What is the height of the frustum? > 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 $λ$. =\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 ^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}}.$$