Friday, February 11, 2011

Problem 578: Circle, Perpendicular Chords, Midpoints, Curvilinear Triangle, Area, 90 Degrees

Geometry Problem
Click the figure below to see the complete problem 578.

Problem 578: Circle, Perpendicular Chords, Midpoints, Curvilinear Triangle, Area, 90 Degrees.
Zoom at: Geometry Problem 578

Posted by Antonio Gutierrez at 6:48 AM
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2 comments:

  1. c .t . e. oFebruary 11, 2011 at 8:10 AM

    draw diameter MN perpendicular to AB,
    diameter KL perpendicular to CD
    A1 - CGOM -2A5 - OFBL = A3 - AKGE + CGOM
    A1 - A3 = 2A5 + 2CGOM + OFBL - AKGE (1)
    AKGE + 2A5 = OFBL (2)
    A2 + EFND = A4 - EFND
    A2 - A4 = - 2EFND (3)
    A1 + A2 - A3 - A4 = 2A5 + (OFBL - AKGE)+ (2CGOM -2EFND)
    A1 + A2 - A3 - A4 = 2A5 + 2A5 + 4A5
    A1 + A2 - A3 - A4 = 8A5

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  2. Peter TranFebruary 11, 2011 at 9:53 AM

    http://img832.imageshack.us/img832/1769/problem578.png

    Let O is the center of the circle.
    Draw segments HFG and ML symmetric of CED and EF over center O ( see picture)
    By observation we have A1-A3= A(EFHC) and A2-A4= -A(EFGD)
    But A(EFGD)=A(CMLH) ( symmetric property)
    So A1-A3+A2-A4 = A(EFHC)-A(CMLH)=A(MEFL)= 8x A5

    Peter Tran

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