Let's do this first

Connect O d, O e, O O, o C.

O, O', C*** line ... (= = This step is difficult to explain, I think)

And O'O must be divided equally ∠AOB.

∫ao and bo are tangents

∴O'D,O'E⊥AO,BO

∠∠AOO '，∠BOO'=30

∴O'D:O'O= 1:2

Then the radius of the circle will be calculated, right? 4π, the radius is 4π divided by π and then the root sign, which is 2.

∴OO'=4

O'C=2

∴0C=6

Okay, there's a sector radius and angle, and then there's a formula.

Transverse area = 6? π×60 /360 =6π

Surface area ... you have to put a circle under it. Okay = =

Arc ACB = 2× 6 π× 60/360 = 2 π.

∴ The circumference of the circle below =2π

Backstepping radius = 1 (with its own formula)

Then come back. The area of the circle below: 1? π=π

∴ Surface area =π+6π=7π

Always asking about height, right?

The radius of the lower circle is 1, and then the sector bus is 6. Come on, Pythagorean theorem.

6? = 1? +h？

36= 1+h？

h？ =35

H= root number 35 (seemingly impossible to divide)

Okay, it's over.