A sector of an annulus has a central angle of $θ$ and a thickness of $λ$. It is curved in such a manner that it forms a frustum of a cone without overlapping.
Find the height of the resultant frustum of a cone in terms of $θ$ and $λ$.
I tried to equate the cuved surface area of a frustum to the area of the sector of the annulus, i.e.
$$πλ(r_1+r_2)=\frac{θ(R^2-r^2)}{2},$$
where $r_1$ and $r_2$ are the radii of the two flat surfaces of the frustum and $R$ and $r$ are the outer and inner radii of the annulus.
Because $r_1=\frac{Rθ}{2π}$ and $r_2=\frac{rθ}{2π}$, then $$\frac{πλRθ}{2π}+\frac{πλrθ}{2π}=\frac{θ(R-r)(R+r)}{2}.$$ $$\frac{λθ(R+r)}{2}=\frac{θλ(R+r)}{2}.$$
Nothing further.....
Does anybody have any other method??
If one creates a conical frustum out of the sector of the annuli then the said radii will have to be calculated. For $r_1$ the circumference of the corresponding circle is given by $r\theta$. That is $$r_1=\frac{r\theta}{2\pi}.$$
Also, $$r_2=\frac{R\theta}{2\pi}.$$ And $$\lambda=R-r.$$
As far as the height. Consider the figure below
So $$h=\sqrt{\lambda^2-(r_2-r_1)^2}=\sqrt{\lambda^2-\left(\frac{R\theta}{2\pi}-\frac{r\theta}{2\pi}\right)^2}=$$
$$=\sqrt{\lambda^2-\frac{\theta^2}{4\pi^2}\lambda^2}=\lambda\sqrt{1-\frac{\theta^2}{4\pi^2}}.$$