In the diagram, square ABCD has side length 3. Find the area of triangle PDQ.
tan (angle ABP) = AP/AB = 1 / 3
Angle ABP + angle BPA = 90
Angle DPQ + angle BPA = 90
So angle ABP = angle DPQ
tan (ABP) = tan ( DPQ)
1/3 = DQ / 2
DQ = 2/3
Area of PDQ = (1/2) (PD)(DQ) = (1/2)(2)(2/3) = 2/3
+1459 
