Last visit was: 24 Apr 2024, 04:37 It is currently 24 Apr 2024, 04:37

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Verbal Expert
Joined: 18 Apr 2015
Posts: 28623
Own Kudos [?]: 33102 [15]
Given Kudos: 25173
Send PM
Most Helpful Community Reply
avatar
Intern
Intern
Joined: 08 Dec 2017
Posts: 40
Own Kudos [?]: 67 [11]
Given Kudos: 0
Send PM
General Discussion
avatar
Intern
Intern
Joined: 01 Sep 2017
Posts: 20
Own Kudos [?]: 31 [0]
Given Kudos: 0
Send PM
avatar
Intern
Intern
Joined: 12 Nov 2017
Posts: 30
Own Kudos [?]: 51 [0]
Given Kudos: 0
Send PM
Re: The circle above has radius 8, and AD is parallel [#permalink]
Can someone explain how he/she is getting the angle DOA (O being the centre of the circle) to be 90 degrees?

Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



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
avatar
Intern
Intern
Joined: 12 Nov 2017
Posts: 30
Own Kudos [?]: 51 [0]
Given Kudos: 0
Send PM
Re: The circle above has radius 8, and AD is parallel [#permalink]
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
Verbal Expert
Joined: 18 Apr 2015
Posts: 28623
Own Kudos [?]: 33102 [1]
Given Kudos: 25173
Send PM
Re: The circle above has radius 8, and AD is parallel [#permalink]
Expert Reply
1
Bookmarks
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
avatar
Manager
Manager
Joined: 15 Feb 2018
Posts: 53
Own Kudos [?]: 34 [0]
Given Kudos: 0
Send PM
Re: The circle above has radius 8, and AD is parallel [#permalink]
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.
avatar
Intern
Intern
Joined: 04 Mar 2018
Posts: 28
Own Kudos [?]: 36 [0]
Given Kudos: 0
GRE 1: Q167 V160
Send PM
Re: The circle above has radius 8, and AD is parallel [#permalink]
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.
Target Test Prep Representative
Joined: 09 May 2016
Status:Head GRE Instructor
Affiliations: Target Test Prep
Posts: 180
Own Kudos [?]: 269 [3]
Given Kudos: 114
Location: United States
Send PM
Re: The circle above has radius 8, and AD is parallel [#permalink]
2
Expert Reply
1
Bookmarks
Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



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
avatar
Intern
Intern
Joined: 27 Jan 2019
Posts: 29
Own Kudos [?]: 53 [0]
Given Kudos: 0
Send PM
Re: The circle above has radius 8, and AD is parallel [#permalink]
1
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.
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 4413
Own Kudos [?]: 68 [0]
Given Kudos: 0
Send PM
Re: The circle above has radius 8, and AD is parallel [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep 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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
[#permalink]
Moderators:
Moderator
1085 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne