Let APQRS be a pyramid, where the base PQRS is a square of side length 20. The total surface area of pyramid APQRS (including the base) is 1200. 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 frustum PQRSWXYZ (including the bases).
I believe that it doesn't make any difference if the pyramid is regular or not.
Total surface area = 1200.
Surface area of the base = 400.
Total surface area of the four sides is 1200 - 400 = 800.
1/2 of the area of each side is below the midline*; therefore 1/2·800 = 400 is below the midline.
Midline value of each side = 10 ---> area of the top base = 100.
Total surface area = 400 + 400 + 100 = 900.