe I've given others the idea on how to move forward ..

(1) Formula 1: The radius of Inscribed circle in a traigle, which has sides length as a, b and c, is: Radius = \(\frac{Area of traingle}{K}\) = \(\frac{\sqrt{K(K-a)(K-b)(K-c)}}{K}\)

(2) Formula 2: The radius of Circumcribed circle for a traingle, which has sides lengths as a, b and c, is:

Radius = \(\frac{abc}{4*Area of traingle}\) = \(\frac{abc}{4*\sqrt{K(K-a)(K-b)(K-c)}}\)

Where, K = \(\frac{a+b+c}{2}\)

So, the radius of a inscribed circle where a=b=c=10 is: K = \(\frac{10+10+10}{2}\) = 15 Radius = \(\frac{\sqrt{15(15-10)(15-10)(15-10)}}{15}\) = \(\frac{5}{\sqrt{3}}\)

Now, the area of a traingle, which has radius \(\frac{5}{\sqrt{3}}\) is: \(pi\) * \(\frac{5}{\sqrt{3}}\) * \(\frac{5}{\sqrt{3}}\) = \(pi\) * \(\frac{25}{3}\) _________________

Cheers! Ravi

If you like my post, consider giving me some KUDOS !!!

Hey nravi549: Thanks for sharing these. Though, let me add a word of caution about formulas: They cannot substitute for conceptual understanding. If you understand the concepts behind a question, knowing a formula can help you save time. That's it. It will not take you anywhere in GMAT as far as increasing your score is concerned. You may use a formula on one question, the next will be similar or higher level and will need understanding. If you answer that and subsequent questions incorrectly, you will get right back to the point from where you started. GMAT does not test you on formulas/theorems. It tests you on your conceptual understanding of the topics and on your application skills. Whether you can twist the question to your advantage, figure out what it is testing you on and apply the basics you have learned. When I took GMAT, I remember thinking during the exam, "After the first 4-5 questions, every question is super interesting. Every question has a trick to it. Not that it is hard to figure out, but it needs thinking and every question is new." And I don't remember using any special formulas/theorems. Just the common ones. _________________

Re: A circle is inside an equalateral triangle ABC. One side of [#permalink]

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14 Mar 2014, 05:10

1

This post received KUDOS

It's a simple question if one knows the formula for calculation of an inscribed circle's radius from the side of an equilateral triangle.(Thanks Bunuel!) Radius of inscribed circle=(sqrt)3*a/6 where a is side of triangle. After calculating radius,we can easily calculate area of inscribed circle.

Re: A circle is inside an equalateral triangle ABC. One side of [#permalink]

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15 Mar 2014, 05:28

Creeper300 wrote:

A circle is inside an equalateral triangle ABC. One side of the triangle is equal to 10, What is the area of the circle?

pi*200/9 Given side of an equilateral triangle is 10.so height is 10*[square_root 2].Also the orthocentre ,centroid coincides and centroid divides in the ratio 2:1 Hence radius of circle is10*squareroot 2/3 Area = pi*r^2=pi*200/9

Re: A circle is inside an equalateral triangle ABC. One side of [#permalink]

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17 Mar 2014, 21:23

Expert's post

thefibonacci wrote:

Creeper300 wrote:

A circle is inside an equalateral triangle ABC. One side of the triangle is equal to 10, What is the area of the circle?

The ques is not framed properly.

It says: A circle is inside an equalateral triangle ABC

What is meant by 'inside' here?

What the question meant to say was that the circle is 'inscribed' inside the triangle. If it is not necessary for it to be inscribed, then it doesn't have a fixed area. _________________

Re: A circle is inside an equalateral triangle ABC. One side of [#permalink]

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16 Jan 2016, 15:11

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