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Truncated Cone (Posted on 2004-01-23) Difficulty: 3 of 5
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?

See The Solution Submitted by Gamer    
Rating: 2.5000 (4 votes)

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solution | Comment 2 of 10 |
The slant height, s, of the truncated cone is s = ((b-a)^2+h^2).

The slant height that the full, untruncated cone (let's call it r, as it will later serve as a radius) would have then is given by

r/s = b/(b-a)
or
r = b*s/(b-a)

Consider the conical surface of the truncated cone as made of paper, and cut along a line perpendicular to the top and bottom. Then lay this developpable surface flat. The arc formed by the top and the arc formed by the bottom will be concentric, as each will be perpendicular to the (now duplicated) cut line at either end of the curved strip. The base arc will be the outer arc, with a radius of curvature equal to r, the slant height of the fully restored cone.

The circumference of the bottom of the cone before it was laid out was 2*pi*b.

The fraction of a circle that the two arcs span about their common center is then 2*pi*b/(2*pi*r) = b/r.

If complete, the outer circle would have area pi*r^2, but that must be multiplied by b/r, resulting in pi*b*r. Likewise the inner sector of a circle to be subtracted out is pi*a*(r-s). After the subtraction we have

pi*(b*r-a*(r-s))

This can be exemplified in a program:

pi = 4 * ATN(1)
DO
INPUT "a,b,h"; a, b, h
s = SQR((b - a) ^ 2 + h * h)
r = b * s / (b - a)
PRINT "Area is"; pi * (b * r - a * (r - s))
LOOP

with some sample output:
a,b,h? 1,2,3
Area is 29.80377
a,b,h? 5,6,7
Area is 244.3586
a,b,h? 7,10,3
Area is 226.587

------
There are alternative formulations for the area of the smaller sector, which are equivalent.

This is the area of the side only. For the whole frustum, add in the area of the top and bottom, pi*(a^2 + b^2).
Edited on January 23, 2004, 12:17 pm
  Posted by Charlie on 2004-01-23 12:14:20
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