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# In the diagram above, A is the center of the circle, angle ABD = 90°

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In the diagram above, A is the center of the circle, angle ABD = 90°  [#permalink]

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09 Apr 2015, 06:20
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Difficulty:

85% (hard)

Question Stats:

52% (02:19) correct 48% (02:23) wrong based on 164 sessions

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In the diagram above, A is the center of the circle, angle BAD = 90°, and the area of triangle ABD is $$24\sqrt{3}$$. What is the area of the circle?

(1) AB = 12

Kudos for a correct solution.

Attachment:

gdrtq_img1.png [ 10.33 KiB | Viewed 4174 times ]

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Re: In the diagram above, A is the center of the circle, angle ABD = 90°  [#permalink]

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09 Apr 2015, 07:33
Bunuel wrote:

In the diagram above, A is the center of the circle, angle BAD = 90°, and the area of triangle ABD is $$24\sqrt{3}$$. What is the area of the circle?

(1) AB = 12

Kudos for a correct solution.

Attachment:
The attachment gdrtq_img1.png is no longer available

if AB is given we can find AD and hence area of circle . option A is sufficient.

option B : please refer the attached image .

we can find all the sides of ACE as BC=AC=R.
finally will get an expression in Something*$$(R^2)$$ = $$24 * \sqrt{3}$$ Sufficient.
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gmatclub.jpg [ 17.33 KiB | Viewed 3493 times ]

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Re: In the diagram above, A is the center of the circle, angle ABD = 90°  [#permalink]

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09 Apr 2015, 23:05
2
1
Bunuel wrote:

In the diagram above, A is the center of the circle, angle BAD = 90°, and the area of triangle ABD is $$24\sqrt{3}$$. What is the area of the circle?

(1) AB = 12

Kudos for a correct solution.

Attachment:
gdrtq_img1.png

1) We know that this is right triangle. We know its area and one of the side AB so we can find AD
$$\frac{(AB * AD )}{2}=24 * \sqrt{3}$$

AB = 12

so $$AD = 4\sqrt{3}$$

Sufficient

Ratio of sides in triangle $$ABD = 1:\sqrt{3}:2$$

So $$AB = R\sqrt{3}$$ and $$AD = R$$

$$\frac{(R\sqrt{3} * R )}{2}=24 * \sqrt{3}$$

$$R^2\sqrt{3}=48 * \sqrt{3}$$

$$R=4 \sqrt{3}$$

R is a radius so we can find area of circle.
Sufficient

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In the diagram above, A is the center of the circle, angle ABD = 90°  [#permalink]

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13 Apr 2015, 03:54
Bunuel wrote:

In the diagram above, A is the center of the circle, angle BAD = 90°, and the area of triangle ABD is $$24\sqrt{3}$$. What is the area of the circle?

(1) AB = 12

Kudos for a correct solution.

Attachment:
gdrtq_img1.png

VERITAS PREP OFFICIAL SOLUTION:

In order to find the area of the circle, we would need the radius, AD.

Statement #1: AB = 12. Notice that AD is the base of triangle ABD. AB is the height, now known, and we know the area. If we know the height and the area, we can find the base by A = 0.5*bh. Therefore, we can find AD, which would allow us to find the area of the circle. This statement, alone and by itself, is sufficient.

Statement #2: We know that AD = AC, because all radii of a circle are equal. If CD also equals these two, then ACD is an equilateral triangle. That means that the angle ADC = 60°, which means that triangle ABD is a 30-60-90 triangle. If AD = x is the base, then this times the square-root of 3 is the height, and we could create an equation because we know the area:

$$24\sqrt{3}=\frac{1}{2}*x*x\sqrt{3}$$

This is an equation we could solve for x, which would allow us to find the area of the circle. This statement, alone and by itself, is sufficient.

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Re: In the diagram above, A is the center of the circle, angle ABD = 90°  [#permalink]

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25 Jun 2018, 08:53
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the diagram above, A is the center of the circle, angle ABD = 90° &nbs [#permalink] 25 Jun 2018, 08:53
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