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13 Dec 2009, 12:49
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50% (00:00) correct
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Got the following question. Wondering how to solve it
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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13 Dec 2009, 13:05
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msunny
Lets say the circle has center C
PCO angle = 70 (by definition since PRO = 35)
Since lines are parallel QCR also has 70
OR = 180
70 + 70 + 180 = 320 therefore PCQ = 40 degrees
circumference = 18pi
40/360 * 18pi = 2pi
17 Dec 2009, 00:34
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Hi lagomez,
when two lines are parallel and there are two transversals cuttting them, then it doesn't necessarily mean that the angles should be equal i.e. anlge PCO is not necessary equal to angle PRO.
My method was :
Let C be the centre of circle
Consider triangle CPR, since CP=CR(radius), angle CPR=angle CRP=35
which means that angle CPQ=70
Consider triangle CPQ.
Since CP=CQ=radius, angle CPQ=ANGLE CQP=70
Therefor,angle PCQ=180-140=40
Hence, length of minor arc PQ=9*(40pi/180)=2pi
when two lines are parallel and there are two transversals cuttting them, then it doesn't necessarily mean that the angles should be equal i.e. anlge PCO is not necessary equal to angle PRO.
My method was :
Let C be the centre of circle
Consider triangle CPR, since CP=CR(radius), angle CPR=angle CRP=35
which means that angle CPQ=70
Consider triangle CPQ.
Since CP=CQ=radius, angle CPQ=ANGLE CQP=70
Therefor,angle PCQ=180-140=40
Hence, length of minor arc PQ=9*(40pi/180)=2pi
lagomez
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
