The formula for volume of a cone, with base radius r and height h, is . Further, since the truncated cone goes from radius 75 to 50 m (losing 1/3 its value) in 35 meters, it will go to 0 in an additional 2(35)= 70 m. Find the volume of two cones, one of radius 75 and height 35+ 70= 105, and the other of radius 50 and height 70, and subtract.

The volume of a square pyramid, with has length s and height h, is . The truncated pyramid goes from edge length 86 to 65, losing 86- 65= 21 cm (losing 21/86 of its value) in slant height 45 cm. If the slant height of the whole pyramid were h, we would have h/86= 45/21 or h= (86)(45)/21= (86)(15)/7= 1290/7= 184.24 cm. A pyramid with slant height 184.24 and base length 86 would have, by the Pythagorean theorem, height = 138.5 cm. A pyramid with slant height 184.24- 45= 139.25 and base length 65 would have height = 104.6. Find the volumes of those two pyramids and subtract.