Since C is the midpoint of ao, AC = Co =1/2bo =1/4ab = 2.

Then, from the intersection theorem (projective, pythagorean and trigonometric functions also become) dc? =ac*cb= 12

So the root number 3 of dc=2

It's easier to say than to do.

Because do=2co

So ∠ aod = 60.

S sector aod= 1/6S circle =8/3π.

S semi-arch acd=S sector aod-S△doc=8/3π-2 root number 3

Then calculate the right-angle sector ced.

= 1/4* 12π=3π

S semicircle =8π

The last big one-two small ones =7/2π+2 radical number 3

(quite annoying, I don't rule out clerical errors)