Surface Area and Volume of a Folded Hexagonal Solid

In a regular hexagonal prism $ABCDEF\text{-}A_1B_1C_1D_1E_1F_1$ (Fig. 1) the base edge length is $1$ and the lateral edge length is $2$. Points $B_2,D_2,F_2$ are given by $\overrightarrow{BB_2}=\lambda\overrightarrow{BB_1}$, $\overrightarrow{DD_2}=\lambda\overrightarrow{DD_1}$, $\overrightarrow{FF_2}=\lambda\overrightarrow{FF_1}$ with $0<\lambda<1$. From the prism, cut away the three tetrahedra $B_2\text{-}ABC$, $D_2\text{-}CDE$, $F_2\text{-}EFA$, then fold triangles $\triangle ACB_2$, $\triangle CED_2$, $\triangle EAF_2$ upward through $180^\circ$ about $AC$, $CE$, $EA$ respectively, so that $B_2,D_2,F_2$ coincide at a single point $P$; this yields the curved-top polyhedron in Fig. 2.

Let $S_1,V_1$ be the surface area and volume of the original prism, and let $S_2,V_2$ be the surface area and volume of the resulting polyhedron when $\lambda=\dfrac14$. Determine, with justification, the relationship between $S_1$ and $S_2$, and between $V_1$ and $V_2$.