In a triangle, the lengths of sides are 6 units, 12 units and x units. How many possible integer values are there for x?

Correct — A. Use the triangle inequality: each side must be less than the sum of the other two and greater than their difference.

x < 6 + 12 = 18
x > 12 − 6 = 6

So 6 < x < 18, and x may be any integer from 7 to 17.

Count them: 17 − 7 + 1 = 11 → option (a).

Why the others are wrong

• (b)
  12 — 12 keeps one endpoint — either 6 or 18. At those values the three lengths lie flat in a straight line, giving zero area, so neither is a triangle.
• (c)
  10 — 10 is 17 − 7, the difference of the endpoints. A run of whole numbers holds one more than that difference, so 7 to 17 is eleven values.
• (d)
  9 — 9 would need a range two shorter than the real one. Nothing excludes 7 or 17 — 6 + 7 > 12 makes 7 valid, and 17 < 18 makes 17 valid.

For three lengths to close into a triangle the largest must be shorter than the sum of the other two. Equality gives a flat, degenerate figure with no area, which does not count.

With two sides fixed at 6 and 12 the third is boxed in at both ends: |12 − 6| < x < 12 + 6.

The marks are lost at the last step. Counting the integers inside an open range means n − m + 1, not n − m.

Key facts

• Triangle inequality: |a − b| < c < a + b for every triangle.
• Sides of 6 and 12 force 6 < x < 18.
• The count of integers from m to n inclusive is n − m + 1.

Common traps

• Including x = 6 or x = 18, which give a straight line rather than a triangle
• Answering 17 − 7 = 10 instead of 17 − 7 + 1 = 11

No directly related past PYQ was found.