How many distinct triangles can be formed with side lengths 4, 8 and 11?

Step-by-step solution

1. 1

   Recognise which existence test applies. All three sides are given, so this is an SSS situation. Unlike SSA, SSS never produces an ambiguous case: if a triangle exists at all it is unique up to congruence. So the answer can only be $0$ or $1$.

2. 2

   State the triangle inequality. Three positive lengths form a triangle exactly when each one is shorter than the sum of the other two:

   $a+b>c,\qquad a+c>b,\qquad b+c>a$

3. 3

   Test all three inequalities with $a=4$, $b=8$, $c=11$.

   $4+8=12>11\qquad\checkmark$
   $4+11=15>8\qquad\checkmark$
   $8+11=19>4\qquad\checkmark$

   In practice only the first needs checking — the inequality involving the longest side is always the binding one.

4. 4

   Note how close the case is. The margin is only $12>11$: had the longest side been $12$ the three lengths would be collinear (a degenerate triangle), and at $13$ no triangle would exist at all.

5. 5

   Conclude. All three inequalities hold and SSS fixes the shape, so exactly

   $\boxed{1}$

   triangle can be formed. As a further check, the largest angle opposite $11$ satisfies $\cos C=\tfrac{16+64-121}{2\cdot 4\cdot 8}=-\tfrac{41}{64}$, a valid cosine, giving $C\approx 129.9^\circ$ — obtuse, as expected when the longest side only just fits.