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

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

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New post 02 Mar 2015, 08:26
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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
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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New post 02 Mar 2015, 10:34
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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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New post 02 Mar 2015, 19:01
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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
Image

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

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

Refer diagram below

Attachment:
c2_img2.png
c2_img2.png [ 14.21 KiB | Viewed 2126 times ]


\(\pi r^2 = 144\pi\)

Radius = 12

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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New post 08 Mar 2015, 14:02
Bunuel wrote:
Image
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 \(24 + 8\pi\).

Answer = 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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New post 13 Feb 2019, 23:03
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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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