All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
A Parallelogram Puzzle (Posted on 2007-05-01) Difficulty: 2 of 5
Consider the parallelogram PQRS with PS parallel to QR and PQ parallel to SR. The bisector of the angle PQR intersects PS at T.

Determine PQ, given that TS = 5, QT = 6 and RT = 6.

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Harder solution | Comment 2 of 5 |

Call the measure of angle PQT by a.  Call the length of PQ=SR=x
Angle PTQ is also a which means PT is also x.
Angle STR is also a.

By applying the law of cosines on triangles PQT and STR we have:

x^2 = x^2 + 6^2 - 2*x*6*cos(a)
x^2 = 5^2 + 6^2 - 2*5*6*cos(a)

Multiply the top equation by 5 and the bottom by -x and adding yields:

5x^2 - x^3 = 5x^2 -25x + 180 - 36x
x^3 - 61x + 180 = 0

This cubic has 3 real roots {-9,4,5}

x cannot equal -9 of course, it also cannot equal 5 because that would make triangles STR and PTQ congruent  which would make PSRQ a rectangle.

So x=4

  Posted by Jer on 2007-05-01 12:13:28
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information