For this case we have that by definition, the volume of a cylinder is given by:
![V = \pi * r ^ 2 * h](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9bbr1apmwdfg5ekvyuzuhy0h4dyqymtih6.png)
Where:
r: It is the radius of the cylinder
h: It is the height of the cylinder
According to the problem data we have:
![V = 90 \pi \ cm ^ 3\\h = 10 \ cm](https://img.qammunity.org/2020/formulas/mathematics/middle-school/fmcgwuesvhz00zsas1tzzqoqoidhkkp84a.png)
We replace the data:
![90 \pi = \pi * r ^ 2 * 10](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qtaljfqc6tdt8kdsnz9fzvjj026a9o6yi8.png)
We simplify:
![90 = 10r ^ 2\\r ^ 2 = \frac {90} {10}\\r ^ 2 = 9\\r = \pm \sqrt {9}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/39v9so1s7uysuali2vmhleq9vvelub4gh3.png)
We choose the positive value:
![r = \sqrt {9}\\r = 3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/tlnan0xz9jpfur7c25iuc8f1aeevvlewdp.png)
Thus, the radius of the cylinder is 3 cm. Therefore the diameter is
![d = 2 (3) = 6 \ cm](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jckz62dnfrc46n0aeuxroaxy1vzre44r27.png)
Answer:
The diameter of the cylinder is
![6 \ cm](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jakzqidqxx67vavqqcnejjsdd9sg3r6hq7.png)