Suppose the sphere has diameter \(\displaystyle d\). 

Then Cube B, the circumscribing cube, has as its edge length the diameter \(\displaystyle d\), and its surface area is \(\displaystyle 6d^{2}\).

Also, Cube A, the inscribed cube, has this diameter as the length of its diagonal. If \(\displaystyle s\) is the length of an edge, then from the three-dimensional extension of the Pythagorean Theorem, 

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

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

The surface area is \(\displaystyle SA = 6s^{2}\), so

\(\displaystyle SA = 6s ^{2 } = 2 \cdot 3s^{2} = 2 d^{2}\).

The ratio of the surface areas is 

\(\displaystyle \frac{6d^{2}}{ 2 d^{2}} = \frac{3}{1}\)

The correct choice is \(\displaystyle 3:1\).

By definition, all sides of a cube are equal in length, so each face is a square, There are six faces on a cube, so its total surface area is six times the area of one of its square faces. If one of its sides is 4 meters, then this will also be the other dimension of one of its square faces, so the total surface area is:

\(\displaystyle SA=6L^2=6(4)^2=96\)  \(\displaystyle m^2\)

To solve, remember that the equation for surface area of a cube is:

\(\displaystyle SA=6s^2=6(4^2)=6*16=96\)

Aperture labs makes a variety of cubes. If each cube has a volume of \(\displaystyle 6ft^3\), what is the surface area of the cube?

Let's work backwards from our goal in this question. 

We know that we need to find surface area. To find surface area of a cube, we can use the following equation:

\(\displaystyle SA=6l^2\)

Where l is the length of one side.

Next, let's look at the volume formula:

\(\displaystyle V_{cube}=l^3\)

So, we can find our length

\(\displaystyle 6ft^3=l^3\)

\(\displaystyle l=(6)^{\frac{1}{3}}\)

Let's leave l like that for the moment, and use it to find our surface area.

\(\displaystyle SA=6((6)^{\frac{1}{3}})^2\approx19.81ft^2\)

\(\displaystyle V=s^3\)

\(\displaystyle V=2^3\)

\(\displaystyle V=8\)

Six feet and seven feet are equal to, respectively, 72 inches and 84 inches. There are three different prime numbers between 72 and 84 - 73, 79, and 83 - so these are the three dimensions of the prism in inches. The volume of the prism is

\(\displaystyle 73 \times 79 \times 83 = 478,661\) cubic inches.

A diagonal of a square has length \(\displaystyle \sqrt{2}\) times that of a side, so each side of each square face of the cube has length \(\displaystyle 5 \sqrt{2} \div \sqrt{2}= 5\). Cube this to get the volume:

\(\displaystyle V= 5^{3} = 125\)

Let \(\displaystyle s\) be the length of one edge of the cube. By the three-dimensional extension of the Pythagorean Theorem, 

\(\displaystyle s^{2}+s^{2}+s^{2}= ( \sqrt{17})^{2}\)

\(\displaystyle 3s^{2}= 17\)

\(\displaystyle s^{2}= \frac{17}{3}\)

\(\displaystyle s =\sqrt{ \frac{17}{3}}\)

Cube the sidelength to get the volume:

\(\displaystyle s^{3} =\left (\sqrt{ \frac{17}{3}} \right )^{3}\)

\(\displaystyle =\frac{ \left (\sqrt{ 17} \right )^{3}}{\left (\sqrt{3} \right )^{3}} \right )\)

\(\displaystyle =\frac{ 17 \sqrt{ 17} }{ 3 \sqrt{3} } \right )\)

\(\displaystyle =\frac{ 17 \sqrt{ 17} \cdot \sqrt{3}}{ 3 \sqrt{3} \cdot \sqrt{3}} \right )\)

\(\displaystyle =\frac{ 17 \sqrt{ 51} }{ 9} \right )\)

The sidelength of the cube is the diameter of the inscribed sphere, which is twice that sphere's radius. The sphere has surface area \(\displaystyle 40 \pi\), so the radius is calculated as follows:

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

\(\displaystyle 4 \pi r^{2} = 40 \pi\)

\(\displaystyle r^{2} = 40 \pi \div 4 \pi = 10\)

\(\displaystyle r = \sqrt{10}\) 

The diameter of the sphere - and the sidelength of the cube - is twice this, or \(\displaystyle 2 \sqrt{10}\).

Cube this sidelength to get the volume of the cube:

\(\displaystyle V= \left ( 2 \sqrt{10} \right )^{3} = 2^{3} \cdot\left ( \sqrt{10 } \right )^{3} = 8 \cdot 10 \sqrt{10} = 80 \sqrt{10}\)

The diameter of the circle - twice its radius - coincides with the length of a diagonal of the inscribed cube. The sphere has volume \(\displaystyle 36 \pi\), so the radius is calculated as follows:

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

\(\displaystyle \frac{4}{3} \pi r^{3} = 36 \pi\)

\(\displaystyle r^{3} = 36\pi \div \left (\frac{4}{3} \pi \right )\)

\(\displaystyle r^{3} = 27\)

\(\displaystyle r = \sqrt[3]{27} = 3\)

The diameter of the sphere - and the length of a diagonal of the cube - is twice this, or 6. 

Now, let \(\displaystyle s\) be the length of one edge of the cube. By the three-dimensional extension of the Pythagorean Theorem, 

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

\(\displaystyle 3s^{2}=36\)

\(\displaystyle s^{2}=12\)

\(\displaystyle s = \sqrt{12} = \sqrt{4} \cdot \sqrt{3} = 2 \sqrt{3}\)

The volume of the cube is the cube of this, or

\(\displaystyle V=\left (2 \sqrt{3} \right )^{3} =2^{3} \left ( \sqrt{3} \right )^{3} = 8 \cdot 3 \sqrt{3} = 24 \sqrt{3}\)