Number of Triangles When Two Sides Are Given (All Sides are Natural Numbers)

🔍 Proof Using Triangle Inequality

For a triangle with sides \( a, b, c \), the third side \( c \) must satisfy:

\[ |a - b| < c < a + b \]

Since \( a \leq b \), this simplifies to:

\[ b - a < c < a + b \]

Now count how many natural numbers lie between \( b - a \) and \( a + b \):

\[ \text{Number of values} = (a + b - 1) - (b - a + 1) + 1 = 2a - 1 \]

📌 Key Takeaway

• Always take the smaller side as a in the formula.
• This shortcut is valid only when all sides are natural numbers (positive integers).
• Very useful in exams like CAT, XAT, SSC CGL, etc. for saving time.