How would you begin to solve this?
Do you use $A = 4\pi r^2$?
$V = \frac43\pi r^3$.
Substitute $180$ for $A$, solve for $r$ and plug into Volume equation(leave in simplified form).
How would you begin to solve this?
Do you use $A = 4\pi r^2$?
$V = \frac43\pi r^3$.
Substitute $180$ for $A$, solve for $r$ and plug into Volume equation(leave in simplified form).
Using the formulas: $$A = 4 \pi r^2$$ and $$V = \frac43 \pi r^3$$ We get that $$V = \frac{A^{\frac32}}{6\sqrt{\pi}}\tag{Quite a known formula}$$ Substitute the values of $A$ and you get: $$\frac{180^{\frac32}}{6\sqrt{\pi}} \approx 227.081$$
Hope it helps :)
Another way is to notice that $A^3 = 64 \pi^3 r^6$ and $V^2 = \frac{16}{9} \pi^2 r^6$.
Therefore
$$\frac{V^2}{A^3} = \frac{1}{36 \pi}$$ and $$V = \frac{A^{3/2}}{6 \sqrt \pi} \approx 227.1$$
Which avoids to solve $r$ from $A$.