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?

1 answer

To determine how many different triangles Joanna can make using the straws of lengths 7, 9, 3, and 10 inches, we need to evaluate whether the combinations of three straws satisfy the triangle inequality theorem. This theorem states that for three sides to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side.

The lengths of the straws are:

  • A = 7 inches
  • B = 9 inches
  • C = 3 inches
  • D = 10 inches

Now, we will evaluate all the combinations of three straws that can be made from these straws.

The combinations of three straws from A, B, C, and D are:

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

Now let's check each combination against the triangle inequality:

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

    • 7 + 3 > 9 → 10 > 9 (True)
    • 7 + 9 > 3 → 16 > 3 (True)
    • 9 + 3 > 7 → 12 > 7 (True)
    • This combination can form a triangle.
  2. Combination A, B, D (7, 9, 10):

    • 7 + 9 > 10 → 16 > 10 (True)
    • 7 + 10 > 9 → 17 > 9 (True)
    • 9 + 10 > 7 → 19 > 7 (True)
    • This combination can form a triangle.
  3. Combination A, C, D (7, 3, 10):

    • 7 + 3 > 10 → 10 > 10 (False)
    • Since one condition fails, this combination cannot form a triangle.
  4. Combination B, C, D (9, 3, 10):

    • 9 + 3 > 10 → 12 > 10 (True)
    • 9 + 10 > 3 → 19 > 3 (True)
    • 3 + 10 > 9 → 13 > 9 (True)
    • This combination can form a triangle.

In summary, the combinations that can form triangles are:

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

Therefore, Joanna can make three different triangles using the straws with the following combinations:

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