Question

triangle abc has the two given side lengths. • ab = 11 units • ac = 6 units select all the inequalities that must be true about the length of bc. □ 6 < bc □ 11 > bc □ 6 + 11 > bc □ 6 + 11 < bc □ 11 − 6 > bc

Explanation:

Step1: Recall Triangle Inequality Theorem

The triangle inequality theorem states that for any triangle with side lengths \(a\), \(b\), and \(c\), the following must hold:
- \(a + b>c\)
- \(a + c>b\)
- \(b + c>a\)
Also, the difference between two sides must be less than the third side, i.e., \(|a - b|\lt c\), \(|a - c|\lt b\), \(|b - c|\lt a\).

For triangle \(ABC\) with \(AB = 11\) (let \(c = 11\)), \(AC=6\) (let \(b = 6\)) and \(BC=a\) (let \(a\) be the length of \(BC\)).

Step2: Analyze each inequality

- **Inequality 1: \(6\lt BC\)**
Using the triangle inequality \(|AB - AC|\lt BC\). \(|11 - 6|=5\), so \(5\lt BC\). But we can also derive from \(AB+AC>BC\) and \(AB - AC<BC\). Also, from \(AC + BC>AB\) (i.e., \(6+BC > 11\)), we get \(BC>11 - 6=5\). But we can also check \(AB>BC - AC\)? Wait, no, the correct form is \(|AB - AC|\lt BC\) and \(BC<AB + AC\) and \(BC>AB - AC\) (since \(AB>AC\)). Wait, \(AB - AC=11 - 6 = 5\), so \(BC>5\). But the inequality is \(6\lt BC\). Is this always true? Let's see, if \(BC = 5.5\), then \(6\lt5.5\) is false. Wait, maybe I made a mistake. Wait, no, let's re - express the triangle inequalities properly.

The three triangle inequalities for sides \(AB = 11\), \(AC = 6\), \(BC=x\) (let \(x = BC\)) are:
1. \(AB+AC>BC\) i.e., \(11 + 6>x\) or \(17>x\)
2. \(AB + BC>AC\) i.e., \(11+x>6\) (which is always true for positive \(x\))
3. \(AC+BC>AB\) i.e., \(6 + x>11\) or \(x>11 - 6=5\)

Now let's analyze each option:
- Option 1: \(6\lt BC\) (i.e., \(x > 6\)): From \(x>5\), \(x\) can be between \(5\) and \(6\) (e.g., \(x = 5.5\)), so this is not always true. Wait, maybe I messed up. Wait, no, let's check again. Wait, \(AC = 6\), \(AB = 11\). If \(BC=5.5\), then \(AC + BC=6 + 5.5 = 11.5>11=AB\), \(AB+BC=11 + 5.5 = 16.5>6 = AC\), \(AB + AC=17>5.5 = BC\). So \(BC = 5.5\) is valid, and \(6\lt5.5\) is false. So this inequality is not always true. Wait, maybe the question has a typo? Or maybe I misread. Wait, no, let's check the other options.

- Option 2: \(11>BC\) (i.e., \(x<11\)): From \(x>5\) and \(x < 17\), \(x\) can be greater than or equal to \(11\) (e.g., \(x = 12\), since \(6+12 = 18>11\), \(11 + 12=23>6\), \(11+6 = 17>12\)). So \(11>x\) is not always true.

- Option 3: \(6 + 11>BC\) (i.e., \(17>x\)): This is the first triangle inequality \(AB + AC>BC\), which is always true for a triangle.

- Option 4: \(6 + 11<BC\) (i.e., \(17<x\)): This violates the triangle inequality \(AB + AC>BC\), so it's false.

- Option 5: \(11-6>BC\) (i.e., \(5>x\)): But from the triangle inequality \(AC + BC>AB\) we have \(x>5\), so this is false.

Wait, maybe I made a mistake in analyzing option 1. Wait, let's re - express the triangle inequalities. The correct set of inequalities from the triangle inequality theorem for sides \(a\), \(b\), \(c\) is \(|a - b|\lt c\lt a + b\). Here, \(a = 11\), \(b = 6\), so \(|11 - 6|\lt BC\lt11 + 6\), i.e., \(5\lt BC\lt17\).

Now let's re - check each option:
- \(6\lt BC\): \(BC\) can be between \(5\) and \(6\) (e.g., \(BC = 5.5\)), so this is not always true.
- \(11>BC\): \(BC\) can be between \(11\) and \(17\) (e.g., \(BC = 12\)), so this is not always true.
- \(6 + 11>BC\) (i.e., \(17>BC\)): This is part of \(5\lt BC\lt17\), so it's always true.
- \(6 + 11<BC\) (i.e., \(17<BC\)): This is false as \(BC\lt17\).
- \(11 - 6>BC\) (i.e., \(5>BC\)): This is false as \(BC>5\).

Wait, but maybe the question has a different approach. Wait, the options are:

1. \(6\lt BC\)
2. \(11>BC\)
3. \(6 + 11>BC\)
4. \(6 + 11<BC\)
5. \(11 - 6>BC\)

Wait, maybe I made a mistake in the first analysis. Let's consider the three triangle inequalities:

1. \(AB+AC>BC\) (11 + 6>BC) → \(17>BC\) (this is true)
2. \(AB + BC>AC\) (11+BC>6) → \(BC>- 5\) (always true since length is positive)
3. \(AC + BC>AB\) (6 + BC>11) → \(BC>5\)

Now, let's check the options:

- \(6\lt BC\): From \(BC>5\), \(BC\) can be \(5.5\) (which is less than 6), so this is not always true.
- \(11>BC\): \(BC\) can be \(12\) (which is greater than 11), so this is not always true.
- \(6 + 11>BC\) (17>BC): This is the first inequality, so it's true.
- \(6 + 11<BC\) (17<BC): False.
- \(11 - 6>BC\) (5>BC): False, since \(BC>5\)

Answer:

The inequality that must be true is \(6 + 11>BC\) (the third option). Also, let's re - check the first option. Wait, if we consider the triangle inequality \(AC+BC>AB\) (6 + BC>11) → BC>5, and \(AB + BC>AC\) (11+BC>6) → BC>- 5, and \(AB + AC>BC\) (17>BC). Now, is \(6\lt BC\) always true? No, because BC can be 5.5. Is \(11>BC\) always true? No, because BC can be 12. The only one that is always true is \(6 + 11>BC\). Also, wait, maybe the first option: \(6\lt BC\). Wait, if BC is 6, then \(AC + BC=6 + 6 = 12>11\), \(AB+BC=11 + 6 = 17>6\), \(AB + AC=17>6\). So BC = 6 is valid? Wait, no, \(AC + BC>AB\) when BC = 6: \(6+6 = 12>11\), which is true. Wait, \(|AB - AC|=5\), so BC must be greater than 5. So BC can be 6. So \(6\lt BC\) is not true when BC = 6. But \(6\leq BC\) would be true? Wait, the option is \(6\lt BC\) (strictly greater). So when BC = 6, \(6\lt BC\) is false. So the only inequality that is always true is \(6 + 11>BC\).