In pyramid APQRS, base PQRS is a square of side length 20. The total surface area of the pyramid (this includes the base) is 2300. Let W, X, Y, and Z be the midpoints of AP, AQ, AR, and AS, respectively. Find the total surface area of frustum* PQRSWXYZ (including the bases).

*Basically a frustum is a part of a cut pyramid, I think

I don't know how to find any volumes or areas here except for PQRS being 400, and that APQRS without the base is 1900. And I don't think that WP, XQ, YR, or ZS are the same. I think that WXYZ is a square though, right? Even so, this is just an assumption.

Guest Mar 30, 2020

#1**+1 **

I believe that it doesn't make any difference if the pyramid is regular or not.

Total surface area = 2300.

Surface area of the base = 400.

Total surface area of the four sides is 2300 - 400 = 1900.

¾s of the area of each side is below the midline^{*}; therefore ¾·1900 = 1425 is below the midline.

Midline value of each side = 10 ---> area of the top base = 100.

Total surface area = 400 + 1425 + 100 = 1925.

^{*}The reason why three-fourths of the area of the side is below the midline.

Draw any triangle and draw the midline.

Find the midpoint of the base.

Connect line segments from the endpoints of the midline to the base.

You will have created three triangles, each congruent to the triangle above the midline.

geno3141 Mar 31, 2020