Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Pedan Trapezium


Date: 05/23/2000 at 03:44:46
From: Cherriblossom
Subject: Pedan trapezium

1) The lengths of the parallel sides p1p3 and p5p4 of the isosceles 
trapezium p1p3p4p5 are 7 and 4 respectively, with the slanted sides 
having length 6. Verify that his trapezium is a Pedan trapezium.

2) We will construct a convex pentagon p1p2p3p4p5 based on the 
trapezium in part (1). The vertex p2 is located by constructing 
triangle p5p1p2 congruent to triangle p1p5p4. Prove that p1p2p3p4 is 
concyclic and in fact that p1p2p4p5 is an isosceles trapezium.

3) Prove that: 
   a) triangle p2p3p4 is congruent to triangle p2p1p5
   b) p1p2p3p4p5 is a Pedan pentagon.

Please help me; I don't understand what the question wants.
Thank you.


Date: 05/24/2000 at 11:22:43
From: Doctor Floor
Subject: Re: Pedan trapezium

>1) The lengths of the parallel sides p1p3 and p5p4 of the isosceles 
>trapezium p1p3p4p5 are 7 and 4 respectively, with the slanted sides 
>having length 6. Verify that his trapezium is a Pedan trapezium.

Let's consider the figure:

              4
         p5------p4
        / |      | \
      6/  |      |  \6    XYp4p5 is a rectangle
      /   |      |   \
     p1---X------Y---p3
       1.5    4   1.5

By Pythagoras' theorem

      Xp5 = sqrt(6^2-1.5^2) = sqrt(33.75)

Again by Pythagoras' theorem 

     p3p5 = sqrt(Xp3^2 + xP5^2)
          = sqrt(5.5^2 + 33.75)
          = sqrt(64)
          = 8

In the same way p1p4 = 8.

By Ptolemy's/Pedan's formula we should have:

     p1p3*p4p5 + p1p5*p3p4 = p3p5*p1p4

Inserting the found values we get:

   7*4 + 6*6 = 8*8

which is correct.


>2) We will construct a convex pentagon p1p2p3p4p5 based on the 
>trapezium in part (1). The vertex p2 is located by constructing 
>triangle p5p1p2 congruent to triangle p1p5p4. Prove that p1p2p3p4 is 
>concyclic and in fact that p1p2p4p5 is an isosceles trapezium.

From the congruence we know that p1p2 = p5p4 = 4 (and also that 
p2p5 = p1p4 = 8).

Consider the perpendicular bisector L of p1p5. Through L we can 
reflect p1 to p5 and conversely p5 to p1. Since triangles p5p1p2 and 
p1p5p4 are congruent, the complete triangles are reflections of each 
other through L. So p2 is the reflection of p4. That means that p2p4 
is perpendicular to L and thus parallel to p1p5.

So p1p2p4p5 is an isosceles trapezium and hence concyclic.


>3) Prove that: 
>   a) triangle p2p3p4 is congruent to triangle p2p1p5
>   b) p1p2p3p4p5 is a Pedan pentagon.

We know that p1p5 = p3p4 = 6 because of (1).

We can assume compute p2p4 by applying Ptolemy's theorem to p2p4p5p1:

     p2p4*p1p5 + p2p1*p4p5 = p2p5*p1p4
     p2p4*6    + 4*4       = 8*8
     p2p4                  = (64-16)/6
                           = 8

So p2p4 = p2p5 = 8

Finally, we can apply Ptolomy's/Pedan's formula to p2p3p4p1, because 
we know that this quadrilateral is cyclic. This gives:

     p2p3*p1p4 + p1p2*p3p4 = p1p3*p2p4
     p2p3*8    + 4*6       = 7*8
     p2p3                  = (56-24)/8
                           = 4

So p2p3 = p1p2 = 4.

This proves the congruence by SSS.

I am not sure what a Pedan pentagon is, but perhaps now you can prove 
it.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/