SOLUTION: A frustum of a regular square pyramid has bases with sides of length 6 and 10. The height of the frustum is 12. Find the surface area

Algebra ->  Bodies-in-space -> SOLUTION: A frustum of a regular square pyramid has bases with sides of length 6 and 10. The height of the frustum is 12. Find the surface area      Log On


   

Question 1137154: A frustum of a regular square pyramid has bases with sides of length 6 and 10. The height of the frustum is 12. Find the surface area
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The areas of the two bases are easy -- 10^2 = 100 and 6^2 = 36.

To find the lateral surface area, you need to know the slant height of each face.

To find that slant height, drop a perpendicular from the middle of one side of the top base to the bottom base. With the side lengths of the two bases 6 and 10, that perpendicular will touch the bottom base 2 units from the edge (half of 10, minus half of 6). Then, since the height of the frustum is 12, the slant height of each face, by the Pythagorean Theorem, is

sqrt%2812%5E2%2B2%5E2%29+=+sqrt%28148%29+=+2%2Asqrt%2837%29

Each face is then a trapezoid with bases 10 and 6 and height 2*sqrt(37); that makes the area of each face 16*sqrt(37).

So the total surface area -- both bases and all four faces -- is

100%2B36%2B4%2816%2Asqrt%2837%29%29+=+136%2B64%2Asqrt%2837%29