Two small identical circles of radii 5 cm and a big circle of radius 8 cm are embedded in rectangle ABCD. The circles are tangential to the rectangle. Given that AB = 34cm, find the area of rectangle ABCD in $cm^2$.
My attempt: Let the gap between the tangents (parallel to each other and perpendicular to ab) for the bigger and smaller circle be x. By symmetry $$ 34 = 10 + 10 + 16 - 2x\\\therefore x=1 $$ I couldn't seem to go further than this. Please help
Sorry for a slightly different labeling of the points.
Triangle POX is rectangle with sides $$OX=8+5=13,\; XP=CD/2=12,\; OP=\sqrt{13^2-12^2}=5.$$ Therefore, the short side of the rectangle is $$MO+OP+XC=8+5+5=18.$$