If you have a truncated cone such that its upper base has a radius of a and the radius of its [larger] lower base is b, and a height h (between bases), how could you figure out its surface area using geometric reasoning?

If we develop the surface of a full cone with radius R and height H we get a "piece of pie" with circular edge of length 2*pi*R cut from the "pie" of radius S=sqrt(R^2+H^2). This piece has area (R/S)*pi*S^2 = pi*R*S. By similar triangles, the given truncated cone is truncated from a cone of radius b and height h*(b/(b-a)) = h + h*(a/(b-a)) = h + h'. The surface area of the truncated cone can thus be found by subtraction of the surface area of a full cone of height h' and radius a from that of a full cone of height h+h' and radius b. Thus the surface area of the truncated cone is pi*(b*t-a*s) where t=sqrt(b^2+(h+h')^2) and s=sqrt(a^2+h'^2).

Posted by Richard on 2004-01-23 13:12:00