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In the circle with center O above, PS = 8. If x = 75, then what is the

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In the circle with center O above, PS = 8. If x = 75, then what is the  [#permalink]

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New post 05 Feb 2019, 03:06
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In the circle with center O above, PS = 8. If x = 75, then what is the perimeter of the shaded region?


A. \(6-2\sqrt{3}\)

B. \(\frac{2\pi}{3}\)

C. \(\frac{2\pi}{3}+8\)

D. \(\frac{2\pi}{3}-2\sqrt{3}+6\)

E. \(\frac{2\pi}{3}+2\sqrt{3}+6\)


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In the circle with center O above, PS = 8. If x = 75, then what is the  [#permalink]

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New post Updated on: 05 Feb 2019, 04:10
Bunuel wrote:
In the circle with center O above, PS = 8. If x = 75, then what is the perimeter of the shaded region?


A. \(6-2\sqrt{3}\)

B. \(\frac{2\pi}{3}\)

C. \(\frac{2\pi}{3}+8\)

D. \(\frac{2\pi}{3}-2\sqrt{3}+6\)

E. \(\frac{2\pi}{3}+2\sqrt{3}+6\)




IMO D

x=75, therefore \(\angle QOT=180-75-75=30\)
Triangle OQT is a 30-60-90 triangle, and with radius = OQ = 4 we have, QT=2 and OT = \(2\sqrt3\)
Thus, TR = OR-OT= \(4-2\sqrt3\)

Also arc QR \(= \frac{30}{360} * 2\pi*4 = \frac{2\pi}{3}\)

And perimeter of shaded region = QT+TR+arc QR = \(2+4-2\sqrt3+\frac{2\pi}{3}\)
\(=\frac{2\pi}{3}-2\sqrt3+6\)

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Originally posted by GMATMBA5 on 05 Feb 2019, 03:55.
Last edited by GMATMBA5 on 05 Feb 2019, 04:10, edited 1 time in total.
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Re: In the circle with center O above, PS = 8. If x = 75, then what is the  [#permalink]

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New post 05 Feb 2019, 04:02
Bunuel wrote:
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In the circle with center O above, PS = 8. If x = 75, then what is the perimeter of the shaded region?


A. \(6-2\sqrt{3}\)

B. \(\frac{2\pi}{3}\)

C. \(\frac{2\pi}{3}+8\)

D. \(\frac{2\pi}{3}-2\sqrt{3}+6\)

E. \(\frac{2\pi}{3}+2\sqrt{3}+6\)


Attachment:
2019-02-05_1402.png


PS = 8 so radius is 4
Since x = 75, x + angle QOR + x = 180

angle QOR = 30 degrees

Note that we get a 30-60-90 triangle (non shaded sector) in which OQ = 4 (radius)
The ratio of sides is \(1:\sqrt{3}:2\) in which 2 corresponds to a value of 4 (hypotenuse)
So the side opposite 30 degrees angle is 2 and the side opposite 60 degrees angle is \(2*\sqrt{3}\).

Length of QR will be (30/360) * Circumference of circle \(= (1/12) * 2*\pi*4 = \frac{2}{3} *\pi\)

Perimeter of shaded region \(= \frac{2}{3}*\pi + 2 + (4 - 2*\sqrt{3}) = \frac{2}{3}*\pi + 6 - 2*\sqrt{3}\)

Answer (D)
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Re: In the circle with center O above, PS = 8. If x = 75, then what is the   [#permalink] 05 Feb 2019, 04:02
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