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18 Nov 2009, 13:40
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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
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Attachment:
perimeter.JPG [ 9.24 KiB | Viewed 10754 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.
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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
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 10555 times ]
18 Nov 2009, 16:22
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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!!!!
