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joanna has four straws of different lengths. The table shows the lengths of the straws.

Straw A B C D
Length in inches 7 9 3 10

Joanna makes as many triangles as she can using combinations of three of these straws. Determine how many different triangles she can make. What combinations of straws does she use? Show your work.(4 points)

1 answer

To determine how many different triangles Joanna can make with the straws, we can use the triangle inequality theorem. This theorem states that, for three lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side.

The straws are as follows:

  • Straw A: 7 inches
  • Straw B: 9 inches
  • Straw C: 3 inches
  • Straw D: 10 inches

Now, let's consider all combinations of three straws from the four available:

  1. Combination A, B, C (7, 9, 3)
  2. Combination A, B, D (7, 9, 10)
  3. Combination A, C, D (7, 3, 10)
  4. Combination B, C, D (9, 3, 10)

Now, let's check which combinations satisfy the triangle inequality:

  1. Combination A, B, C (7, 9, 3)

    • Check:
      \(7 + 9 > 3\) (16 > 3) ✅
      \(7 + 3 > 9\) (10 > 9) ✅
      \(9 + 3 > 7\) (12 > 7) ✅
    • This combination can form a triangle.

  2. Combination A, B, D (7, 9, 10)

    • Check:
      \(7 + 9 > 10\) (16 > 10) ✅
      \(7 + 10 > 9\) (17 > 9) ✅
      \(9 + 10 > 7\) (19 > 7) ✅
    • This combination can form a triangle.

  3. Combination A, C, D (7, 3, 10)

    • Check:
      \(7 + 3 > 10\) (10 > 10) ❌
      \(7 + 10 > 3\) (17 > 3) ✅
      \(3 + 10 > 7\) (13 > 7) ✅
    • This combination cannot form a triangle.

  4. Combination B, C, D (9, 3, 10)

    • Check:
      \(9 + 3 > 10\) (12 > 10) ✅
      \(9 + 10 > 3\) (19 > 3) ✅
      \(3 + 10 > 9\) (13 > 9) ✅
    • This combination can form a triangle.

Summary of Valid Combinations:

  • The valid combinations that can form triangles are:
    • A, B, C (7, 9, 3)
    • A, B, D (7, 9, 10)
    • B, C, D (9, 3, 10)

Total Different Triangles:

Joanna can make 3 different triangles with the following combinations:

  1. (7, 9, 3)
  2. (7, 9, 10)
  3. (9, 3, 10)