A triangle consisting of a vertex and the midpoint of two opposite sides on the bottom surface is shown in the figure.

Through a triangular relationship

Sinα = a/ (under root number 2 (square of h+square of a))

and

sinα=r/(h-r)

solve

R = h/( 1+ (under root number 2 (square of h+square of a)) /a)

Volume of inscribed sphere

V=4πr cubic /3