The ancient Greeks, being masters of geometric manipulation, often tried their hand at "squaring" various shapes. This involved using only the most fundamental rules of geometry to construct a square whose area equals the area of the original shape.

Can you follow in their footsteps and square a simple triangle?

The solution must hold for all types of triangles.

(In reply to Solution by Old Original Oskar!)

  P      A          Q
  
                      
          C

I get it! The crux is that Angle PCQ is 90 degrees. With PA=b, AQ=h/2 = c (say), and AC=x, we have PC=sqrt(b^2+x^2), CQ=sqrt(c^2+x^2) so that PQ=b+c=sqrt(b^2+c^2+2*x^2).  Squaring both sides of the last equation and cancelling, x^2=bc=bh/2 as desired.

Nice uncluttered solution!

Posted by Richard on 2004-09-22 15:09:30