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Circles - I don't know OA

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Circles - I don't know OA [#permalink] New post 18 Nov 2009, 12:40
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Attachment:
perimeter.JPG
perimeter.JPG [ 9.24 KiB | Viewed 1582 times ]


Three circles with centers O, P and Q and radii 1 cm each are as shown in the figure above. What is the perimeter of the shaded region?
a) 2(Pi/3)
b) 6(Pi/3)
c) 10(Pi/3)
d) 4(Pi/3)
e) 8(Pi/3)
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Re: Circles - I don't know OA [#permalink] New post 18 Nov 2009, 14:33
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Refer to the figure I have attached.

Since line OA = Line AP = 1, A line drawn from A perpendicular to OP will bisect it.
Hence, RP = 0.5 cm.

Now, RP = 0.5 cm and AP = 1 cm

Let x be the angle formed between APR

Therefore, cos(x) = RP/AP = 0.5/1 = \(60^{\circ}\) = \(\pi/3\)

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
perimeter.png [ 27.18 KiB | Viewed 1448 times ]


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Last edited by sriharimurthy on 19 Nov 2009, 04:51, edited 1 time in total.
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Re: Circles - I don't know OA [#permalink] New post 18 Nov 2009, 15:22
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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: Circles - I don't know OA [#permalink] New post 18 Nov 2009, 15:50
:thanks
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!
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Re: Circles - I don't know OA [#permalink] New post 18 Nov 2009, 15:58
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! :-D
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compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

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

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Re: Circles - I don't know OA [#permalink] New post 19 Nov 2009, 04:19
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? :oops:

Thanks!
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Re: Circles - I don't know OA [#permalink] New post 19 Nov 2009, 04:52
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? :oops:

Thanks!


Done.

Let me know if any particular step needs clarification.

Cheers.
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compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

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

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Re: Circles - I don't know OA [#permalink] New post 13 May 2011, 07:16
2 * (2pi - 120/360 * 2pi) + 2pi - (2* 120/360 * 2 pi)

= 4pi(1 - 1/3) + 2pi - 4pi/3

= 8pi/3 + 2pi/3

= 10pi/3

Answer - C
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Re: Circles - I don't know OA [#permalink] New post 13 May 2011, 11:05
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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
Ques3.jpg [ 6.8 KiB | Viewed 1026 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.

Required perimeter = 3*2*pi - 4*(1/3)*2*pi = (10/3)pi
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Re: Circles - I don't know OA [#permalink] New post 13 May 2011, 11:54
It is some what distorted , but Just showing how I saw the diagram to calculate the Perimeter.

I did not calculated any degrees just counted the parts
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Re: Circles - I don't know OA   [#permalink] 13 May 2011, 11:54
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