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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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New post 09 Apr 2015, 07:20
1
8
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A
B
C
D
E

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  75% (hard)

Question Stats:

53% (01:52) correct 47% (01:40) wrong based on 160 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

(2) AD = CD


Kudos for a correct solution.

Attachment:
gdrtq_img1.png
gdrtq_img1.png [ 10.33 KiB | Viewed 3874 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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New post 09 Apr 2015, 08:33
Bunuel wrote:
Image
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

(2) AD = CD


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

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New post 10 Apr 2015, 00:05
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Bunuel wrote:
Image
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

(2) AD = CD


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 \(AD = 4\sqrt{3}\)

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

2) AD = CD so triangle ADC equalateral and angle ADC = 60 degrees
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

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

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New post 13 Apr 2015, 04:54
Bunuel wrote:
Image
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

(2) AD = CD


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.

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

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New post 25 Jun 2018, 09:53
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Re: In the diagram above, A is the center of the circle, angle ABD = 90° &nbs [#permalink] 25 Jun 2018, 09:53
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