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Let $MA=a$ and $MD=d.$ It follows that $MC=a+2$ and $MB=a+4.$ As shown below, note that $△MAD$ and $△MBC$ are both right triangles. By the Pythagorean Theorem, we have \begin{alignat}{6} AD^2 &= MA^2 - MD^2 &&= a^2 - d^2, \\ BC^2 &= MB^2 - MC^2 &&= (a+4)^2 - (a+2)^2. \end{alignat} Since $AD=BC$ in rectangle $ABCD,$ we equate the expressions for $A{D}^2$ and $B{C}^2,$ then rearrange and factor: $$\begin{array}{rl}{a}^2-d^2& =(a+4{)}^2-(a+2{)}^2\\ a^2-d^2& =4a+12\\ a^2-4a-d^2& =12\\ (a-2{)}^2-d^2& =16\\ (a+d-2)(a-d-2)& =16.\end{array}$$ As $a+d-2$ and $a-d-2$ have the same parity, we get $a+d-2=8$ and $a-d-2=2,$ from which $(a,d)=(7,3).$ Applying the Pythagorean Theorem to right $△MAD$ and right $△MCD,$ we obtain $AD=2\sqrt{10}$ and $CD=6\sqrt{2},$ respectively. Let the brackets denote areas. Together, the volume of pyramid $MABCD$ is $$\frac{1}{3}\cdot [ABCD]\cdot MD=\frac{1}{3}\cdot (AD\cdot CD)\cdot MD=\boxed{24\sqrt{5}}.$$