It is known that the circle O and the circle K are the big circle and the small circle of the ball o, and their common chord length is equal to the radius of the ball o, OK = 3 / 2, and the angle between the circle O and the plane of the circle K is 60 degrees, Then the surface area of ball 0 is?

It is known that the circle O and the circle K are the big circle and the small circle of the ball o, and their common chord length is equal to the radius of the ball o, OK = 3 / 2, and the angle between the circle O and the plane of the circle K is 60 degrees, Then the surface area of ball 0 is?

The surface area is 4 π

AB and CD are two strings on ⊙ o respectively. The distances from the center O to them are OM and on respectively. If AB > CD, what is the relationship between the size of OM and on?
Be more detailed and urgent

Link OA, OC, then
OM²=OA²-AM²
ON²=OC²-CN²
OA=OC
AM=AB/2 CN=CD/2
AB>CD
∴OM²

AB and CD are two strings on the circle O respectively. The distances from the center O to them are OM and on respectively. If AB > CD, what is the relationship between the size of OM and on? Why?

OM＜ON
prove:
Connect OA, OC
∵ OM and on are chord centers of strings AB and CD respectively
That is om ⊥ AB, on ⊥ CD
‖ am = 1 / 2Ab, CN = 1 / 2CD (vertical diameter theorem)
∵AB＞CD
∴AM＞CN
According to Pythagorean theorem:
OM^2=OA^2-AM^2
ON^2=OC^2-CN^2
∵ OA = OC = ⊙ o radius
∴OM＜ON