Answer :
To solve this problem, we need to use the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. For a triangle with sides \(a\), \(b\), and \(x\) (where \(x\) is the length of the third side), the inequalities are as follows:
1. \(a + b > x\)
2. \(a + x > b\)
3. \(b + x > a\)
Given the specific side lengths \(a = 200\) units and \(b = 300\) units, we can substitute these values into the inequalities.
1. \(200 + 300 > x\)
2. \(200 + x > 300\)
3. \(300 + x > 200\)
We simplify these inequalities step by step:
1. \(500 > x\) or \(x < 500\)
2. \(200 + x > 300 \implies x > 300 - 200 \implies x > 100\)
3. \(300 + x > 200 \implies x > 200 - 300 \implies x > -100\) (but this is always true since \(x\) must be positive)
Thus, combining the useful inequalities from steps 1 and 2, we get:
[tex]\[ 100 < x < 500 \][/tex]
So, the range of the possible lengths for the third side, [tex]\(x\)[/tex], is [tex]\(100 < x < 500\)[/tex].
1. \(a + b > x\)
2. \(a + x > b\)
3. \(b + x > a\)
Given the specific side lengths \(a = 200\) units and \(b = 300\) units, we can substitute these values into the inequalities.
1. \(200 + 300 > x\)
2. \(200 + x > 300\)
3. \(300 + x > 200\)
We simplify these inequalities step by step:
1. \(500 > x\) or \(x < 500\)
2. \(200 + x > 300 \implies x > 300 - 200 \implies x > 100\)
3. \(300 + x > 200 \implies x > 200 - 300 \implies x > -100\) (but this is always true since \(x\) must be positive)
Thus, combining the useful inequalities from steps 1 and 2, we get:
[tex]\[ 100 < x < 500 \][/tex]
So, the range of the possible lengths for the third side, [tex]\(x\)[/tex], is [tex]\(100 < x < 500\)[/tex].
