GMAT Prep Question
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- jayhawk2001
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The key here is that the minor arc OP is twice the measure
of the inscribed angle (PRO). So, arc OP = 70 / 360 * 2* pi * r
Similarly, if you draw a line OQ, it will subtend the same angle and hence
QR will be of the same measure i.e. 70 / 360 * 2* pi * r.
So, arc OP + arc QR = 70 / 360 * 2*pi*9 * 2 = 7*pi
Arc length OR = 9*pi
So, minor arc PQ = 9*pi - 7*pi = 2*pi
This is actually a question in OG-11.
of the inscribed angle (PRO). So, arc OP = 70 / 360 * 2* pi * r
Similarly, if you draw a line OQ, it will subtend the same angle and hence
QR will be of the same measure i.e. 70 / 360 * 2* pi * r.
So, arc OP + arc QR = 70 / 360 * 2*pi*9 * 2 = 7*pi
Arc length OR = 9*pi
So, minor arc PQ = 9*pi - 7*pi = 2*pi
This is actually a question in OG-11.
- jayhawk2001
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We are given that line PQ is parallel to the diameter. So, the corresponding angles PRO and ROQ will be of the same value.moneyman wrote:Hi jayhawk..
Just wanted to know..How can you be sure that arc QR wold have the same inscribed angle??
Put another way, if you draw a line from P to center of the circle (lets
call it C) you will get a triangle PRC. Since PC = CR = radius, angles
RPC and angles CRP will be equal = 35 degrees. So, angle PCR = 110
degrees and so angle PCO = 70 degrees.
Similarly, angle QCR = 70 degrees. The remaining 180 - 70 - 70 is the
angle subtended by arc PQ which is 40 degrees.
We can hence calculate 40 / 360 * 2 * pi * 9 = 2 * pi