A cubical cake with edge length $2$ inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where $M$ is the midpoint of a top edge. The piece whose top is triangle $B$ contains $c$ cubic inches of cake and $s$ square inches of icing. What is $c+s$? $\text{(A) } \frac{24}{5} \qquad$ $ \text{(B) } \frac{32}{5} \qquad$ $ \text{(C) } 8+\sqrt5 \qquad$ $ \text{(D) } 5+\frac{16\sqrt5}{5} \qquad$ $ \text{(E) } 10+5\sqrt5$
Solution
Set the top face as the square with corners $(0,0),(2,0),(2,2),(0,2)$ and $M=(0,1)$ the midpoint of the left edge. Cut $1$ is the segment from $M$ to the far corner $(2,2)$.
Cut $2$ drops from $(2,0)$ perpendicular to cut $1$, meeting it at $P=\left(\frac65,\frac85\right)$.
Triangle $B$ has vertices $P,(2,2),(2,0)$; its (top) area is $\frac45$, so the piece's volume is $c=\frac45\cdot2=\frac85$.
Its icing is the top ($\frac45$) plus the one exterior side face, the right edge of length $2$ times height $2$: $s=\frac45+4=\frac{24}5$.
So $c+s=\frac85+\frac{24}5=\frac{32}5$. The answer is (\textbf{B}).
A cubical cake with edge length $6$ inches is iced on the top and the four sides. Label the corners of its top face $P,Q,R,S$ in order, and let $M$ be the midpoint of $\overline{PS}$. The cake is cut straight down along $\overline{MQ}$ and $\overline{MR}$, giving three pieces. The piece whose top is triangle $PMQ$ contains $c$ cubic inches of cake and $s$ square inches of icing. What is $c+s$?
$\text{(A) } \frac{24}{5} \qquad$ $ \text{(B) } \frac{32}{5} \qquad$ $ \text{(C) } 8+\sqrt5 \qquad$ $ \text{(D) } 5+\frac{16\sqrt5}{5} \qquad$ $ \text{(E) } 10+5\sqrt5$ 
