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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.)

 Dec 26, 2023
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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:

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,

\[ \frac{40 }{ 3}PQ + 40\cdot \frac{4}{3}QS+ PS^2 = \frac{40}{3}( 2\sqrt{2}\cdot 10)+(5\sqrt{2})^2 = 40 \cdot 10 + 50 = \boxed{450}.\]

 Dec 26, 2023

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