Find the range of values for the third side of the triangle x when one side is 7 and the other sides aren't shown

1 answer

To find the range of values for the third side of a triangle when one side is 7, we can use the triangle inequality theorem. According to the theorem, for any triangle with sides \(a\), \(b\), and \(c\):

  1. \(a + b > c\)
  2. \(a + c > b\)
  3. \(b + c > a\)

Let's denote the sides of the triangle as follows:

  • one side \(a = 7\)
  • let the second side be \(b\)
  • let the third side be \(x\)

We'll analyze our inequalities using the known side \(a = 7\):

  1. \(7 + b > x\) (i)
  2. \(7 + x > b\) (ii)
  3. \(b + x > 7\) (iii)

From inequality (i): \[ x < 7 + b \]

From inequality (ii): \[ x > b - 7 \]

From inequality (iii): \[ x > 7 - b \]

Now we can summarize the inequalities. To find the valid range for \(x\), we need to combine the results from these inequalities. Without loss of generality, we consider varying values of \(b\).

  1. If we know or assume certain values for \(b\), we can analyze the inequalities to find bounds for \(x\).
  2. For instance, setting some hypothetical values for \(b\), we can summarize:

Example with possible ranges for \(b\):

  • If \(b\) is positive and less than or equal to 7, i.e., \(0 < b \leq 7\):
    • From \(x > 7 - b\), since \(b\) can be at most 7, \(x\) must be positive.
    • From \(x < 7 + b\), as \(b\) approaches the upper limit, \(x\) should always be less than \(14\) when \(b = 7\).

This gives a general bound of: \[ x > 0 , \text{and} , x < 14 \]

Conclusion:

Thus, combining all the results of the triangle inequality: \[ x \in (|b - 7|, b + 7) \] If not further specifying \(b\), the range for any general third side \(x\) must fulfill \((0, 14)\).

Final conclusion: \[ x \in (0, 14) \] for a triangle with one side being 7.