Thursday, December 23, 2010

Problem 552: Parallel lines, Transversal, Angles

Geometry Problem
Click the figure below to see the complete problem 552.

Problem 552: Parallel lines, Transversal, Angles.
Zoom
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 8:15 PM
Labels: , ,

2 comments:

  1. c .t . e. oDecember 24, 2010 at 1:02 PM

    name A, B, C, D, E, F vertexes of a, b, c, d, e, f
    extend AB to P, P on L2, FE to Q, Q on L1
    ED to G, G on AP, BC to H, H on FQ
    a + 180 - f + e + PGE = 180 - a + b + 180 - f + BHQ
    2a + e + PGE = 2f + b + BHQ (1)
    PGE = 180 + c - b - d
    BHQ = 180 - c - e + b
    substitute to (1)
    2c + 2e + 2a = 2d + 2b + 2f
    c + e + a = d + b + f

    ReplyDelete
  2. Peter TranDecember 24, 2010 at 10:18 PM

    http://img843.imageshack.us/img843/3136/problem552.png

    Make polygon with 7 sides per attached pictures
    Summation all internal angles=5*180=900
    90-f+e+360-d+c+360-b+a+90=900
    So a+c+e=b+d+f

    Peter Tran

    ReplyDelete

Subscribe to: Post Comments (Atom)