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In the image shown in the attachment, lines d and l are para

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bhandariavi
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In the image shown in the attachment, lines d and l are para [#permalink] New post   Updated on: 23 Jun 2013, 01:03
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In the image shown in the attachment, lines d and l are parallel, and d is a diameter of the circle. d and l intersect the circle at points A and B, and C and D respectively. If the ratio of minor arc CD to arc AB is 1:3, what is the ratio of the length of line l to line d?

A. 1/4
B. 1/3
C. 2/5
D. 1/2
E. π/6
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D

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Originally posted by bhandariavi on 09 Feb 2011, 08:35.
Last edited by Bunuel on 23 Jun 2013, 01:03, edited 2 times in total.
Renamed the topic and edited the question.
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Re: This one is really important! [#permalink] New post   09 Feb 2011, 08:52
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bhandariavi wrote:
In the image shown in the attachment, lines d and l are parallel, and d is a diameter of the circle. d and l intersect the circle at points A and B, and C and D respectively. If the ratio of minor arc CD to arc AB is 1:3, what is the ratio of the length of line l to line d?
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A. 1/4
B. 1/3
C. 2/5
D.1/2
E. π/6
Attachment:
untitled.PNG
untitled.PNG [ 8.6 KiB | Viewed 4863 times ]


As AB is a diameter then central angle AOB is straight line = 180 degrees (where O is the center);
Now, as (arc CD)/(arc AB)=1/3 then central angle COD=1/3*180=60 degrees --> triangle COD is equilateral (because CO=DO=radius and <COD=60 degrees) --> CD=r=d/2.

Answer: D.
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Re: This one is really important! [#permalink] New post   09 Feb 2011, 08:57
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Arc CD = ϴ/360 x 2πR

Arc AB = 180/360 x 2πR

Arc CD/Arc AB = ϴ/180

ϴ/180 = 1/3 or ϴ = 60

Now triangle OCD (where O is the centre)
OC = OD = R [Radius]

Thus angle OCD = angle CDO = x
In Triangle OCD
x + x + 60 = 180
or x = 60

Yipeee! The triangle is an equilateral triangle! So CD = R

AB = 2R

So CD/AB = 1/2
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Re: This one is really important! [#permalink] New post   09 Feb 2011, 08:58
Bunuel always beats me to it :(
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Re: This one is really important! [#permalink] New post   09 Feb 2011, 09:00
Once again great explanation from Bunuel..thanks !
Thanks Entwistle
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Re: This one is really important! [#permalink] New post   09 Feb 2011, 11:09
Once again Bunuel beats me to it :P
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Re: This one is really important! [#permalink] New post   09 Feb 2011, 11:46
Thanks to both of you for the image and a nice explanation.

Ans: "D"
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Re: This one is really important! [#permalink] New post   10 Feb 2011, 20:19
Interesting question. Getting to the 60degree part was easy. I took a minute to arrive at 1 :2 damn!
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Re: This one is really important! [#permalink] New post   10 Feb 2011, 20:56
Interesting question.. Great explanation too by bunuel....
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Re: In the image shown in the attachment, lines d and l are para [#permalink] New post   18 Jul 2014, 03:50
As the arc length is 1/3 of the semicircle , divide the semicircle in to 3 equal regions. You will now see that the angle at the center is 180/3 = 60 and also the triangle in the center will be equilateral as the two sides are equal as they are both radius of the circle.

You can now get the ratio as radius/diameter = 1/2 (Option D)
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Re: In the image shown in the attachment, lines d and l are para [#permalink] New post   05 Jan 2018, 18:33
Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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).

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Re: In the image shown in the attachment, lines d and l are para [#permalink]
05 Jan 2018, 18:33
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