# How did they arrive at the formula for lateral area of a frustum

Steps to arrive at the formula of a lateral surface area of a frustum (source):

If $$r = \frac{1}{2} (r_1 + r_2)$$, lateral area $$= 2 \pi r s$$

(Proof →)

Since $$\frac{\bar s}{r_2} = \frac{s + \bar s}{r_1}$$ or $$r_1 \bar s = r_2 s + r_2 \bar s$$, lateral area of frustum
$$= \pi r_1 (s + \bar s) - \pi r_2 \bar s$$
$$= \pi (r_1 + r_2) s$$

I tried to solve for $$r_1$$ from the ratio equation and substitute the resulting expression in the equation for the difference of the small frustum from the large frustum but it didn't work.I got this question from George F Simmon's precalculus in a nutshell

• Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. Feb 24, 2022 at 9:42
• I'm not well versed with MathJax I also can't draw the diagrams and it seems my mathJax isn't applying I and tried it but its not reflecting on the question Feb 24, 2022 at 9:59
• It's the third time I see this exact question. Why are you posting this same question again and again?
– ACB
Feb 24, 2022 at 10:26
• As LaTeX maths notation is not so easy for beginners to grasp, even with the linked tutorial on meta, I have added the text with equations to the question myself. I hope that this will help the asker to learn by example, as now they can compare the TeX commands (right click equation → Show Math As) to the equations. Feb 24, 2022 at 14:26
• Thanks Tim Pederick for your help I've edited the question to include the name of the text book Ill practice on my MathJax but thanks alot for your help youve been very helpful Feb 24, 2022 at 15:20

So we know that the lateral surface area of a cone, with (base) radius $$r$$ and slant height $$s$$, is $$\pi r s$$.

In the diagram, the frustum is taken from a cone with radius $$r_1$$ and slant height $$(s+\bar s)$$, slicing off a smaller cone with radius $$r_2$$ and slant height $$\bar s$$.[1] So the frustum’s lateral surface area is the difference between those of the two cones: $$A = \pi r_1 (s+\bar s) - \pi r_2 \bar s$$

So far, so good. Now, the text introduces a radius $$r$$ that is halfway between $$r_1$$ and $$r_2$$,[2] i.e. $$r = \frac{1}{2}(r_1 + r_2)$$. It then asserts that the frustum’s lateral surface area is $$2 \pi r s$$ and provides a proof.

The proof uses the following reasoning, but omits a number of steps. I hope that by adding these extra steps, I have answered your question and made it clearer!

The ratio slant height ∶ radius is equal between the two cones.[3] Hence:

\begin{aligned} \frac{\bar s}{r_2} &= \frac{s + \bar s}{r_1}\\ \therefore r_1 \bar s &= r_2 (s + \bar s)\\ &= r_2 s + r_2 \bar s \\ \therefore r_2 \bar s &= r_1 \bar s - r_2 s \end{aligned}

Substituting this last line into the equation for the difference between the lateral areas, we have:

\begin{aligned} A &= \pi r_1 (s+\bar s) - \pi r_2 \bar s \\ &= \pi r_1 (s+\bar s) - \pi (r_1 \bar s - r_2 s) \\ &= \pi r_1 s + \pi r_1 \bar s - \pi r_1 \bar s + \pi r_2 s \\ &= \pi r_1 s + \pi r_2 s \\ &= \pi (r_1 + r_2) s \end{aligned}

Recall that $$r$$ was defined thus: $$r = \frac{1}{2}(r_1 + r_2)$$. Hence $$2r = (r_1 + r_2)$$, which we can substitute into the last line above to obtain their formula: $$A = 2 \pi r s$$. ∎

[1]: $$s$$ is thus the slant height of the frustum.

[2]: To be more exact, it is the radius at the midpoint of the frustum’s slant height. It makes a certain intuitive sense that this radius should be halfway between $$r_1$$ and $$r_2$$, but it’s not difficult to prove more rigorously.

[3]: No proof of this is given—or at least, it isn’t visible in the image of the text—but it follows from the similarity of the two solids.

• Hi isn't it supposed to be $r_1\bar s =r_2(s+\bar s)$ Sep 29, 2022 at 11:12
• @Gonja *facepalm* Right, right. Fixed. Sep 30, 2022 at 4:01