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Note : An easier way to approach this is that since AO = OP = AP = 1 cm, Triangle AOP will be equilateral and will have all sides equal to \(60^{\circ}\). Thus angle APR will be equal to \(60^{\circ}\) or \(\pi/3\)

Now, arc OA = l*x = 1*\(\pi/3\) = \(\pi/3\)

Now, perimeter of the shaded region = perimeter of all 3 circles - 8*(length of arc OA)

Note: we have to compensate for 8 arcs which are similar in length to OA

Perimeter of shaded region = 3*(2\(\pi\)r) - 8*\(\pi/3\) = \(10(\pi/3)\)

Answer : C

Attachments

perimeter.png [ 27.18 KiB | Viewed 2657 times ]

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Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

Last edited by sriharimurthy on 19 Nov 2009, 05:51, edited 1 time in total.

Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.

Re: Three circles with centers O, P and Q and radii 1 cm each are as shown [#permalink]

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18 Nov 2009, 16:22

1

This post received KUDOS

sriharimurthy wrote:

Refer to the figure I have attached.

Answer : C

Figure was missing but, Gotcha!!! Very very impressive!!!!

AOP is equilateral triangle since AP = OA = OP = 1cm. Same is the case downwards, let's call it B. Angle AOB will be 120 degree and so also angle APB will be 120 degree. Again the same with center Q as well.

So we see 120 degrees of circle with center O is in intersection region. 120 degrees of circle with center Q is in intersection region and 2 of 120 degrees of circle with center P are in intersection regions.

Therefore, 240 degrees of arc in circle O + 240 degrees of arc in circle Q + 120 degrees of arc in circle P will be the perimeter of the shaded region.

srihari you are just awesome in quant buddy!!!!
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Re: Three circles with centers O, P and Q and radii 1 cm each are as shown [#permalink]

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18 Nov 2009, 16:50

Sorry about the figure.. forgot to attach it. Didn't really seem necessary though! You seem to have got a pretty good hang of the question yourself now! Anyway, the more you practice, the easier these questions will become since you'll know what to look for. All the best!
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

Re: Three circles with centers O, P and Q and radii 1 cm each are as shown [#permalink]

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18 Nov 2009, 16:58

Quote:

srihari you are just awesome in quant buddy!!!!

I'm guessing you haven't come across Bunuel yet... He makes most of us seem like amateurs!
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

Re: Three circles with centers O, P and Q and radii 1 cm each are as shown [#permalink]

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19 Nov 2009, 05:52

h2polo wrote:

Nice question SensibleGuy! +!

sriharimurthy... great explaination! +1

Do you think you can still attach your figure for the rest of us that seem to have trouble following?

Thanks!

Done.

Let me know if any particular step needs clarification.

Cheers.
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

You can solve it quickly if you realize that the required perimeter is 3*perimeter of each circle - 4*arc in red (in the figure)

Attachment:

Ques3.jpg [ 6.8 KiB | Viewed 2234 times ]

Each circle is identical with radius 1 so perimeter of a circle is 2*pi

Triangle APQ is equilateral because each side is 1 (each side starts at the center of a circle and touches the circumference) so angle AQP is 60 which means angle AQB is 120. Hence each arc is a third of the perimeter of the circle.

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