Solve A=2pir^2+2rrh

$$A = 2 \pi r ^ { 2 } + 2 r r h$$

Solve for h

$$\left\{\begin{matrix}h=-\pi +\frac{A}{2r^{2}}\text{, }&r\neq 0\\h\in \mathrm{R}\text{, }&A=0\text{ and }r=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply r and r to get r^{2}.

$$A=2\pi r^{2}+2r^{2}h$$

Swap sides so that all variable terms are on the left hand side.

$$2\pi r^{2}+2r^{2}h=A$$

Subtract 2\pi r^{2} from both sides.

$$2r^{2}h=A-2\pi r^{2}$$

Divide both sides by 2r^{2}.

$$\frac{2r^{2}h}{2r^{2}}=\frac{A-2\pi r^{2}}{2r^{2}}$$

Dividing by 2r^{2} undoes the multiplication by 2r^{2}.

$$h=\frac{A-2\pi r^{2}}{2r^{2}}$$

Divide A-2\pi r^{2} by 2r^{2}.

$$h=-\pi +\frac{A}{2r^{2}}$$

Solve for A

$$A=2\left(h+\pi \right)r^{2}$$