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An A-frame restaurant is shaped as a triangle with two side lengths of 20 m and 30 m. Complete the inequality below to describe the range of possible lengths [tex]\( x \)[/tex] of the third side of the restaurant.

[tex]\[ \square \ \textless \ x \ \textless \ \square \][/tex]

Sagot :

To solve this problem, we need to apply the triangle inequality theorem. This theorem states that for any triangle, the sum of any two sides must be greater than the length of the third side.

Let's denote the sides of the triangle as follows:
- [tex]$a = 20 \, \text{m}$[/tex]
- [tex]$b = 30 \, \text{m}$[/tex]
- [tex]$x$[/tex] is the unknown third side.

According to the triangle inequality theorem, we have three conditions that must be satisfied:

1. [tex]\( a + b > x \)[/tex]
2. [tex]\( a + x > b \)[/tex]
3. [tex]\( b + x > a \)[/tex]

Substituting the known values:

1. [tex]\( 20 + 30 > x \)[/tex]
[tex]\[ 50 > x \][/tex]
[tex]\[ x < 50 \][/tex]

2. [tex]\( 20 + x > 30 \)[/tex]
[tex]\[ x > 10 \][/tex]

3. [tex]\( 30 + x > 20 \)[/tex]
[tex]\[ x > -10 \][/tex]

The third condition, [tex]\( x > -10 \)[/tex], is always true for positive lengths of [tex]\( x \)[/tex], so it is redundant in this context. Therefore, we consider only the effective constraints:

[tex]\[ 10 < x < 50 \][/tex]

Hence, the range of possible lengths for the third side [tex]\( x \)[/tex] of the restaurant is:

[tex]\[ 10 < x < 50 \][/tex]

So the correct answer to fill in the blanks is:
[tex]\[ \boxed{10} < x < \boxed{50} \][/tex]