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

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

If two sides of a triangle are given as a and b, where \( a \leq b \), and the third side is a natural number, then the number of triangles that can be formed is:

Number of triangles = \( 2a - 1 \)

Condition: \( a, b \in \mathbb{N} \) (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 \( a + b \):

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

✅ Example

If two sides of a triangle are 5 and 8, how many triangles are possible?

Let \( a = 5 \), \( b = 8 \)

Then, the number of triangles = \( 2 \times 5 - 1 = 9 \)

Possible third sides: 4, 5, 6, 7, 8, 9, 10, 11, 12 (9 values)

📌 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.