Saturday, September 29, 2012

Problem 808: Parallelogram, Perpendicular, Right Triangle, Semicircle Area, Diameter

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 808.

Online Geometry Problem 808: Parallelogram, Perpendicular, Right Triangle, Semicircle Area, Diameter.

Posted by Antonio Gutierrez at 3:59 PM
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2 comments:

  1. Jacob HA (EMK2000)September 29, 2012 at 5:20 PM

    Let the area of segment FG be k.

    From Pythagoras theorem,
    AE^2 + DE^2 = AD^2 = BC^2

    Multiply both side by π/4,
    (π/4)AE^2 + (π/4)DE^2 = (π/4)BC^2

    S1 + (S2 + k) = S + k
    S1 + S2 = S

    ReplyDelete
  2. Peter TranSeptember 29, 2012 at 6:54 PM

    http://imageshack.us/a/img141/4056/problem808.png
    Let S3 is the red area ( see sketch)
    In right triangle AED we have pi/4(AD^2)=pi/4(AE^2+ED^2)
    So S+S3=S1+S2+S3 => S=S1+S2

    ReplyDelete

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