The problem describes a box filled with 12 identical balls. The box has dimensions $3x + 3$ cm and $2x + 4$ cm. The area of the box is 432 $cm^2$. We need to find the diameter of one ball in cm.

1. Problem Description

The problem describes a box filled with 12 identical balls. The box has dimensions $3x + 3$ cm and $2x + 4$ cm. The area of the box is 432 $cm^2$. We need to find the diameter of one ball in cm.

2. Solution Steps

First, we set up an equation for the area of the box:

$(3x + 3)(2x + 4) = 432$

Expanding the expression gives:

$6x^2 + 12x + 6x + 12 = 432$

$6x^2 + 18x + 12 = 432$

$6x^2 + 18x - 420 = 0$

Divide by 6:

$x^2 + 3x - 70 = 0$

Now we factor the quadratic equation:

$(x + 10)(x - 7) = 0$

The possible values for $x$ are $x = -10$ or $x = 7$.

Since the dimensions of the box must be positive, we discard the negative value $x = -10$.

Therefore, $x = 7$.

Now we can find the dimensions of the box:

Length = $3x + 3 = 3(7) + 3 = 21 + 3 = 24$ cm

Width = $2x + 4 = 2(7) + 4 = 14 + 4 = 18$ cm

The box contains 3 balls along the length and 4 balls along the width, so there are 12 balls in total.

Let $d$ be the diameter of each ball.

Then, 3d = length = 24

And, 4d = width = 18 (typo in question - should be 3 x 4 rather than 4 x 3.)

(The diagram shows 3 rows of 4 balls - hence the length and width relationships given.)

We use length to determine d.

$3d = 24$

$d = \frac{24}{3} = 8$

3. Final Answer

The diameter of one ball is 8 cm.