Problem Solving

Please explain how to solve this.

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.

Hi jayhawk..

Just wanted to know..How can you be sure that arc QR wold have the same inscribed angle??

moneyman wrote:Hi jayhawk..

Just wanted to know..How can you be sure that arc QR wold have the same inscribed angle??

We are given that line PQ is parallel to the diameter. So, the corresponding angles PRO and ROQ will be of the same value.

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

Thanks Jay..Was very helpful..Your explanation rocks..