2011 AMC 12B Problem 18

Let the apex be $A$ and the base be square $BCDE.$ Then $AB=AD=1$ and $BD=\sqrt2,$ so $\triangle BAD$ is an isosceles right triangle. Let the cube have edge length $x.$ Its intersection with the plane of $\triangle BAD$ is a rectangle of height $x$ and width $\sqrt2\,x,$ whose top corners lie on $AB$ and $AD.$ Because the legs $AB$ and $AD$ meet the base at $45^\circ,$ each portion of $BD$ outside the rectangle has length $x,$ so $$\sqrt2=BD=\sqrt2\,x+2x,$$ which reduces to $x=\dfrac{\sqrt2}{2+\sqrt2}=\sqrt2-1.$ The volume is $$(\sqrt2-1)^3=5\sqrt2-7.$$ Thus, the correct answer is A.