Question

The number of triangles with integral sides that can be made which have perimeter of 14, are:

• A. 6
• B. 5
• C. 4
• D. 3

Hint:

When counting integer-sided triangles with a fixed perimeter, fix an order ($a \leq b \leq c$) and apply the triangle inequality $a+b>c$ to limit cases quickly.

Answer 1

The Correct Option is C

We are told that the perimeter of the triangle is \(14\). Let the three sides be \(a\), \(b\), and \(c\) such that: \[ a + b + c = 14 \] Also, \(a, b, c\) are positive integers and must satisfy the triangle inequalities: \[ a + b > c,\quad b + c > a,\quad c + a > b \]

Step 1: Assume ordering to avoid repeats

Let us assume \(a \leq b \leq c\). This ensures that we count each triangle only once.

Step 2: Apply the largest side restriction

If \(c\) is the largest side, then \(a + b > c\). From \(a + b = 14 - c\), we have: \[ 14 - c > c \quad \Rightarrow \quad 14 > 2c \quad \Rightarrow \quad c < 7 \] Thus, the largest side \(c\) can be at most \(6\).

Step 3: List all possibilities

For each \(c\), find \(a\) and \(b\) (integers, \(a \leq b \leq c\)):

• \(c = 6\): \(a + b = 8\) Possibilities: \((2,6)\) invalid (\(2 + 6 = 8\) not \(> 6\)), \((3,5)\) valid, \((4,4)\) valid. Triangles: \((3,5,6)\), \((4,4,6)\).
• \(c = 5\): \(a + b = 9\) Possibilities: \((4,5)\) valid, \((3,6)\) invalid (\(b > c\) ordering fail), \((2,7)\) invalid. Triangle: \((4,5,5)\).
• \(c = 4\): \(a + b = 10\), but with \(b \leq 4\) this is impossible unless \(a \geq 6\), which breaks ordering.

Step 4: Total triangles

Unique sets: \((3,5,6)\), \((4,4,6)\), \((4,5,5)\). \((2,6,6)\) invalid, and \((5,5,4)\) is the same as \((4,5,5)\). Thus: \[ \boxed{\text{Number of triangles} = 3} \]