The sum of the lengths of the edges of a cube is equal to half the perimeter of a square. If the numerical value of the volume of the cube is equal to one-sixth of the numerical value of the area of the square, then the length of one side of the square is:
36 units
This problem involves two common geometric shapes: a cube and a square. We are given two relationships connecting properties of the cube (specifically its edges and volume) with properties of the square (its perimeter and area). Our goal is to use these relationships to find the length of one side of the square.
Let's define variables for the dimensions of the shapes:
The problem provides two key pieces of information that we can translate into mathematical equations.
The first relationship states that the sum of the lengths of the edges of a cube is equal to half the perimeter of a square.
So, the equation for Relationship 1 is:
\(12a = \frac{1}{2} \times 4s\)
\(12a = 2s\)
We can simplify this equation:
\(s = 6a \quad (Equation 1)\)
This equation tells us that the side length of the square is 6 times the edge length of the cube.
The second relationship states that the numerical value of the volume of the cube is equal to one-sixth of the numerical value of the area of the square.
So, the equation for Relationship 2 is:
\(a^3 = \frac{1}{6} \times s^2 \quad (Equation 2)\)
Now we have a system of two equations with two variables ('a' and 's'):
We can use substitution to solve this system. Substitute Equation 1 into Equation 2:
\(a^3 = \frac{1}{6} (6a)^2\)
Simplify the right side of the equation:
\(a^3 = \frac{1}{6} (36a^2)\)
\(a^3 = 6a^2\)
To solve for 'a', move all terms to one side:
\(a^3 - 6a^2 = 0\)
Factor out \(a^2\):
\(a^2(a - 6) = 0\)
This equation holds true if either \(a^2 = 0\) or \(a - 6 = 0\).
So, the edge length of the cube is 6 units.
Now that we have the value of 'a', we can find the value of 's' using Equation 1 (\(s = 6a\)):
\(s = 6 \times 6\)
\(s = 36\)
The length of one side of the square is 36 units.
Based on the given relationships between the cube and the square, the calculated length of one side of the square is 36 units.
| Shape | Dimension | Formula | Value (Calculated) |
|---|---|---|---|
| Cube | Edge Length (a) | - | 6 units |
| Cube | Sum of Edge Lengths | \(12a\) | \(12 \times 6 = 72\) units |
| Cube | Volume | \(a^3\) | \(6^3 = 216\) cubic units |
| Square | Side Length (s) | - | 36 units |
| Square | Perimeter | \(4s\) | \(4 \times 36 = 144\) units |
| Square | Area | \(s^2\) | \(36^2 = 1296\) square units |
Let's verify the original conditions using these values:
The calculated side length of the square, 36 units, satisfies both conditions.
| Shape | Property | Formula |
|---|---|---|
| Cube (edge 'a') | Sum of Edge Lengths | \(12a\) |
| Cube (edge 'a') | Volume | \(a^3\) |
| Square (side 's') | Perimeter | \(4s\) |
| Square (side 's') | Area | \(s^2\) |
Let's review some basic properties of cubes and squares that are relevant to this problem.
Understanding these basic properties is crucial for setting up the correct equations when solving geometry problems like this one.
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