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The circle above has radius 8, and AD is parallel [#permalink]
05 Aug 2017, 01:55
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48% (03:45) correct
51% (02:30) wrong based on 33 sessions
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#GREpracticequestion The circle above has radius 8, and AD is parallel to BC.jpg [ 21.65 KiB  Viewed 668 times ]
The circle above has radius 8, and AD is parallel to BC. If the length of arc AYD is twice the length of arc BXC, what is the length of arc BXC? A. \(2 \pi\) B. \(\frac{8\pi}{3}\) C. \(3 \pi\) D. \(4 \pi\) E. \(\frac{16\pi}{3}\)
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Re: The circle above has radius 8, and AD is parallel [#permalink]
18 Oct 2017, 13:18
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Because lines XB and AD are parallel The Angle AB is equal to the angle BC.
Now we have 2 inscribed angles of 45 degrees. Inscribed Angles = 1/2 central angles. SO we know that we have two central angles of 90 degrees, which make up 160/360 or 1/2 of the circumference of the circle.
(ARC AD + ARC BC) => we know Arc AD=2*ArcBC so combining these we have 3 Arc BC
To calculate the arcs we have to solve the following equation where we need to solve for a (ARC BC)
2(8)pi=3a + (1/2)(16pi) = 8/3pi
I hope this helped



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Re: The circle above has radius 8, and AD is parallel [#permalink]
07 Feb 2018, 00:09
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The circumference of the circle is 16pi. The arc length of CD is 90/360=CD/16pi or, CD=4pi The arc length of AB is 4pi. Let, the arc length of BXC is x, so the arc length of AYD is 2x So, x+2x+8pi=16pi or, x= 8pi/3 So the answer is B.



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Re: The circle above has radius 8, and AD is parallel [#permalink]
15 Feb 2018, 04:30
Can someone explain how he/she is getting the angle DOA (O being the centre of the circle) to be 90 degrees? Carcass wrote: Attachment: circle (2).jpg The circle above has radius 8, and AD is parallel to BC. If the length of arc AYD is twice the length of arc BXC, what is the length of arc BXC? A. 2 π B. 8 π/3 C. 3 π D. 4 π E. 16π/3



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Re: The circle above has radius 8, and AD is parallel [#permalink]
15 Feb 2018, 09:03
AB and AD are not parallel. You just copied the explanation from the end of the book. I have now officially wasted an entire day on a stupid question. Thanks to this WRONG explanation. I drew the actual circle and the chords and this question is incorrect! simon1994 wrote: Because lines XB and AD are parallel The Angle AB is equal to the angle BC.
Now we have 2 inscribed angles of 45 degrees. Inscribed Angles = 1/2 central angles. SO we know that we have two central angles of 90 degrees, which make up 160/360 or 1/2 of the circumference of the circle.
(ARC AD + ARC BC) => we know Arc AD=2*ArcBC so combining these we have 3 Arc BC
To calculate the arcs we have to solve the following equation where we need to solve for a (ARC BC)
2(8)pi=3a + (1/2)(16pi) = 8/3pi
I hope this helped



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Re: The circle above has radius 8, and AD is parallel [#permalink]
19 Feb 2018, 01:58
This is the OE. Quote: Because the AD is parallel to BC, the measure of angle ACB is also 450. Angles CAD and ACB are both inscribed angles of the circle. The measures of the corresponding central angles are twice 450, or 90° each. Therefore, taken together, minor arcs AB and CD make up 180° of the entire circle, leaving 180° for arcs BXC and AYD. Because arc AYD is twice the length of arc BXC, arc BXC must correspond to a 6o° central angle and arc AYD to a 120° central angle. Therefore, arc BXC is \(\frac{60}{360} = \frac{1}{6}\) of the entire circumference of the circle, which equals \(2\pi = 16\pi\). The length of arc BXC is thus \(\frac{16\pi}{r} = \frac{8\pi}{3}\) Hope this helps Regards
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Re: The circle above has radius 8, and AD is parallel [#permalink]
22 Feb 2018, 23:36
Isn't the arc length of CD = 45/360 * 16 pi = 2 pi? YMAkib wrote: The circumference of the circle is 16pi. The arc length of CD is 90/360=CD/16pi or, CD=4pi The arc length of AB is 4pi. Let, the arc length of BXC is x, so the arc length of AYD is 2x So, x+2x+8pi=16pi or, x= 8pi/3 So the answer is B.



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Re: The circle above has radius 8, and AD is parallel [#permalink]
08 Mar 2018, 00:14
Here, the chords AD and BC are parallel, angle DAC is given 45, therefore, angle BCA is also 45. Now we cant directly calculate the length of given segments, we have to apply indirect method, total circumference= seg. BXC + seg. CD + seg. DYA + seg. AB, i.e. 16pi= x+ 2x+ CD+ AB( i assumed seg BXC to be x). Now, seg CD and seg AB are equal. Reason mark the center of the circle, now angle of the seg AB will be twice of the angle subtended by its chord AB(45), therefore, the angle is 90. Same goes for the other one and since they have the same angle, their length of the segment will be 2*pi*r*(90/360)=16pi/4=4pi. Therefore, 16pi=3x+4pi+4pi, this gives us x=8pi/3. Hence lenght of seg BXC is 8pi/3.



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Re: The circle above has radius 8, and AD is parallel [#permalink]
02 May 2018, 09:42
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Carcass wrote: Attachment: circle (2).jpg The circle above has radius 8, and AD is parallel to BC. If the length of arc AYD is twice the length of arc BXC, what is the length of arc BXC? A. 2 π B. 8 π/3 C. 3 π D. 4 π E. 16π/3 We see that since arc CD corresponds to a 45 degree inscribed angle (angle CAD), arc CD is twice the measure of angle CAD. The measure of arc CD is 90 degrees, thus its arc length is 90/360 = 1/4 of the circle. We also can conclude that angle BCA is also 45 degrees (since BC is parallel to AD) and hence arc BA is also 90/360 = 1/4 of the circle. Since arc AYD is twice the length of arc BXC, we can let arc BXC = x (where x represents the fraction the arc length of BXC as of the circumference of the circle) and thus arc AYD = 2x. We can create the equation: 1/4 + 1/4 + x + 2x = 1 3x = 1/2 x = 1/6 Thus, arc BXC is 1/6 of the circumference of the circle, and arc AYD is 1/3 of the circumference of the circle. Since the radius is 8, the circumference is 16π and thus arc BXC is 1/6 x 16π = 16π/6 = 8π/3. Answer: B
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Re: The circle above has radius 8, and AD is parallel [#permalink]
09 May 2019, 08:28
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The two arcs making 45 degree angle make up half the circumference. the remaining half will be shared by two arcs AYD and BXC in 2:1 ratio. So half the circumference is 8pi. which divided by 3 gives 8pi/3 the length of arc AXC. Answer B.




Re: The circle above has radius 8, and AD is parallel
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09 May 2019, 08:28





