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First, draw line PO to obtain triangle OPR. We know that a triangle inscribed a half circle with one side being the diameter of a circle is always a right triangle. This means angle OPR is a right angle. Given that angle PRO is 35, we know that angle POR is 180-90-35 = 55
Set center point of circle = C Since POR = POC = 55, this means OPC = 55, and PCO = 180 - 55 - 55 = 70. Using symmetry, we know that angle PCQ is 180 - 70 - 70 = 40
To find the length of minor arc PQ, we need to get the angle extended by the arc at the center of the circle. Name the center as C. We need to find the angle PCQ and substitute in the formula (θ/360)*2πr where θ is the angle subtended by the arc at the center.
Since the lines OR and PQ are parallel and Angle ORP is 35 degrees, as per alternate angles rule, angle RPQ is 35 degrees.
Draw a line from P to C. Line PC is equal to CR, which is the radius i.e 9. Now consider the triangle RPC. In this triangle, 2 sides are equal PC and CR and the angle CRP is 35 degrees. As per the rules of isosceles triangle, angle RPC must also be 35 degrees.
So, the total angle CPQ is 70 degrees.
Draw a line from Q to C. So, line PC is equal to line QC. If angle CPQ is 70 degrees, then angle PQC will also be 70 degrees.
Finally, angle PCQ is 180-140 = 40 degrees
Substitute 40 in the formula (θ/360)*2πr = (40/360)*(2*π*9) = 2π
The answer is (A)
Re: Circle Question
16 May 2011, 21:20