Question:
What is the number of distinct triangles with integral valued sides and perimeter 14?
What is the number of distinct triangles with integral valued sides and perimeter 14?
Show Hint
Order the sides as \(a \le b \le c\). The triangle inequality with perimeter 14 forces the longest side \(c\) to be less than 7.
Updated On: Nov 4, 2026
- 6
- 5
- 4
- 3
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Set up the condition.
Let the sides be integers \(a \le b \le c\) with \(a+b+c=14\). A triangle exists only if \(a+b>c\), i.e. \(14-c>c\), so \(c<7\), i.e. \(c \le 6\). Also \(c \ge 5\) because \(3c \ge 14\).
Step 2: Enumerate by the longest side.
\(c=6\): \(a+b=8\) with \(a \le b \le 6\): \((2,6),(3,5),(4,4)\), so 3 triangles.
\(c=5\): \(a+b=9\) with \(a \le b \le 5\): \((4,5)\), so 1 triangle.
\(c=4\): \(a+b=10\) with \(b \le 4\) is impossible.
Step 3: Total.
The triangles are \((2,6,6),(3,5,6),(4,4,6),(4,5,5)\), a total of 4.
Final Answer:
The correct option is (3): 4
Let the sides be integers \(a \le b \le c\) with \(a+b+c=14\). A triangle exists only if \(a+b>c\), i.e. \(14-c>c\), so \(c<7\), i.e. \(c \le 6\). Also \(c \ge 5\) because \(3c \ge 14\).
Step 2: Enumerate by the longest side.
\(c=6\): \(a+b=8\) with \(a \le b \le 6\): \((2,6),(3,5),(4,4)\), so 3 triangles.
\(c=5\): \(a+b=9\) with \(a \le b \le 5\): \((4,5)\), so 1 triangle.
\(c=4\): \(a+b=10\) with \(b \le 4\) is impossible.
Step 3: Total.
The triangles are \((2,6,6),(3,5,6),(4,4,6),(4,5,5)\), a total of 4.
Final Answer:
The correct option is (3): 4
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