Question

The longest side of a triangle is 10 cm, which group of line segments could form the other 2 sides of the triangle?

Answer 1

To determine which groups of line segments can form the other two sides of a triangle when the longest side (the hypotenuse if it's a right triangle) is 10 cm, we can use the triangle inequality theorem. This theorem states that for any triangle with sides \(a\), \(b\), and \(c\) (where \(c\) is the largest side), the following conditions must be satisfied:

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

In this case, \(c = 10\) cm. We need to find values for \(a\) and \(b\) such that:

1. \(a + b > 10\)
2. \(a + 10 > b\) (which will always be true if \(a, b < 10\))
3. \(b + 10 > a\) (which will also always be true if \(a, b < 10\))

From the first inequality, \(a + b > 10\), we can derive that both \(a\) and \(b\) must each be less than 10 cm, and the sum of \(a\) and \(b\) must be greater than 10 cm.

For example, potential pairs for \(a\) and \(b\) could be:

- \(a = 6\), \(b = 5\): \(6 + 5 = 11 > 10\) (valid)
- \(a = 7\), \(b = 4\): \(7 + 4 = 11 > 10\) (valid)
- \(a = 8\), \(b = 3\): \(8 + 3 = 11 > 10\) (valid)
- \(a = 9\), \(b = 2\): \(9 + 2 = 11 > 10\) (valid)
- \(a = 5\), \(b = 6\): \(5 + 6 = 11 > 10\) (valid)

And so forth, as long as both line segments \(a\) and \(b\) remain positive and their combined length exceeds 10 cm.

So a valid pair of segments could be any combination where both numbers are less than 10 cm and their sum exceeds 10 cm. For example:
- \( (7, 5) \)
- \( (6, 6) \)
- \( (8, 3) \)
- \( (9, 2) \)

In conclusion, any two line segments whose lengths satisfy the condition \(a + b > 10\) while being less than 10 cm would form valid sides of the triangle with the longest side being 10 cm.