I have realized a thing about cones. If we want to find which angle lateral surface spans and extends itself to, we have to compute

either 2\pi the ratio between the base circle (which happens to be the length of the circular base of the surface area) and the length of the circle having the slang height as radius, so


2\pi the ratio of the lateral surface and the surface of the entire circle that has the slant height as radius, hence

$$\alpha=2\pi\frac{S_l}{S_{360}}$$, with ${S_l}$ as curved surface and the other label as the circle having the slant height as radius

or (this is the interesting thing),2\pi the ratio between the base surface and the curved surface or 2\pi the ratio between the radius and the slant height.

$$\alpha=2\pi\frac{S_b}{S_l}$$, with ${S_l}$ as the curved surface, and I obtained this interesting result by doing some manipulations as below


I've cancelled out the 2 and multiplied by r, but despite having obtained an algebraic implication, I still can't understand the geometric one. The first formulae above make sense to me geometrically , because they simply use proportions and similarities with regards to a circle and the 360-angle, but how can the last one be demonstrated GEOMETRICALLY?

Labels:With $a$ I mean slant height, with ${C_{360}}$ I mean the circumference of the circle containing all the curved area of the cone, with ${S_{360}}$ I mean the area of the circle containing all the curved area of the cone , with ${C_{b}}$ I mean the base circumference, with ${S_l}$ I mean the curved surface, and with $ {S_b}$ I mean the base area. And of course, $h$ is the height and $r$ is the radius.


1 Answer 1


$\alpha = … = 2 \pi \dfrac{r}a$ is equivalent to $a \times \alpha = 2 \pi r$.

The RHS is the circumference of the base circle.

The LHS is the arc length of the cut-and-flattened cone in the form of a sector whose radius is $a = \sqrt{r^2+h^2}$ and central angle $= \alpha$ (in radian).

Note:- $\arc{AB} = 2 \pi a (\dfrac {\alpha}{360})$ if $\alpha $ is in degree; and $\arc{AB} = a \alpha $ if $\alpha$ is in radian.

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  • $\begingroup$ But I do not get the real geometrical implication, or at least not in this form. It is just a simple multiplication of both RHS and LHS by $2\pi$ How can I show geometrically that $ \frac{\alpha}{2\pi}=\frac{r}{a}$ or $\frac{\alpha}{2\pi}=\frac{S_b}{S_l}$? How can I show that the base circle is analogue to the angle $\alpha$ while the curved surface is analogue to the whole $2\pi$ angle? $\endgroup$
    – us er
    Commented Sep 6, 2019 at 18:11
  • $\begingroup$ @user See added note and diagram. One way to find out the relationship is to cut a conical drinking paper cup and flatten it. $\endgroup$
    – Mick
    Commented Sep 7, 2019 at 16:10

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