Alternative Approach for 'Hard' Geometry Q

Greg solved this problem using the choose numbers method. I’m wondering if any folks can show/explain the algebraic way? I tried it, applying 45-45-90 rules, but didn’t arrive seem to arrive at anything resembling the answers. I’m sure there is something simple I’m missing.


  • AB = x
  • BC = kx
  • CD = x

So the area of the shaded region would be 1 - (Area of the small triangles) - (Area of the large triangles)

Small triangles are ABL, GHF. Large triangles are JLH, FDB.

  • Area of ABL = (1/2)* base* height = \frac{x^2}{2}
  • Area of JLH = (1/2)* base* height = \frac{(x+kx)^2}{2} = \frac{x^2*(1+k)^2}{2}

So finally, area = 1 - \frac{x^2}{2} - \frac{x^2}{2} - \frac{x^2*(1+k)^2}{2} - \frac{x^2*(1+k)^2}{2}

Simplify and voila!

Thank you so much, I see what I did now!