Cyclic TrapezoidDate: 01/17/2002 at 03:12:13 From: varghese Subject: Circles PQ is a diameter; AB is a chord parallel to PQ. If PQ=50cm and AB= 14cm, find PB. I used the similarity of triangles, i.e. triangle AOB and triangle POQ are similar: PQ/AB = 50/14 = 25/7 = PO/OB . PO+OB = PB PB = 25+7 = 32 I am not satisfied with this way of doing it. Can you please help? Date: 01/17/2002 at 06:04:08 From: Doctor Floor Subject: Re: Circles Hi, Thanks for your question. I am not sure what you mean by O, but I think it is the center of the circle. In that case I cannot see that triangles AOB and POQ are similar. To find the correct answer, we have to see that PQBA is an isosceles trapezoid: A_---14--_B /90`-._.-'90\ /y _.-' `-._ y\ /_.'x x`._\ P--------50------Q Explanation of the figure: In this figure angles QPB, QAB, PBA, PQA are all congruent (indicated by x, which is meant to fit in the angles QAB and PBA), because they are inscribed angles on congruent chords. The same goes for angles APB and AQB (indicated by y). Angles PAQ and PBQ are 90 degrees, since they are inscribed angles on diameter PQ. In triangle PQB we know that the sum of the angles is 180 degrees, so that y = 90-2x. Let PB = t. In triangle PBQ we have: cos x = t/50 ..... [1] Using the Law of Sines in triangle APB we find t/sin(90+x) = 40/sin(90-2x) t/cos x = 40/cos(2x) Now we substitute [1] in the left-hand side: 50 = 40/cos(2x) cos(2x) = 4/5 2cos^2(x) - 1 = 4/5 cos^2(x) = 9/10 Again we substitute [1] (t/50)^2 = 9/10 t^2 = 2250 t = 15sqrt(10) So we conclude that PB = t = 15sqrt(10). If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ Date: 01/19/2002 at 06:23:04 From: varghese Subject: Circles Sir, The question was misunderstood, I think. Since PQ is a diameter of a circle, and AB is a chord parallel to PQ, ABPQ becomes a cyclic trapezoid. O is the point of intersection of the diagonals PB and QA of the trapezoid, so by the AA corollary triangles AOB and POQ are similar. Thank you, sir. Date: 01/19/2002 at 13:45:23 From: Doctor Floor Subject: Re: circles Hi, again, Thanks for your reaction. I did not misunderstand the question, except for the point O. My calculations were correct. More specifically, it is not difficult to see that the isosceles trapezoid is also a cyclic trapezoid. Let PQBA be the cyclic trapezoid. Draw diagonal AQ. A_--------B / `-._ \ / `-._ \ / `._\ P-----------------Q Then angles PQA and QAB are congruent ("Z-angles"), so the arcs QB and AP of the circle are congruent, and thus the chords QB and AP are congruent as well. I hope for the rest you can do with the answer I already sent. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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