Right prism with octagon base

Relevant wiki: Volume - Problem Solving - Medium

Consider my diagram, the area of the base of this right prism is equal to the area of the 10 cm x 10 cm square minus the area of the shaded region which is also equivalent to two squares of side length $x$ . Considering one shaded part and by applying pythagorean theorem, we get

$10-2x=x\sqrt{2}$

$10=2x+x\sqrt{2}$

$x=\dfrac{10}{2+\sqrt{2}}$

$shaded~area=2\left(\dfrac{10}{2+\sqrt{2}}\right)^2=2\left(\dfrac{100}{4+4\sqrt{2}+2}\right)=\dfrac{200}{6+4\sqrt{2}}$

$base~area=10^2-\dfrac{200}{6+4\sqrt{2}}=100-\dfrac{200}{6+4\sqrt{2}}$

Therefore, the desired volume is

$v=base~area~\times~height=\left(100-\dfrac{200}{6+4\sqrt{2}}\right)(15)\approx \color{#D61F06}\boxed{1242.6~cm^3}$