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

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Joined: 02 Sep 2009
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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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05 Feb 2019, 02:06
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Difficulty:

35% (medium)

Question Stats:

69% (03:11) correct 31% (02:44) wrong based on 16 sessions

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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 [ 32.6 KiB | Viewed 164 times ]

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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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Updated on: 05 Feb 2019, 03:10
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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2019-02-05_1402.png [ 33.97 KiB | Viewed 115 times ]

Originally posted by GMATMBA5 on 05 Feb 2019, 02:55.
Last edited by GMATMBA5 on 05 Feb 2019, 03:10, edited 1 time in total.
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Joined: 16 Oct 2010
Posts: 8895
Location: Pune, India
Re: In the circle with center O above, PS = 8. If x = 75, then what is the  [#permalink]

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05 Feb 2019, 03:02
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. $$\frac{2\pi}{3}+2\sqrt{3}+6$$

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 $$= \frac{2}{3}*\pi + 2 + (4 - 2*\sqrt{3}) = \frac{2}{3}*\pi + 6 - 2*\sqrt{3}$$

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Veritas Prep GMAT Instructor

Re: In the circle with center O above, PS = 8. If x = 75, then what is the   [#permalink] 05 Feb 2019, 03:02
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