The cube with the greater sidelength has the greater volume, so we need only calculate and compare sidelengths.

(a) \(\displaystyle A=6s^{2}\), so the sidelength of the first cube can be found as follows:

\(\displaystyle A=6s^{2}\)

\(\displaystyle 6s^{2}= 864\)

\(\displaystyle 6s^{2} \div 6= 864 \div 6\)

\(\displaystyle s^{2} = 144\)

\(\displaystyle s = \sqrt{144 }= 12\) inches

(b) \(\displaystyle d^{2} = s^{2}+ s^{2}+ s^{2} = 3 s^{2}\) by an extension of the Pythagorean Theorem, so the sidelength of the second cube can be found as follows:

\(\displaystyle 3 s^{2}= d^{2}\)

\(\displaystyle 3 s^{2}= 21^{2}= 441\)

\(\displaystyle 3 s^{2}\div 3= 441 \div 3\)

\(\displaystyle s^{2}= 147\)

\(\displaystyle s=\sqrt{ 147}\)

Since \(\displaystyle 147 > 144\), \(\displaystyle \sqrt{147 }> \sqrt{144}\). The second cube has the greater sidelength and, subsequently, the greater volume. This makes (b) greater.

The sidelengths of Cubes 1, 2, 3, and 4 can be given values \(\displaystyle s, 2s, 4s, 8s\), respectively.

Then the volumes of the cubes are as follows:

Cube 1: \(\displaystyle V= s^{3}\)

Cube 2: \(\displaystyle V= (2s)^{3} = 8s^{3}\)

Cube 3: \(\displaystyle V= (4s)^{3} = 64s^{3}\)

Cube 4: \(\displaystyle V= (8s)^{3} = 512s^{3}\)

In both answer choices ask for a mean, so we can determine which answer (mean) is greater simply by comparing the sums of volumes.

(a) The sum of the volumes of Cubes 1 and 4 is \(\displaystyle s^{3}+ 512^{3} = 513s^{3}\).

(b) The sum of the volumes of Cubes 2 and 3 is \(\displaystyle 8s^{3}+ 64^{3} = 72s^{3}\).

Regardless of \(\displaystyle s\), the sum of the volumes of Cubes 1 and 4 is greater, and therefore, so is their mean.

This question is relatively straightforward. The equation for the volume of a cube is:

\(\displaystyle V = s^3\)

(It is like doing the area of a square, then adding another dimension!)

Now, for our data, we merely need to "plug and chug:"

\(\displaystyle v=7.236^3 =378.874760256\)

One of the faces of the cube could be drawn like this:

Notice that this makes a \(\displaystyle 45-45-90\) triangle.

This means that we can create a proportion for the sides. On the standard triangle, the non-hypotenuse sides are both \(\displaystyle 1\), and the hypotenuse is \(\displaystyle \sqrt{2}\).  This will allow us to make the proportion:

\(\displaystyle \frac{1}{\sqrt{2}} = \frac{x}{2}\)

Multiplying both sides by \(\displaystyle 2\), you get:

\(\displaystyle x=\frac{2}{\sqrt{2}}\)

Recall that the formula for the volume of a cube is:

\(\displaystyle V = s^3\)

Therefore, we can compute the volume using the side found above:

\(\displaystyle V = (\frac{2}{\sqrt{2}})^3=\frac{8}{(\sqrt{2})^3}=\frac{8}{2\sqrt{2}}=\frac{4}{\sqrt2}\)

Now, rationalize the denominator:

\(\displaystyle \frac{4}{\sqrt2} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{2}= 2\sqrt{2}\)