Let $APQRS$ be a pyramid, where the base $PQRS$ is a square. The total surface area of pyramid $APQRS$ (including the base) is $800$. Let $W$, $X$, $Y$, and $Z$ be the midpoints of $\overline{AP}$, $\overline{AQ}$, $\overline{AR}$, and $\overline{AS}$, respectively. Find the total surface area of solid $PQRSWXYZ$ (including the bases). (This solid is called a frustum.)
Since $\overline{XW}\perp\overline{PQ}$ and $\overline{PQ}\perp\overline{QS}$, then $\overline{XW}\parallel\overline{QS}$.
[asy]
import three;
currentprojection = perspective(6,3,2);
triple W, X, Y, Z;
triple P = (0,0,0);
triple Q = (0,1,0);
triple R = (1,1,0);
triple S = (1,0,0);
triple A = (1,1,1);
W = midpoint(A--P);
X = midpoint(A--Q);
Y = midpoint(A--R);
Z = midpoint(A--S);
draw(P--Q--R--S--cycle);
draw(A--P--Q--R--S--cycle);
draw(W--X--Y--Z--cycle, dashed);
label("$P$",P,dir(-4,0,0));
label("$Q$",Q,N);
label("$R$",R,SE);
label("$S$",S,dir(-4,0,0));
label("$A$",A,NW);
draw(A--W);
label("$W$",W,E);
draw(A--X);
label("$X$",X,E);
draw(A--Y);
label("$Y$",Y,NE);
draw(A--Z);
label("$Z$",Z,E);
[/asy]
And, $\overline{YX}\parallel\overline{PQ}$ and $\overline{PQ}\perp\overline{QR}$, so $\overline{YX}\perp\overline{QR}$. Therefore $\triangle AYX \cong \triangle ARQ$ and $AY = \frac{1}{4}AR$.
Similarly, we can show that $\overline{WZ}\parallel\overline{QR}$ and $AZ = \frac{1}{4}AQ$.
Hence, the lengths of line segments $AY, AZ, BW$ and $BZ$ are one-fourth the lengths of line segments $QR, QS, PQ$ and $PS$ respectively.
Since $\overline{AY}\parallel\overline{QS}$ and $\triangle AYX\cong\triangle ARQ$, then the Pentagon $AYXZW$ is similar to the Pentagon $ARQPS$. And, they also $\angle YAW\cong\angle QAR$, etc. This implies that $\triangle AYW\sim\triangle QSR$ and $\triangle AXZ\sim\triangle SPQ$.
Finally, to find the surface area of the frustum, we sum the surface areas of two congruent trapezoids and that of the square:
\begin{align*}
800/2 + 800/2 + 1 &=40(XW + ZY) + PS^2\\
&=40(((4/3)(QR))+((1/3)(PQ ))) + PS^2\\
&=\frac{40 }{ 3}PQ + 40\cdot \frac{4}{3}QS+ PS^2.
\end{align*}
However, knowing that each side of the square has length $10$ implies that $\frac{PS}{QS} = \frac{4}{3}$. Let $QS = k$ so that $PS=k\cdot \frac{4}{3}$. Then, using the Pythagorean Theorem on the two diagonals in base $PQRS$, we have $k^2+k^2 = 10^2$ and so $QS= PS = k = \frac{10}{\sqrt{2}} = 5\sqrt{2}$.
Thus,
\[ \frac{40 }{ 3}PQ + 40\cdot \frac{4}{3}QS+ PS^2 = \frac{40}{3}( 2\sqrt{2}\cdot 10)+(5\sqrt{2})^2 = 40 \cdot 10 + 50 = \boxed{450}.\]