Let ABCD be a square and X a point such that A and X are on opposite sides of CD. The lines AX and BX intersect CD in Y and Z respectively. IF the area of ABCD is one and area XYZ = 2/3 what is the length of YZ.
I worked the area of trapezium ABYZ to equal YZ: Area of square not covered by triangle = (1-YZ)(1) reason (rectangle Length*Breadth) Therefore area of trapezium = 1-(Area of rectangle) = 1-(1-YZ) = YZ. Area of trapezium = [(a+b)/2]*h Therefore: YZ = (1+YZ)/2 (h=1) YZ = 1.
Where did i go wrong.
EDIT please do not give the answer