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# In the diagram above, O is the center of the circle and ACDE is a squa

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Joined: 02 Sep 2009
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In the diagram above, O is the center of the circle and ACDE is a squa  [#permalink]

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15 Oct 2019, 23:41
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Difficulty:

25% (medium)

Question Stats:

77% (01:19) correct 23% (02:03) wrong based on 26 sessions

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In the diagram above, O is the center of the circle and ACDE is a square. What is the area of the square?

(1) the circle has a radius of 2

(2) angle ACB = 90°

Attachment:

GMAT_DS_Magoosh_161.png [ 7.08 KiB | Viewed 241 times ]

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Re: In the diagram above, O is the center of the circle and ACDE is a squa  [#permalink]

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15 Oct 2019, 23:54
In the diagram above, O is the center of the circle and ACDE is a square. What is the area of the square?

(1) the circle has a radius of 2

(2) angle ACB = 90°

Answer : The Correct answer should be (A) -> Option (1) is sufficient

- AB is the diameter, so Angle ACB must be 90degree

- OA=OB=2, Thus, Dia AOB =4, BC =3, Angle ACB =90Degree, So we can find the value of AC

- All sides are equal for Square. Hence Area can be evaluated from a^2 ( a - length of each side)
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Re: In the diagram above, O is the center of the circle and ACDE is a squa  [#permalink]

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16 Oct 2019, 01:20
Bunuel wrote:

In the diagram above, O is the center of the circle and ACDE is a square. What is the area of the square?

(1) the circle has a radius of 2

(2) angle ACB = 90°

Attachment:
GMAT_DS_Magoosh_161.png

Angle in a semi-circle is 90 deg
--> Triangle ACB is right angled
--> $$AB^2 = AC^2 + BC^2$$
--> $$AC^2 = AB^2 - 9$$
--> $$AC = \sqrt{AB^2 - 9}$$

(1) the circle has a radius of 2
--> $$AB = 4$$
--> $$AC = \sqrt{4^2 - 9} = \sqrt{7}$$
--> Area of square = $$AC^2 = 7$$ --> Sufficient

(2) angle ACB = 90°
This is the property of triangle insid a semi-circle
--> Nothing can be said about the value of side AC --> Insufficient

IMO Option A
Re: In the diagram above, O is the center of the circle and ACDE is a squa   [#permalink] 16 Oct 2019, 01:20
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# In the diagram above, O is the center of the circle and ACDE is a squa

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