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

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Math Expert
Joined: 02 Sep 2009
Posts: 58410
If the circle with center O has area 9π, what is area of equilateral t  [#permalink]

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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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19 Mar 2015, 07:36
Area = 9pi
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 $$12\sqrt{3}$$
Math Expert
Joined: 02 Sep 2009
Posts: 58410
If the circle with center O has area 9π, what is area of equilateral t  [#permalink]

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23 Mar 2015, 04:41
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Bunuel wrote:

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.

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.

Attachment:

originals.jpg [ 17.01 KiB | Viewed 10943 times ]

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

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