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Cyclic Trapezoid


Date: 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/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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