Question

How many differently shaped triangles exist in which no two sides are of the same length, each side is of integral unit length and the perimeter of the triangle is less than 14 units?

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

List scalene integer-sided triangles a less than b less than c with a + b greater than c and perimeter under 14; check systematically starting from the smallest side.

Answer 1

The Correct Option is C

Step 1: Set up the conditions.
We need sides a less than b less than c, all positive integers and all different, satisfying the triangle inequality \(a + b > c\), with perimeter \(a + b + c < 14\).

Step 2: Try the smallest side a = 1.
If a = 1, the triangle inequality needs \(1 + b > c\), that is \(c < b + 1\). Since c > b already (sides are increasing and all different), no integer c can be both greater than b and less than \(b+1\). So a = 1 gives no valid triangle.

Step 3: Try a = 2.
Here \(2 + b > c\), so \(c < b + 2\), meaning c can only be \(b+1\).
For b = 3: c = 4, perimeter = 2+3+4 = 9, less than 14. Valid.
For b = 4: c = 5, perimeter = 2+4+5 = 11, less than 14. Valid.
For b = 5: c = 6, perimeter = 2+5+6 = 13, less than 14. Valid.
For b = 6: c = 7, perimeter = 2+6+7 = 15, not less than 14. Stop.
This gives 3 triangles: (2,3,4), (2,4,5), (2,5,6).

Step 4: Try a = 3.
Now \(3 + b > c\), so c can be \(b+1\) or \(b+2\).
For b = 4: c = 5 gives perimeter 12, valid; c = 6 gives perimeter 13, valid.
For b = 5: c = 6 gives perimeter 14, which is not less than 14, so this fails, and any larger c only makes the perimeter bigger.
This gives 2 more triangles: (3,4,5), (3,4,6).

Step 5: Try a = 4 and beyond.
The smallest possible triangle with a = 4 is (4,5,6), whose perimeter is already 15, not less than 14. So no triangles exist for a of 4 or more.

Final Answer:
Total valid triangles are (2,3,4), (2,4,5), (2,5,6), (3,4,5), (3,4,6), which is \[ \boxed{5} \]