Select the correct answer.

A triangle has one side of length 29 units and another of length 40 units. Determine the range in which the length of the third side must lie.

A. [tex]-11\ \textless \ x\ \textless \ 69[/tex]

B. [tex]11 \leq x \leq 69[/tex]

C. [tex]11\ \textless \ x\ \textless \ 69[/tex]

D. [tex]-11 \leq x \leq 69[/tex]

To find the range in which the length of the third side of a triangle must lie, we can use the triangle inequality theorem. This theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Let's denote the sides of the triangle as follows:
- Side 1: 29 units
- Side 2: 40 units
- Side 3: [tex]$ x $[/tex] units (the length we need to find)

Using the triangle inequality theorem, we set up the following inequalities:

1. [tex]$ 29 + 40 > x $[/tex]
2. [tex]$ 29 + x > 40 $[/tex]
3. [tex]$ 40 + x > 29 $[/tex]

Now let's solve each inequality:

1. [tex]$ 29 + 40 > x $[/tex]
- [tex]$ 69 > x $[/tex] or [tex]$ x < 69 $[/tex]

2. [tex]$ 29 + x > 40 $[/tex]
- Subtract 29 from both sides: [tex]$ x > 11 $[/tex]

3. [tex]$ 40 + x > 29 $[/tex]
- Subtract 40 from both sides: [tex]$ x > -11 $[/tex]

Since the requirement from the third inequality [tex]$ x > -11 $[/tex] is always satisfied when [tex]$ x > 11 $[/tex], we only need to consider the range from inequalities 1 and 2. Hence, combining these two results, we have:

[tex]$ 11 < x < 69 $[/tex]

Therefore, the third side must lie strictly between 11 and 69 units. So the correct answer is:

C. [tex]$ 11 < x < 69 $[/tex]