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)

Question

3,10,11

3,8,10

8,10,11

3,8,11

Answer 1

To determine which combinations of toothpick lengths can form a triangle according to the triangle inequality theorem, we need to ensure that the sum of the lengths of any two sides of the triangle is greater than the length of the third side. That is, for three sides \(a\), \(b\), and \(c\), the following must hold true:

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

Now let's evaluate each combination:

1. Combination: 3, 10, 11

   • \(3 + 10 = 13 > 11\) (True)
   • \(3 + 11 = 14 > 10\) (True)
   • \(10 + 11 = 21 > 3\) (True)
   • This combination can form a triangle.

2. Combination: 3, 8, 10

   • \(3 + 8 = 11 > 10\) (True)
   • \(3 + 10 = 13 > 8\) (True)
   • \(8 + 10 = 18 > 3\) (True)
   • This combination can form a triangle.

3. Combination: 8, 10, 11

   • \(8 + 10 = 18 > 11\) (True)
   • \(8 + 11 = 19 > 10\) (True)
   • \(10 + 11 = 21 > 8\) (True)
   • This combination can form a triangle.

4. Combination: 3, 8, 11

   • \(3 + 8 = 11 \not> 11\) (False)
   • \(3 + 11 = 14 > 8\) (True)
   • \(8 + 11 = 19 > 3\) (True)
   • This combination cannot form a triangle.

Based on this analysis, the combinations that can form triangles are:

• 3, 10, 11
• 3, 8, 10
• 8, 10, 11

So the valid combinations are: 3, 10, 11; 3, 8, 10; 8, 10, 11.