The length of the internal diagonal of a cube can be calculated using the Pythagorean theorem. Through two steps, we can find a formula for the diagonal of cubes. The first step requires us to calculate the diagonal of one of the faces of the cube using the lengths of the sides. In the second step, we use the diagonal of a face and one of the sides to find the final formula for the internal diagonal of a cube.
Here, we will look at these steps in detail. We will derive the formula for the diagonal of a cube and use it to solve some problems.
Formula to find the diagonal of a cube
There are two diagonals of a cube, the internal diagonal of the cube and the diagonal of a face. In the following image, we can see that both are different. The internal diagonal starts at one vertex and extends to the opposite vertex. The diagonal of a face begins at one vertex and ends at another vertex so that it is completely on one face.
To calculate the diagonal of a face of the cube, we have to use the Pythagorean theorem once. However, to calculate the internal diagonal, we need to use the Pythagorean theorem twice. So, we do this in two steps:
Step 1: We calculate the diagonal of a face. All the faces of a cube are square, so we have the following:
To find this distance, we have to use the Pythagorean theorem. If we use c to represent the diagonal, we have the following:
$latex {{c}^2}={{a}^2}+{{a}^2}$
$latex {{c}^2}=2{{a}^2}$
$latex c=\sqrt{2{{a}^2}}$
| $latex c=\sqrt{2}~a$ |
Therefore, we have found a formula for the diagonal of the face.
Step 2: To find an expression for the internal diagonal, we have to use another right triangle. We use the diagonal of a face as the base of the triangle, one side of the cube as the height of the triangle, and the internal diagonal as the hypotenuse:
We have to use the Pythagorean theorem again to find the length of the diagonal represented by d, so we have:
$latex {{d}^2}={{a}^2}+{{(\sqrt{2}~a)}^2}$
$latex {{d}^2}={{a}^2}+2{{a}^2}$
$latex {{d}^2}=3{{a}^2}$
$latex d=\sqrt{3{{a}^2}}$
| $latex d=\sqrt{3}~a$ |
We have found the formula for the internal diagonal of a cube.
Diagonal of a cube – Examples with answers
The following cube diagonals examples can be used to practice using the formula derived above. It is recommended that you try to solve the problems yourself before looking at the solution.
EXAMPLE 1
What is the diagonal of a cube that has sides of length 5 m?
Solution
We can use the diagonal formula with $latex a=5$. Therefore, we have:
$latex d=\sqrt{3}~a$
$latex d=\sqrt{3}(5)$
$latex d= 8.66$
The diagonal measures 8.66 m.
EXAMPLE 2
If a cube has sides with a length of 6 m, what is its diagonal?
Solution
We use the value $latex a=6$ in the diagonal formula. Therefore, we have:
$latex d=\sqrt{3}~a$
$latex d=\sqrt{3}(6)$
$latex d= 10.4$
The diagonal has a length of 10.4 m.
EXAMPLE 3
A cube has sides that are 21 cm long. What is its diagonal?
Solution
We have the length $latex a=21$, so we use this value in the diagonal formula:
$latex d=\sqrt{3}~a$
$latex d=\sqrt{3}(21)$
$latex d= 36.37$
The diagonal measures 36.37 cm.
EXAMPLE 4
If the diagonal of a cube is 10 m, what is the length of its sides?
Solution
In this case, we have the length of the diagonal and we want to find the length of the sides, so we use $latex d=10$ and solve for a:
$latex d=\sqrt{3}~a$
$latex 10=\sqrt{3}~a$
$latex a= \frac{10}{\sqrt{3}}$
$latex a=5.77$
The sides are 5.77 m long.
EXAMPLE 5
What is the length of the sides of a cube that has a diagonal of 20 m?
Solution
We use the length of the diagonal $latex d=20$ and solve for a:
$latex d=\sqrt{3}~a$
$latex 20=\sqrt{3}~a$
$latex a= \frac{20}{\sqrt{3}}$
$latex a=11.55$
The sides are 11.55 m long.
Diagonal of a cube – Practice problems
Practice using the cube diagonal formula to solve the following problems. Select an answer and check it to see if you got the correct answer.
See also
Interested in learning more about cubes? Take a look at these pages:
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
