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 [ 27.18 KiB | Viewed 1894 times ]

_________________

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