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

\(\displaystyle V=s^3\)

Since the sides of a cube are merely squares, the surface area equation is just \(\displaystyle 6\) times the area of one of those squares:

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

So, for our two quantities:

Quantity A

\(\displaystyle 343=s^3\)

Use your calculator to estimate this value (since you will not have a square root key). This is \(\displaystyle s=7\).

Quantity B

\(\displaystyle 294=6s^2\)

First divide by \(\displaystyle 6\):

\(\displaystyle s^2=49\)

Therefore, \(\displaystyle s=7\)

Therefore, the two quantities are equal.

The surface area of a cube is made up of \(\displaystyle 6\) squares. Therefore, the equation is merely \(\displaystyle 6\) times the area of one of those squares.  Since the sides of a square are equal, this is:

\(\displaystyle SA=6s^2\), where \(\displaystyle s\) is the length of one side of the square.

For our data, we know:

\(\displaystyle 1350=6s^2\)

This means that:

\(\displaystyle s^2=225\)

Now, while you will not have a calculator with a square root key, you do know that \(\displaystyle 15^2=225\). (You can always use your calculator to test values like this.) Therefore, we know that \(\displaystyle s=15\). This is the length of one side

There are 6 sides to a cube. If the total surface area is 54 square inches, then each face must have an area of 9 square inches.

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

\(\displaystyle 54=6s^2\)

\(\displaystyle 9=s^2\)

Every face of a cube is a square, so if the area is 9 square inches, each edge must be 3 inches.

First, we must ascertain the length of each side.  Based on our initial data, we know that the 6 faces of the cube will have a surface area of 6x². This yields the equation:

6x² = 486, which simplifies to: x² = 81; x = 9.

Therefore, each side has a length of 9.  Imagine the cube is centered on the origin.  This means its "front-left-bottom corner" will be at (–4.5, –4.5, 4.5) and its "rear-right-top corner" will be at (4.5, 4.5, –4.5).  To find the distance between these, we use the three-dimensional distance formula:

d = √((x₁ – x₂)² + (y₁ – y₂)² + (z₁ – z₂)²)

For our data, this will be:

√( (–4.5 – 4.5)² + (–4.5 – 4.5)² + (4.5 + 4.5)²) =

√( (–9)² + (–9)² + (9)²) = √(81 + 81 + 81) = √(243) =

√(3 * 81) = √(3) * √(81) = 9√(3)

The shortest length between any two non-adjacent corners will be the diagonal of the smallest face of the rectangular box. The smallest face of the rectangular box is a six-inch by six-inch square. The diagonal of a six-inch square is \(\displaystyle \dpi{100} \small 6\sqrt{2}\).

The diagonal length of a cube is found by a form of the distance formula that is akin to the Pythagorean Theorem, though with an additional dimension added to it. It is:

\(\displaystyle \sqrt{s^2+s^2+s^2}\), or \(\displaystyle \sqrt{3s^2}\), or \(\displaystyle s\sqrt{3}\)

Now, if the the value of \(\displaystyle s\) is \(\displaystyle 4\), we get simply \(\displaystyle 4\sqrt{3}\)

To begin, the best thing to do is to find the length of a side of the cube. This is done using the formula for the surface area of a cube. Recall that a cube is made up of \(\displaystyle 6\) squares. Therefore, its surface area is:

\(\displaystyle SA=6s^2\), where \(\displaystyle s\) is the length of a side.

Therefore, for our data, we have:

\(\displaystyle 726=6s^2\)

Solving for \(\displaystyle s\), we get:

\(\displaystyle 121=s^2\)

This means that \(\displaystyle s=11\)

Now, the diagonal length of a cube is found by a form of the distance formula that is akin to the Pythagorean Theorem, though with an additional dimension added to it. It is:

\(\displaystyle \sqrt{s^2+s^2+s^2}\), or \(\displaystyle \sqrt{3s^2}\), or \(\displaystyle s\sqrt{3}\)

Now, if the the value of \(\displaystyle s\) is \(\displaystyle 11\), we get simply \(\displaystyle 11\sqrt{3}\)

The box is 2 times as long as it is high, so H = L/2. It is also 4 times as wide as it is long, so W = 4L. Now we need volume = L * W * H = L * 4L * L/2 = 2L³.

The surface area of a cube is merely the sum of the surface areas of the \(\displaystyle 6\) squares that make up its faces. Therefore, the surface area equation understandably is:

\(\displaystyle SA = 6*s^2\), where \(\displaystyle s\) is the side length of any one side of the cube. For our values, we know:

\(\displaystyle 150=6s^2\)

Solving for \(\displaystyle s\), we get:

\(\displaystyle 25=s^2\) or \(\displaystyle s=5\)

Now, the volume of a cube is defined by the simple equation:

\(\displaystyle V=s^3\)

For \(\displaystyle s=5\), this is:

\(\displaystyle V=5^3=125\:in^3\)

Recall that the volume of a cube is defined by the equation:

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

Therefore, if we know that \(\displaystyle V=64\), we can solve:

\(\displaystyle 64=s^3\)

This means that \(\displaystyle s=4\).

Now, if we triple \(\displaystyle s\) to \(\displaystyle 12\), the new volume of our cube will be:

\(\displaystyle V=12^3 = 1728\:in^3\)