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

Answer 1

To determine whether Sarah can make a triangle with combinations of toothpicks of different lengths, we can use the triangle inequality theorem. According to this theorem, for any triangle with side lengths \( a \), \( b \), and \( c \), the following must hold true:

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

Given the toothpick lengths: 3 inches, 8 inches, 10 inches, and 11 inches, we can check combinations of 3 lengths at a time to see if they can form a triangle.

Let's evaluate all the combinations:

1. **3, 8, 10**:
- \( 3 + 8 = 11 \) (not > 10)
- This combination cannot form a triangle.

2. **3, 8, 11**:
- \( 3 + 8 = 11 \) (not > 11)
- This combination cannot form a triangle.

3. **3, 10, 11**:
- \( 3 + 10 = 13 > 11 \) (holds)
- \( 3 + 11 = 14 > 10 \) (holds)
- \( 10 + 11 = 21 > 3 \) (holds)
- This combination can form a triangle.

4. **8, 10, 11**:
- \( 8 + 10 = 18 > 11 \) (holds)
- \( 8 + 11 = 19 > 10 \) (holds)
- \( 10 + 11 = 21 > 8 \) (holds)
- This combination can form a triangle.

So, based on the evaluations:

- The valid combinations that can form triangles are **(3, 10, 11)** and **(8, 10, 11)**.
- Hence, Sarah can make **2 triangles** with the toothpicks of given lengths.