How many triangles have sides a = 4, b = 7 and c = 11?

Step-by-step solution

1. 1

   State the triangle inequality precisely. Three positive lengths form a triangle exactly when the sum of every two of them is strictly greater than the third. In practice only one check is needed: the two shorter sides against the longest.

2. 2

   Identify the longest side and test. The longest is $c=11$, and

   $4+7=11$

3. 3

   Interpret the equality case. The sum is equal to, not greater than, the third side. Equality means the three segments lay end to end exactly along a straight line: the "triangle" collapses to a segment with zero area and zero height. Such a degenerate figure is not counted as a triangle.

4. 4

   Conclude.

   $\boxed{0\ \text{triangles}}$

5. 5

   Contrast with a nearby case. Changing $b$ from $7$ to $8$ gives $4+8=12>11$, and the other two checks ($4+11>8$, $8+11>4$) hold as well, so sides $4,8,11$ do determine exactly one triangle — three fixed side lengths always give at most one triangle up to congruence. The count in this family of questions is therefore always $0$ or $1$; a count of $2$ arises only in SSA (side-side-angle) problems, not SSS.