The top and base are parallel and the trapezoid can be split into 3 figures: two triangles and a rectangle. The triangular "wings" and the rectangle all have the same height, h. The area of the rectangle is 22h and the sum of the areas of the triangles is (1/2)hx+(1/2)h(10-x)=(1/2)h(x+10-x)=5h, where x is the base of one triangle and 32-22-x the base of the other triangle. Note how x cancels out. The total area is therefore 22h+5h=27h. We bring in x again and use Pythagoras on both triangles. h^2=6^2-x^2 and h^2=8^2-(10-x)^2, so 36-x^2=64-(100-20x+x^2)=-36+20x-x^2. The x^2 terms cancel out, and 72=20x, so x=72/20=18/5. h^2=36-x^2=36-324/25=576/25, so h=24/5 (alternatively, h^2=64-1024/25=576/25) and the whole area=27*24/5=648/5=129.6 sq units.