Pedan TrapeziumDate: 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/ |
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