Question

Sarah has toothpicks that are 4 different lengths, 3 in, 8in, 10in, 11in. She is trying to make as many triangles as she can. How many of the combinations below will make triangles according to the triangle inequality theorem? Pick ALL that apply(3 points) Responses 3,8,11 3,8,11 8,10,11 8,10,11 3,10,11 3,10,11 3,8,10

Answer 1

To determine whether a set of three lengths can form a triangle, they must satisfy the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the third side. We will check each combination:

1. **3, 8, 11**
- \(3 + 8 = 11\) (not greater than 11, so this combination **does not form a triangle**)

2. **3, 8, 11** (duplicate)
- Same computation as above, so this combination **does not form a triangle**.

3. **8, 10, 11**
- \(8 + 10 = 18 > 11\) (valid)
- \(8 + 11 = 19 > 10\) (valid)
- \(10 + 11 = 21 > 8\) (valid)
- All inequalities hold, so this combination **forms a triangle**.

4. **8, 10, 11** (duplicate)
- Same as above, so this combination **forms a triangle**.

5. **3, 10, 11**
- \(3 + 10 = 13 > 11\) (valid)
- \(3 + 11 = 14 > 10\) (valid)
- \(10 + 11 = 21 > 3\) (valid)
- All inequalities hold, so this combination **forms a triangle**.

6. **3, 10, 11** (duplicate)
- Same as above, so this combination **forms a triangle**.

7. **3, 8, 10**
- \(3 + 8 = 11 > 10\) (valid)
- \(3 + 10 = 13 > 8\) (valid)
- \(8 + 10 = 18 > 3\) (valid)
- All inequalities hold, so this combination **forms a triangle**.

Based on the evaluations:
- The combinations that form triangles are **8, 10, 11** (counted twice), **3, 10, 11** (counted twice), and **3, 8, 10**.

Thus, the combinations that will make triangles are:
- 8, 10, 11
- 3, 10, 11
- 3, 8, 10

In total, **3 unique combinations form triangles**.