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If the circle with center O has area 9π, what is area of equilateral t

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If the circle with center O has area 9π, what is area of equilateral t [#permalink]

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New post 19 Mar 2015, 06:09
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If the circle with center O has area 9π, what is area of equilateral triangle ABC?

A. \(9\sqrt{3}\)
B. 18
C. \(12\sqrt{3}\)
D. 24
E. \(16\sqrt{3}\)

Kudos for a correct solution.

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Re: If the circle with center O has area 9π, what is area of equilateral t [#permalink]

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New post 19 Mar 2015, 07:36
Area = 9pi
radius= 3
diameter = 6 = height

Halfof an equilateral is a 30 60 90 triangle

So the base of the equilateral is
\(4\sqrt{3}\)

(1/2)× \(4 \sqrt{3}\) ×6 = \(12\sqrt{3}\)

Answer: C \(12\sqrt{3}\)
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If the circle with center O has area 9π, what is area of equilateral t [#permalink]

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New post 23 Mar 2015, 04:41
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Bunuel wrote:
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If the circle with center O has area 9π, what is area of equilateral triangle ABC?

A. \(9\sqrt{3}\)
B. 18
C. \(12\sqrt{3}\)
D. 24
E. \(16\sqrt{3}\)

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

We are told the circle has an area of 9 times pi. From Archimedes' famous formula …

\(A=\pi{r^2}\)

… we know that the radius of the circle must equal r = 3. This means AO = OD = 3, which means AD = 6. Now, we have the length of the side of triangle ABD.

ABD is the kind of triangle we get when we bisect an equilateral. The angle B is still 60°, the angle at D is a right angle, and the angle at A has been bisected, so that angle DAB = 30°. This is a 30-60-90 triangle, which has special properties. You can read about these special properties at this GMAT blog.

Suppose you don't remember all those properties. Look at side DB --- that's exactly half the side of the equilateral triangle. If we call DB = x, then a full side, like BC or AB, must equal 2x. Given that AD = 6, BD = x, and AB = 2x, we can use the Pythagorean Theorem in right triangle ABD.

Image

This means that a full side of the equilateral is twice this:
\(AB=BC=4\sqrt{3}\)

Incidentally, if you remembered your 30-60-90 triangle properties, you could have found:
\(\frac{AB}{BD}=\frac{2}{\sqrt{3}}\) --> \(\frac{AB}{6}=\frac{2}{\sqrt{3}}\) --> \(AB=4\sqrt{3}\)


If you are unfamiliar with roots, and particular with the procedure of rationalizing a denominator (used here to divide 12 by root 3), then see this GMAT blog where more is explained.

Either way, now we have a height, BD = 6, and a base,

\(AB = BC = 4\sqrt{3}\)

Well, now we can find the area.

Area = 1/2*bh = \(\frac{1}{2}*6*4\sqrt{3}=12\sqrt{3}\)

That's the area of the equilateral triangle ABC.

Answer = C

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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If the circle with center O has area 9π, what is area of equilateral t [#permalink]

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New post 10 Jan 2018, 19:36
Area =9*pi => r=3
Altitude of the triangle = 2r
side * cos 30 = 2r
side * (sqrt (3) /2) = 6
side = 4 * sqrt(3)
area of equilateral triangle = (sqrt(3)/4)*side*side
Re: If the circle with center O has area 9π, what is area of equilateral t   [#permalink] 10 Jan 2018, 19:36
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