The volume of a pyramid or a cone with height \(\displaystyle h\) and base of area \(\displaystyle B\) is 

\(\displaystyle V = \frac{1}{3} Bh\), 

so in both cases, the area of the base is

\(\displaystyle \frac{1}{3} Bh = V\)

\(\displaystyle \frac{3}{h} \cdot \frac{1}{3} Bh =\frac{3}{h} \cdot V\)

\(\displaystyle B = \frac{3V}{h}\)

Since the pyramid and the cone have the same volume and height, their bases has the same area \(\displaystyle B\).

The length of one side of the square base of the pyramid is the square root of this, or \(\displaystyle \sqrt{B}\), and the perimeter is four times this, or \(\displaystyle P = 4 \sqrt{B}\).

The radius and the area of the base of the cone are related as follows:

\(\displaystyle \pi r ^{2} = B\)

\(\displaystyle \sqrt{\pi r ^{2} }= \sqrt{B}\)

\(\displaystyle \sqrt{\pi } \cdot r = \sqrt{B}\)

Multiply both sides by \(\displaystyle 2 \sqrt{\pi}\) to get:

\(\displaystyle 2 \sqrt{\pi} \cdot \sqrt{\pi } \cdot r =2 \sqrt{\pi} \cdot \sqrt{B}\)

\(\displaystyle 2\pi r =2 \sqrt{\pi} \cdot \sqrt{B}\)

\(\displaystyle C =2 \sqrt{\pi} \cdot \sqrt{B}\)

\(\displaystyle 2 \sqrt{\pi} < 2 \sqrt{4} = 2 \times 2 = 4\), so 

\(\displaystyle 2 \sqrt{\pi} \cdot \sqrt{B} < 4 \sqrt{B}\), and

\(\displaystyle C < P\)

The perimeter of the base of the pyramid, which is (A), is greater than the circumference of the base of the cone.

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:

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

The volume of a cube is defined by the formula

\(\displaystyle V=s^{3}\)

where \(\displaystyle s\) is the length of one side.

If \(\displaystyle V=343\), then 

\(\displaystyle s^{3} = 343\)

and 

\(\displaystyle s = \sqrt[3]{343} = 7\)

So one side measures 7 inches. 

The surface area of a cube is defined by the formula

\(\displaystyle A = 6s^{2}\) , so

\(\displaystyle A = 6 \cdot 7^{2} = 294\)

The surface area is 294 square inches.

Recall that the formula for the surface area of a cube is:

\(\displaystyle SA = 6 * s^2\), where \(\displaystyle s\) is the length of a side of the cube. This equation is easy to memorize because it is merely a multiplication of a single side (\(\displaystyle s^2\)) by \(\displaystyle 6\) because a cube has \(\displaystyle 6\) equal sides.

For our data, we know that \(\displaystyle s = 5.7\); therefore, our equation is:

\(\displaystyle SA = 6 * 5.7^2 = 6 * 32.49 = 194.94\)

To solve this, first calculate the side length based on the volume given. Recall that the equation for the volume of a cube is:

\(\displaystyle V = s^3\), where \(\displaystyle s\) is the side length.

For our data, this gives us:

\(\displaystyle s^3 = 512\)

Taking the cube-root of both sides, we get:

\(\displaystyle s = 8\)

Now, use the surface area formula to compute the total surface area:

\(\displaystyle SA = 6 * s^2\), where \(\displaystyle s\) is the length of a side of the cube. This equation is easy to memorize because it is merely a multiplication of a single side (\(\displaystyle s^2\)) by \(\displaystyle 6\) because a cube has \(\displaystyle 6\) equal sides.

For our data, this gives us:

\(\displaystyle SA = 6*8^2 = 6*64=384\)

To solve this, first calculate the side length based on the volume given. Recall that the equation for the volume of a cube is:

\(\displaystyle V = s^3\), where \(\displaystyle s\) is the side length.

For our data, this gives us:

\(\displaystyle s^3 = 274.625\)

Taking the cube-root of both sides, we get:

\(\displaystyle s = 6.5\)

(You will need to use a calculator for this.  If your calculator gives you something like \(\displaystyle 6.599999999999\) . . . it is okay to round. This is just the nature of taking roots!).

Now, use the surface area formula to compute the total surface area:

\(\displaystyle SA = 6 * s^2\), where \(\displaystyle s\) is the length of a side of the cube. This equation is easy to memorize because it is merely a multiplication of a single side (\(\displaystyle s^2\)) by \(\displaystyle 6\) because a cube has \(\displaystyle 6\) equal sides.

For our data, this gives us:

\(\displaystyle SA = 6*6.5^2 = 6*42.25=253.5\)

Now, this could look like a difficult problem; however, think of the equation for finding the length of the diagonal of a cube. It is like the Pythagorean Theorem, just adding an additional dimension:

\(\displaystyle D = \sqrt{3s^2}\)

(It is very easy, because the three lengths are all the same: \(\displaystyle s\)).

So, we know this, then:

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

To solve, you can factor out an \(\displaystyle s\) from the root on the right side of the equation:

\(\displaystyle 3\sqrt{3}=s\sqrt{3}\)

Just by looking at this, you can tell that the answer is:

\(\displaystyle s=3\)

Now, use the surface area formula to compute the total surface area:

\(\displaystyle SA = 6 * s^2\), where \(\displaystyle s\) is the length of a side of the cube. This equation is easy to memorize because it is merely a multiplication of a single side (\(\displaystyle s^2\)) by \(\displaystyle 6\) because a cube has \(\displaystyle 6\) equal sides.

For our data, this is:

\(\displaystyle SA = 6 * 3^2 = 6*9=54\)