Solve the equaiton for the surface area of a sphere for the radius and plug in the values:

\(\displaystyle SA=4{\pi}r^2\rightarrow r=\sqrt{\frac{SA}{4\pi}}= \sqrt{\frac{64\pi}{4\pi}}=\sqrt16=4\)

Recall that the equation for the volume of a sphere is:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

For our data, we know:

\(\displaystyle 2304\pi=\frac{4}{3}\pi r^3\)

Solve for \(\displaystyle r\). First, multiply both sides by \(\displaystyle \frac{3}{4}\):

\(\displaystyle \frac{3}{4}2304\pi=\pi r^3\)

\(\displaystyle 1728\pi=\pi r^3\)

Now, divide out the \(\displaystyle \pi\):

\(\displaystyle 1728= r^3\)

Using your calculator, you can solve for \(\displaystyle r\). Remember, if need be, you can raise \(\displaystyle 1728\) to the power of \(\displaystyle \frac{1}{3}\) if your calculator does not have a variable-root button.

This gives you:

\(\displaystyle r=12\)

If you get something like \(\displaystyle 11.999999....\), just round up. This is a rounding issue with some calculators.

Recall that the equation for the volume of a sphere is:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

For our data, we know:

\(\displaystyle 972\pi = \frac{4}{3}\pi r^3\)

Solve for \(\displaystyle r\). Begin by dividing out the \(\displaystyle \pi\) from both sides:

\(\displaystyle 972 = \frac{4}{3} r^3\)

Next, multiply both sides by \(\displaystyle \frac{3}{4}\):

\(\displaystyle \frac{3}{4} * 972 =r^3\)

\(\displaystyle 729=r^3\)

Using your calculator, solve for \(\displaystyle r\). Recall that you can always use the \(\displaystyle \frac{1}{3}\) power if you don't have a variable-root button.  

You should get:

\(\displaystyle r=9\)  If you get \(\displaystyle r=8.99999999...\), just round up to \(\displaystyle 9\). This is a general rounding problem with calculators. Since you are looking for the diameter, you must double this to \(\displaystyle 18\).

Recall that the surface area of a sphere is found by the equation:

\(\displaystyle SA = 4\pi r^2\)

For our data, this means:

\(\displaystyle 169\pi = 4\pi r^2\)

Solve for \(\displaystyle r\). First, divide by \(\displaystyle 4\pi\):

\(\displaystyle 42.25=r^2\)

Take the square root of both sides:

\(\displaystyle r = 6.5\)

Given the volume of the sphere, \(\displaystyle 36\prod \textup{units}^{3}\), you need to use the formula for volume of a sphere \(\displaystyle \left ( \left ( \frac{4}{3} \right )\cdot \prod \cdot r^{3}\right )\) and work backwards to find the radius. I would multiply both sides by \(\displaystyle \frac{3}{4}\) to get rid of the \(\displaystyle \frac{4}{3}\) in the formula. You then have \(\displaystyle 27\prod=\prod r^{3}\). Next, divide both sides by \(\displaystyle \prod\) so that all vyou have left is \(\displaystyle 27=r^{3}\). Finally take the cube root of \(\displaystyle 27\), to get \(\displaystyle 3\) units for the radius.

Solving this problem requires recognizing that since the cube is circumscribed by the sphere, both solids share the same center. Now it is just a matter of finding the diagonal of the cube, which will double as the diameter of the sphere (by definition, any straight line which passes through the center of the sphere). The formula for the diagonal of a cube is \(\displaystyle D = s\sqrt{3}\), where \(\displaystyle s\) is the length of the side of a cube. (This occurs because you must use the Pythagorean theorem once for each 2-dimensional "corner" you travel to find the diagonal for a 3-dimensional shape, but for the ACT it's much faster to memorize the formula.)

In this case:

\(\displaystyle D = s\sqrt3 = 25\sqrt3\)

Since the radius is half the diameter, divide the result in half:

\(\displaystyle r = \frac{D}{2} = \frac{25\sqrt{3}}{2}\)

\(\displaystyle r = \frac{25\sqrt{3}}{2}\)