Answer:
![r = \sqrt{(a)/(\pi h )}\\\\](https://img.qammunity.org/2023/formulas/mathematics/high-school/p6wi128lbbefhkouqcmviaqvh3ykf7qt6c.png)
Work Shown:
![h = (a)/(\pi r^2)\\\\h r^2 = (a)/(\pi )\\\\ r^2 = (a)/(\pi h )\\\\ r = \pm \sqrt{(a)/(\pi h )}\\\\ r = \sqrt{(a)/(\pi h )}\\\\](https://img.qammunity.org/2023/formulas/mathematics/high-school/vu4hkwd1d1qyw8eeq0mfeq2no5f1rcqdso.png)
I multiplied both sides by r^2 in the second step. Then I divided both sides by h (step 3)
Apply the square root to fully isolate r. The plus minus is dropped if r is referring to the radius. A negative radius value is not possible.