+400 Square $ABCD,$ shown here, has side length $6$ units. Points $P$ and $Q$ are located on sides $AD$ and $BC,$ respectively, with $AP = BQ = 1$ unit. Triangles $ACP$ and $BDQ$ overlap in the square to form the shaded quadrilateral. What is the area of the shaded quadrilateral? Express your answer as a common fraction.
+129850 
EF parallel to QB
So triangle DEF is similar to triangle DQB
F is the midpoint of DB (the diagonal of the square)
So
DF / DB = EF / QB
(1/2)DB / DB = EF / QB
(1/2) = EF / 1
EF =1/2
And EF is the diagonal of the square formed by the overlap
So.....the side of the squre = (1/2)(1/sqrt2) = 1/ sqrt 8
Area of the quadrialeral = (1/sqrt8)^2 = 1/8
