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# Circles - I don't know OA

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Circles - I don't know OA [#permalink]

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18 Nov 2009, 13:40
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0% (00:00) correct 100% (00:58) wrong based on 3 sessions

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Attachment:

perimeter.JPG [ 9.24 KiB | Viewed 2005 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]

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18 Nov 2009, 15:33
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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*$$\pi/3$$ = $$10(\pi/3)$$

Attachments

perimeter.png [ 27.18 KiB | Viewed 1868 times ]

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Last edited by sriharimurthy on 19 Nov 2009, 05:51, edited 1 time in total.
Senior Manager
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Re: Circles - I don't know OA [#permalink]

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18 Nov 2009, 16:22
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sriharimurthy wrote:
Refer to the figure I have attached.

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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Manager
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Re: Circles - I don't know OA [#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!
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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

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

Manager
Joined: 29 Oct 2009
Posts: 211
GMAT 1: 750 Q50 V42
Followers: 99

Kudos [?]: 1216 [0], given: 18

Re: Circles - I don't know OA [#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!
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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

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

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Re: Circles - I don't know OA [#permalink]

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19 Nov 2009, 05:19
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?

Thanks!
Manager
Joined: 29 Oct 2009
Posts: 211
GMAT 1: 750 Q50 V42
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Re: Circles - I don't know OA [#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

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

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Re: Circles - I don't know OA [#permalink]

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13 May 2011, 08:16
2 * (2pi - 120/360 * 2pi) + 2pi - (2* 120/360 * 2 pi)

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

= 8pi/3 + 2pi/3

= 10pi/3

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Re: Circles - I don't know OA [#permalink]

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13 May 2011, 12:05
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Expert's post
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 1445 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]

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13 May 2011, 12:54
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
Attachments

file.jpg [ 15.99 KiB | Viewed 1413 times ]

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Re: Circles - I don't know OA   [#permalink] 13 May 2011, 12:54
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