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What is the area of the portion of the circle above created by arc ACB

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What is the area of the portion of the circle above created by arc ACB  [#permalink]

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New post 12 Sep 2018, 01:03
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What is the area of the portion of the circle above created by arc ACB and cord AB if the radius of the circle is 6 and the length of arc ACB is 2π?


A. \(6\pi\)

B. \(6\pi - 6\)

C. \(6\pi - 9\)

D. \(6\pi - \frac{25}{2}\)

E. \(6\pi - 9\sqrt{3}\)



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What is the area of the portion of the circle above created by arc ACB  [#permalink]

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New post 12 Sep 2018, 01:26
Bunuel wrote:
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What is the area of the portion of the circle above created by arc ACB and cord AB if the radius of the circle is 6 and the length of arc ACB is 2π?


A. \(6\pi\)

B. \(6\pi - 6\)

C. \(6\pi - 9\)

D. \(6\pi - \frac{25}{2}\)

E. \(6\pi - 9\sqrt{3}\)
Attachment:
image035.jpg


Let the centre of circle is at O. Join OA and OB.
Arc ACB=\(2\pi*6(\frac{AngleAOB}{2*\pi})\)
Or, \(2*\pi=6*AngleAOB\) Or, Angle AOB=60 degree

Now, OAB is an equilateral triangle. OA=OB=AB=6

Required area= Are aof sector AOBCA- Area of triangle AOB
=\(\pi*6^2(60/360)-\sqrt{3}/4 * 6^2\)
=\(6\pi - 9\sqrt{3}\)
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What is the area of the portion of the circle above created by arc ACB   [#permalink] 12 Sep 2018, 01:26
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What is the area of the portion of the circle above created by arc ACB

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