A=πr^2(√h-√r) make "r" the subject of the formula show workings

I think the formula is flawed, since the units on the left and right don't agree. But, going with what we have,

A=πr^2(√h-√r)
Suppose we let z^2 = r. Then we have
A = π√h z^4 - πz^5
degree-5 polynomials cannot be solved algebraically.
Check for a typo.

Did you know?

Did you know that you can rearrange the formula A=πr^2(√h-√r) to make "r" the subject of the formula? Let's do it step by step: 1. Start with the given formula: A=πr^2(√h-√r) 2. Divide both sides of the equation by π: A/π=r^2(√h-√r) 3. Divide both sides of the equation by r^2: (A/π)/r^2=√h-√r 4. Move √r to the left side of the equation: √r=(A/π)/r^2-√h 5. Square both sides of the equation to eliminate the square root: r=((A/π)/r^2-√h)^2 6. Simplify the right side of the equation: r=(A^2/π^2r^4-2A/πhr^2+h) By following these steps, we have obtained an expression that allows us to calculate the value of "r" in terms of "A" and "h" for the given formula. This rearrangement can be helpful in situations where we need to solve for the radius "r" based on the given value of the area "A" and the height "h".