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

Posted by Antonio Gutierrez at 8:11 PM
Labels: , ,

4 comments:

  1. c .t . e. oAugust 21, 2010 at 12:45 PM

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

    ReplyDelete
  2. Peter TranAugust 21, 2010 at 1:02 PM

    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

    ReplyDelete
  3. AnonymousAugust 26, 2010 at 5:03 PM

    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.

    ReplyDelete
  4. APOSTOLIS MANOLOUDISJune 21, 2016 at 2:14 AM

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

    ReplyDelete

Subscribe to: Post Comments (Atom)