b) since r is the radius of the base of the cone, the circuference of that base is equal to \(\displaystyle{2}\pi\cdot{r}\).

From answer a) that circuference is also equal to \(\displaystyle{8}\pi-{x}\) so:

\(\displaystyle{8}\pi-{x}={2}\pi{r}\)

\(\displaystyle{r}={\frac{{{8}\pi-{x}}}{{{2}\pi}}}={4}-{\frac{{{x}}}{{{2}\pi}}}\)

c) since h, r and 4 make up a right triangle, we can use the pythagorean theorem:

\(\displaystyle{r}^{{2}}+{h}^{{2}}={16}\)

next we substitute the answer from b) for r and solve for h:

\(\displaystyle{h}=\sqrt{{{16}-{\left({4}-{\frac{{{x}}}{{{2}\pi}}}\right)}^{{2}}}}\)

d) the volume of a cone is:

\(\displaystyle{V}={\frac{{{1}}}{{{3}}}}\pi{r}^{{2}}{h}\)

which means:

\(\displaystyle{V}={\frac{{\pi}}{{{3}}}}{\left({4}-{\frac{{{x}}}{{{2}\pi}}}\right)}^{{2}}\sqrt{{{16}-{\left({4}-{\frac{{{x}}}{{{2}\pi}}}\right)}^{{2}}}}\)