How is the Formula for Surface Area of a Frustum Verified in Exercise 62?

[ElijahRockers]

Ok, so I have problems understanding where they came up with the formula
S=2∏rL, where r = (1/2)(r₁+r₂).

They say the formula is verified in exercise 62. I did the exercise and showed that S=∏rL.

Correct me if I'm wrong, but isn't this contradictory? I can see how they're using the average radius of the frustum, but I'm not sure what allows them to do that. I guess I just need to see a geometric proof before I really get it.

Any ideas? Thanks in advance.

 

[lippyka]

The formula S=2πrL can be understood by looking at the geometry of the frustum. First, consider the two circles at the ends of the frustum. The circumference of each circle is 2πr, where r is the radius of each circle. Since the circumference of each circle is the same, and they are connected by a circular arc, the total circumference of the frustum is 2πr. Now, since the total circumference of the frustum is the same as the perimeter of the rectangle (formed by the length L and the circumference of the frustum), it follows that the total circumference of the frustum is equal to 2πrL. By substituting the average radius (1/2)(r1+r2) into the formula, we get S=2π(1/2)(r1+r2)L, which is equivalent to S=2πrL.