2009 AMC 10B Problem 22

Set the top face as a $2\times2$ square. The cut from $M$ toward the far corner creates the top triangle $A$ with legs $1$ and $2,$ so area $1$ and hypotenuse $\sqrt5.$ Triangle $B$ is similar to $A$ but with hypotenuse $2,$ so its area is $\left(\dfrac{2}{\sqrt5}\right)^2\cdot1=\dfrac45.$ Since the cake has height $2,$ the volume is $c=\dfrac45\cdot2=\dfrac85.$ The icing on this piece is its top ($\dfrac45$) plus the full cube side face it borders ($2\times2=4$), so $s=\dfrac45+4=\dfrac{24}{5}.$ Therefore $c+s=\dfrac85+\dfrac{24}{5}=\dfrac{32}{5}.$ Thus, the correct answer is B.