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# In the figure shown below, line segments AC and BD are diameters of th

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In the figure shown below, line segments AC and BD are diameters of th  [#permalink]

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05 Oct 2018, 00:41
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Difficulty:

15% (low)

Question Stats:

84% (01:32) correct 16% (01:33) wrong based on 32 sessions

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In the figure shown below, line segments AC and BD are diameters of the circle. If the distance between A and D is 6/√2, what is the area of the circle?

(A) 4π
(B) 6π
(C) 9π
(D) 18π
(E) 32π

Attachment:

2018-10-05_1139.png [ 6.61 KiB | Viewed 550 times ]

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In the figure shown below, line segments AC and BD are diameters of th  [#permalink]

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Updated on: 05 Oct 2018, 00:59
In Triangle AOD, (AD)^2=(AO)^2 + (OD)^2 ,since AO=OD=r, on substituting value and calculating we will get , r=3.
Area of circle will be πr^2=9π

Originally posted by Navishek on 05 Oct 2018, 00:56.
Last edited by Navishek on 05 Oct 2018, 00:59, edited 1 time in total.
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Re: In the figure shown below, line segments AC and BD are diameters of th  [#permalink]

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05 Oct 2018, 00:57
Bunuel wrote:

In the figure shown below, line segments AC and BD are diameters of the circle. If the distance between A and D is 6/√2, what is the area of the circle?

(A) 4π
(B) 6π
(C) 9π
(D) 18π
(E) 32π

Attachment:
2018-10-05_1139.png

Since AD=$$\frac{6}{\sqrt{2}}$$, $$\sqrt{2}$$ x R=$$\frac{6}{\sqrt{2}}$$
Area = $$\pi$$ $$R^2$$ = 9$$\pi$$
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Re: In the figure shown below, line segments AC and BD are diameters of th  [#permalink]

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06 Oct 2018, 03:18
and we know that AD= 6/√2
Therefore r=3 and Area= 9π
Re: In the figure shown below, line segments AC and BD are diameters of th   [#permalink] 06 Oct 2018, 03:18
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