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

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Math Expert
Joined: 02 Sep 2009
Posts: 50610
What is the area of the portion of the circle above created by arc ACB  [#permalink]

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12 Sep 2018, 00:03
00:00

Difficulty:

45% (medium)

Question Stats:

75% (01:28) correct 25% (01:32) wrong based on 16 sessions

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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 [ 1.85 KiB | Viewed 275 times ]

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

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12 Sep 2018, 00:26
Bunuel wrote:

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 &nbs [#permalink] 12 Sep 2018, 00:26
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