Friday, August 20, 2010

Problem 508: Triangle with three Squares, Equal Area

Geometry Problem
Click the figure below to see the complete problem 508 about Triangle with three Squares, Equal Area.

Problem 508. Triangle with three Squares, Equal Area


See also:
Complete Problem 508

Level: High School, SAT Prep, College geometry

4 comments:

  1. Paralleogram BGTD => ▲BGT = ▲A3 = ▲A1 see P502
    ( ▲M3TG = ▲M3BD, M3 midpoint of DG )
    In the same way A2 = A1, A4 = A1

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  2. A1=area(ABC)=1/2AB.BC.sin(ABC)=1/2.AB.AC.sin(BAC)=1/2.BC.AC.sin(BCA)
    but sin(ABC)=sin(DBG) (ABC) supplement to (DBG)
    and sin (BAC)=sin(EAJ)
    sin(BCA)=sin(FCH)
    from above we can conclude that A1=A2=A3=A4

    Peter Tran

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  3. Rotate triangle CFH 90 deg. around point C.
    Point F moves on Point B, H moves to H'.
    A, C, H' are collinear, and AC=CH', so A1=A4.

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  4. Area ABC is (ABC).But <ABC+<DBG=180 so (ABC)/(DBG)=(AB.BC)/(BD.BG)=1 so A1=A3.
    Similar A1=A2=A4.

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