Answer:
In this problem, we have three solids, namely:
1. A cylinder.
2. Cone.
3. sphere.
These three figures have the same radius. The cylinder and cone have the same height with [tex]h=r[/tex]. We also know that:
[tex]Volume \ of \ the \ cone \ is \ 36units^3[/tex]
Also, some formulas are provided:
[tex]Volume \ of \ a \ cylinder: V=\pi r^2h \\ \\ Volume \ of \ a \ cone: V=\frac{1}{3}\pi r^2h \\ \\ Volume \ of \ a \ sphere: V=\frac{4}{3}\pi r^3[/tex]
Since [tex]h=r[/tex] then our new formulas are:
[tex]Volume \ of \ a \ cylinder: V=\pi r^3 \\ \\ Volume \ of \ a \ cone: V=\frac{1}{3}\pi r^3 \\ \\ Volume \ of \ a \ sphere: V=\frac{4}{3}\pi r^3[/tex]
For the cone, we know that [tex]V=36units^3[/tex] then we can get the radius [tex]r[/tex]:
[tex]V=\frac{1}{3}\pi r^3 \\ \\ 36=\frac{1}{3}\pi r^3 \\ \\ 36(3)=\pi r^3 \\ \\ r^3=\frac{108}{\pi}[/tex]
VOLUME OF THE CYLINDER:
[tex]V=\pi r^3 \\ \\ V=\pi \left(\frac{108}{\pi}\right) \\ \\ \boxed{V=108units^3}[/tex]
VOLUME OF THE SPHERE:
[tex]V=\frac{4}{3}\pi r^3 \\ \\ V=\frac{4}{3}\pi \left(\frac{108}{\pi}\right) \\ \\ \boxed{V= 144units^3}[/tex]
