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# In the shaded region above, ∠KOL = 120°, and the area of the entire ci

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Joined: 02 Sep 2009
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In the shaded region above, ∠KOL = 120°, and the area of the entire ci  [#permalink]

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02 Mar 2015, 08:26
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5% (low)

Question Stats:

91% (01:28) correct 9% (01:42) wrong based on 64 sessions

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In the shaded region above, ∠KOL = 120°, and the area of the entire circle is $$A = 144\pi$$. The perimeter of the shaded region is

(A) 12 + 8{pi}
(B) 12 + 16{pi}
(C) 24 + 8{pi}
(D) 24 + 16{pi}
(E) 24 + 24{pi}

Kudos for a correct solution.

Attachment:

c2_img2.png [ 9.13 KiB | Viewed 2168 times ]

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Re: In the shaded region above, ∠KOL = 120°, and the area of the entire ci  [#permalink]

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02 Mar 2015, 10:34
1
Bunuel wrote:
In the shaded region above, ∠KOL = 120°, and the area of the entire circle is $$A = 144\pi$$. The perimeter of the shaded region is

(A) 12 + 8{pi}
(B) 12 + 16{pi}
(C) 24 + 8{pi}
(D) 24 + 16{pi}
(E) 24 + 24{pi}

Kudos for a correct solution.

pi r^2=144 pi
r=12

2sides of sector =24(2 radii of circle)
length of arc=Q/360*2pi*r
=120/360*2 pi r
=8 pi
thereforeperimitere is 24+8pi (c)
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Re: In the shaded region above, ∠KOL = 120°, and the area of the entire ci  [#permalink]

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02 Mar 2015, 19:01
1
Bunuel wrote:
In the shaded region above, ∠KOL = 120°, and the area of the entire circle is $$A = 144\pi$$. The perimeter of the shaded region is

(A) 12 + 8{pi}
(B) 12 + 16{pi}
(C) 24 + 8{pi}
(D) 24 + 16{pi}
(E) 24 + 24{pi}

Kudos for a correct solution.

Here is the original image, courtesy of Magoosh

Area=144$${pi}$$, so radius =12

Perimeter of the shaded portion of the circle is C=(2*{pi}*12)/3 = 8$${pi}$$
8$${pi}$$+2 radii = 8$${pi}$$ + 24

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Re: In the shaded region above, ∠KOL = 120°, and the area of the entire ci  [#permalink]

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02 Mar 2015, 22:35
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Answer (C) $$24 + 8{pi}$$

Refer diagram below

Attachment:

c2_img2.png [ 14.21 KiB | Viewed 2126 times ]

$$\pi r^2 = 144\pi$$

Perimeter of the shaded curve (only) $$= \frac{2\pi*12}{3} = 8\pi$$

Total perimeter $$= 12+12+8\pi = 24+8\pi$$
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Re: In the shaded region above, ∠KOL = 120°, and the area of the entire ci  [#permalink]

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08 Mar 2015, 14:02
Bunuel wrote:

In the shaded region above, ∠KOL = 120°, and the area of the entire circle is $$A = 144\pi$$. The perimeter of the shaded region is

(A) 12 + 8{pi}
(B) 12 + 16{pi}
(C) 24 + 8{pi}
(D) 24 + 16{pi}
(E) 24 + 24{pi}

Kudos for a correct solution.

Attachment:
c2_img2.png

MAGOOSH OFFICIAL SOLUTION:

The area $$144\pi=\pi{r^2}$$, so r = 12. This means KO = 12 and OL = 12, so those two sides together are 24. The remaining side is arc KL. The whole circumference is $$C=2\pi{r}=24\pi$$. The angle of 120° is 1/3 of a circle, so the arclength is 1/3 of the circumference. This means, $$arclength=8\pi$$, and therefore the entire perimeter is $$24 + 8\pi$$.

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Re: In the shaded region above, ∠KOL = 120°, and the area of the entire ci  [#permalink]

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13 Feb 2019, 23:03
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the shaded region above, ∠KOL = 120°, and the area of the entire ci   [#permalink] 13 Feb 2019, 23:03
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