Question 1137151


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. 

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(12^2+2^2) = sqrt(148) = 2*sqrt(37)}}} 

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+36+4(16*sqrt(37)) = 136+64*sqrt(37)}}}