Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
May 2512:01 AM PDT
-11:59 PM PDT
On the 25th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PST
May 2110:00 AM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.
May 2211:00 AM EDT
-11:59 PM EDT
Take 30% off all study plans with code PICNIC - Expires on Monday, May 25th- Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
18 Nov 2009, 13:40
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
60% (02:39) correct
40%
(02:34)
wrong
based on 58
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
perimeter.JPG [ 9.24 KiB | Viewed 10714 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)
Updated on: 19 Nov 2009, 05:51
Originally posted by sriharimurthy on 18 Nov 2009, 15:33.
Last edited by sriharimurthy on 19 Nov 2009, 05:51, edited 1 time in total.
Last edited by sriharimurthy on 19 Nov 2009, 05:51, edited 1 time in total.
Kudos
Bookmarks
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
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 10514 times ]
18 Nov 2009, 16:22
Kudos
Bookmarks
sriharimurthy
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!!!!
