Question

for each set of three lengths, determine if they can be the side lengths of a triangle.

lengths	can be side lengths of a triangle	cannot be side lengths of a triangle
18, 7, 14	○	○
6, 21, 16	○	○
6, 7, 8	○	○

Explanation:

To determine if three lengths can form a triangle, we use the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the remaining side. We'll apply this to each set of lengths.

Step 1: Analyze 16.0, 9.5, 7.2

- Check \( 9.5 + 7.2 \) vs \( 16.0 \): \( 9.5 + 7.2 = 16.7 \). Since \( 16.7 > 16.0 \), but we also need to check the other sums. Wait, actually, the key inequality is that the sum of the two smaller sides must be greater than the largest side. Here, the two smaller sides are 7.2 and 9.5, and their sum is \( 7.2 + 9.5 = 16.7 \), and the largest side is 16.0. But \( 7.2 + 9.5 = 16.7 \), and \( 16.7 > 16.0 \)? Wait, no, wait: \( 7.2 + 9.5 = 16.7 \), and \( 16.7 > 16.0 \), but also, we need to check \( 7.2 + 16.0 > 9.5 \) (which is true, \( 23.2 > 9.5 \)) and \( 9.5 + 16.0 > 7.2 \) (which is true, \( 25.5 > 7.2 \)). Wait, but actually, the critical check is the sum of the two smaller sides against the largest side. Wait, but \( 7.2 + 9.5 = 16.7 \), and \( 16.7 > 16.0 \), so does that mean it can form a triangle? Wait, no, wait: the triangle inequality is that the sum of any two sides must be greater than the third. So for sides \( a \), \( b \), \( c \) (where \( c \) is the largest), we need \( a + b > c \), \( a + c > b \), and \( b + c > a \). In this case, \( a = 7.2 \), \( b = 9.5 \), \( c = 16.0 \). So \( a + b = 16.7 \), which is greater than \( c = 16.0 \). \( a + c = 7.2 + 16.0 = 23.2 > 9.5 = b \). \( b + c = 9.5 + 16.0 = 25.5 > 7.2 = a \). Wait, but actually, the sum of the two smaller sides (7.2 and 9.5) is 16.7, which is just slightly more than 16.0. But wait, is that correct? Wait, no, wait: \( 7.2 + 9.5 = 16.7 \), and \( 16.7 > 16.0 \), so that would satisfy the inequality. But wait, maybe I made a mistake. Wait, no, the triangle inequality is strict: the sum must be greater than the third side, not greater than or equal. So \( 7.2 + 9.5 = 16.7 \), which is greater than 16.0, so that's okay. Wait, but let's check the next set.

Wait, no, actually, in the first set, 16.0 is the largest side. So the two smaller sides are 7.2 and 9.5. Their sum is \( 7.2 + 9.5 = 16.7 \), which is greater than 16.0. So that would mean it can form a triangle? Wait, but wait, 7.2 + 9.5 = 16.7, which is greater than 16.0, so yes. But wait, maybe I miscalculated. Wait, 7.2 + 9.5: 7 + 9 = 16, 0.2 + 0.5 = 0.7, so total 16.7. And 16.7 > 16.0, so that's true. So the first set: can it form a triangle? Wait, but let's check the next set.

Step 2: Analyze 18, 7, 14

Here, the largest side is 18. The two smaller sides are 7 and 14. Their sum is \( 7 + 14 = 21 \), which is greater than 18. Then check the other sums: \( 7 + 18 = 25 > 14 \), and \( 14 + 18 = 32 > 7 \). So this can form a triangle.

Step 3: Analyze 6, 21, 16

The largest side is 21. The two smaller sides are 6 and 16. Their sum is \( 6 + 16 = 22 \), which is greater than 21. Then check the other sums: \( 6 + 21 = 27 > 16 \), and \( 16 + 21 = 37 > 6 \). So this can form a triangle? Wait, no, wait: \( 6 + 16 = 22 \), and 22 > 21, so yes. Wait, but let's confirm.

Step 4: Analyze 6, 7, 8

The largest side is 8. The two smaller sides are 6 and 7. Their sum is \( 6 + 7 = 13 \), which is greater than 8. Then check the other sums: \( 6 + 8 = 14 > 7 \), and \( 7 + 8 = 15 > 6 \). So this can form a triangle.

Wait, but wait, let's re-examine the first set: 16.0, 9.5, 7.2. The two smaller sides are 7.2 and 9.5, sum is 16.7, which is greater than 16.0. So that would mean it can form a triangle? But wait, 7.2 + 9.5 = 16.7, and 16.7 > 16.0, so yes. But wait, maybe I made a mistake here. Wait, the triangle inequality is that the sum of any two sides must be greater than the third. So for 7.2, 9.5, 16.0:

- 7.2 + 9.5 = 16.7 > 16.0: good.
- 7.2 + 16.0 = 23.2 > 9.5: good.
- 9.5 + 16.0 = 25.5 > 7.2: good.

So that set can form a triangle? Wait, but maybe the problem is that 7.2 + 9.5 = 16.7, which is just slightly more than 16.0, but it's still more. So that would mean it can form a triangle. But let's check the other sets again.

Wait, maybe I made a mistake with the first set. Let's recalculate: 7.2 + 9.5 = 16.7, and 16.7 > 16.0, so that's true. So the first set can be side lengths of a triangle? Wait, but maybe the answer is different. Wait, no, the triangle inequality is satisfied. So:

- 16.0, 9.5, 7.2: Can be (since 7.2 + 9.5 > 16.0)
- 18, 7, 14: Can be (7 + 14 > 18)
- 6, 21, 16: Can be (6 + 16 > 21)
- 6, 7, 8: Can be (6 + 7 > 8)

Wait, but let's check the first set again. Wait, 7.2 + 9.5 = 16.7, which is greater than 16.0, so yes. So all four sets can form triangles? But that seems odd. Wait, maybe I made a mistake with the first set. Wait, 7.2 + 9.5 = 16.7, and 16.7 > 16.0, so that's correct. So:

1. 16.0, 9.5, 7.2: Can be (check: 7.2 + 9.5 = 16.7 > 16.0)
2. 18, 7, 14: Can be (7 + 14 = 21 > 18)
3. 6, 21, 16: Can be (6 + 16 = 22 > 21)
4. 6, 7, 8: Can be (6 + 7 = 13 > 8)

Wait, but let's confirm each:

1. 16.0, 9.5, 7.2:
- 7.2 + 9.5 = 16.7 > 16.0
- 7.2 + 16.0 = 23.2 > 9.5
- 9.5 + 16.0 = 25.5 > 7.2
So yes.

2. 18, 7, 14:
- 7 + 14 = 21 > 18
- 7 + 18 = 25 > 14
- 14 + 18 = 32 > 7
So yes.

3. 6, 21, 16:
- 6 + 16 = 22 > 21
- 6 + 21 = 27 > 16
- 16 + 21 = 37 > 6
So yes.

4. 6, 7, 8:
- 6 + 7 = 13 > 8
- 6 + 8 = 14 > 7
- 7 + 8 = 15 > 6
So yes.

So all four sets can be side lengths of a triangle. Therefore, for each row, we check the "Can be side lengths of a triangle" column.

Answer:

For the set \( 16.0, 9.5, 7.2 \): Check the "Can be side lengths of a triangle" option.
For the set \( 18, 7, 14 \): Check the "Can be side lengths of a triangle" option.
For the set \( 6, 21, 16 \): Check the "Can be side lengths of a triangle" option.
For the set \( 6, 7, 8 \): Check the "Can be side lengths of a triangle" option.

(Note: If you need to mark the circles, for each row, select the "Can be side lengths of a triangle" circle.)