Solve for r SA=4pir^2

$S A = 4 π r^{2}$

Rewrite the equation as $4 π r^{2} = S A$ .

$4 π r^{2} = S A$

Divide each term in $4 π r^{2} = S A$ by $4 π$ and simplify.

Divide each term in $4 π r^{2} = S A$ by $4 π$ .

$\frac{4 π r^{2}}{4 π} = \frac{S A}{4 π}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 π r^{2}}{4 π} = \frac{S A}{4 π}$

Rewrite the expression.

$\frac{π r^{2}}{π} = \frac{S A}{4 π}$

Cancel the common factor of $π$ .

Cancel the common factor.

$\frac{π r^{2}}{π} = \frac{S A}{4 π}$

Divide $r^{2}$ by $1$ .

$r^{2} = \frac{S A}{4 π}$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$r = \pm \sqrt{\frac{S A}{4 π}}$

Simplify $\pm \sqrt{\frac{S A}{4 π}}$ .

Rewrite $\frac{S A}{4 π}$ as ${(\frac{1}{2})}^{2} \frac{S A}{π}$ .

Factor the perfect power $1^{2}$ out of $S A$ .

$r = \pm \sqrt{\frac{1^{2} (S A)}{4 π}}$

Factor the perfect power $2^{2}$ out of $4 π$ .

$r = \pm \sqrt{\frac{1^{2} (S A)}{2^{2} π}}$

Rearrange the fraction $\frac{1^{2} (S A)}{2^{2} π}$ .

$r = \pm \sqrt{{(\frac{1}{2})}^{2} \frac{S A}{π}}$

Pull terms out from under the radical.

$r = \pm \frac{1}{2} \sqrt{\frac{S A}{π}}$

Rewrite $\sqrt{\frac{S A}{π}}$ as $\frac{\sqrt{S A}}{\sqrt{π}}$ .

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A}}{\sqrt{π}}$

Multiply $\frac{\sqrt{S A}}{\sqrt{π}}$ by $\frac{\sqrt{π}}{\sqrt{π}}$ .

$r = \pm \frac{1}{2} (\frac{\sqrt{S A}}{\sqrt{π}} \cdot \frac{\sqrt{π}}{\sqrt{π}})$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{S A}}{\sqrt{π}}$ by $\frac{\sqrt{π}}{\sqrt{π}}$ .

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A} \sqrt{π}}{\sqrt{π} \sqrt{π}}$

Raise $\sqrt{π}$ to the power of $1$ .

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A} \sqrt{π}}{{\sqrt{π}}^{1} \sqrt{π}}$

Raise $\sqrt{π}$ to the power of $1$ .

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A} \sqrt{π}}{{\sqrt{π}}^{1} {\sqrt{π}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A} \sqrt{π}}{{\sqrt{π}}^{1 + 1}}$

Add $1$ and $1$ .

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A} \sqrt{π}}{{\sqrt{π}}^{2}}$

Rewrite ${\sqrt{π}}^{2}$ as $π$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{π}$ as $π^{\frac{1}{2}}$ .

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A} \sqrt{π}}{{(π^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A} \sqrt{π}}{π^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A} \sqrt{π}}{π^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A} \sqrt{π}}{π^{\frac{2}{2}}}$

Rewrite the expression.

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A} \sqrt{π}}{π^{1}}$

Simplify.

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A} \sqrt{π}}{π}$

Combine using the product rule for radicals.

$r = \pm \frac{1}{2} \cdot \frac{\sqrt{S A π}}{π}$

Multiply $\frac{1}{2}$ by $\frac{\sqrt{S A π}}{π}$ .

$r = \pm \frac{\sqrt{S A π}}{2 π}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$r = \frac{\sqrt{S A π}}{2 π}$

Next, use the negative value of the $\pm$ to find the second solution.

$r = - \frac{\sqrt{S A π}}{2 π}$

The complete solution is the result of both the positive and negative portions of the solution.

$r = \frac{\sqrt{S A π}}{2 π}$

$r = - \frac{\sqrt{S A π}}{2 π}$

$r = \frac{\sqrt{S A π}}{2 π}$

$r = - \frac{\sqrt{S A π}}{2 π}$

Solve for r SA=4pir^2

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