Question

Which of the following combinations of side lengths will form a triangle with vertices X, Y, and Z?

(A.) XY = 18 mm , YZ = 20 mm , XZ = 39 mm

(B. ) XY = 17 mm , YZ = 20 mm , XZ = 36 mm

(C.) XY = 16 mm , YZ = 20 mm , XZ = 37 mm

(D.) XY = 17 mm , YZ = 19 mm , XZ = 37 mm

Answer 1

(A.) A triangle will be formed with vertices X, Y, and Z with side lengths XY = 18 mm, YZ = 20 mm, and XZ = 39 mm. This is because the sum of the lengths of any two sides of a triangle must be greater than the length of the third side according to the triangle inequality theorem. In this case, XY + YZ = 18 mm + 20 mm = 38 mm, which is greater than XZ = 39 mm, satisfying the triangle inequality theorem.

(B.) A triangle will be formed with vertices X, Y, and Z with side lengths XY = 17 mm, YZ = 20 mm, and XZ = 36 mm. This is because XY + YZ = 17 mm + 20 mm = 37 mm, which is greater than XZ = 36 mm, satisfying the triangle inequality theorem.

(C.) A triangle will not be formed with vertices X, Y, and Z with side lengths XY = 16 mm, YZ = 20 mm, and XZ = 37 mm. This is because XY + YZ = 16 mm + 20 mm = 36 mm, which is not greater than XZ = 37 mm, violating the triangle inequality theorem.

(D.) A triangle will be formed with vertices X, Y, and Z with side lengths XY = 17 mm, YZ = 19 mm, and XZ = 37 mm. This is because XY + YZ = 17 mm + 19 mm = 36 mm, which is greater than XZ = 37 mm, satisfying the triangle inequality theorem.

Explanation:

Answer 2

Answer B.
a+b>c
17+20>36
37>36