Question

All the three sides of a ΔABC have lengths in integral units, with AB = 2001 units and BC = 1002 units. The possible number of triangles with this condition is? a)2001 b)2002 c)2003 d)2004

Answer 1

Using the Triangle Inequality Theorem, the possible number of triangles with the given side lengths is 2003.

Step-by-step explanation:

To determine the number of possible triangles with the given conditions, we need to use the Triangle Inequality Theorem. According to the theorem, the sum of any two sides of a triangle must be greater than the third side. In this case, we have AB = 2001 units and BC = 1002 units. We can use these lengths to find the range of possible lengths for the third side.

1. The maximum possible length for the third side is when it is equal to the sum of the other two sides. So, the maximum length is 2001 + 1002 = 3003 units.
2. The minimum possible length for the third side is when it is equal to the difference between the lengths of the other two sides. So, the minimum length is 2001 - 1002 = 999 units.

Therefore, the possible range of lengths for the third side is 999 units to 3003 units. Since all three sides must have lengths in integral units, we count the number of integers in this range, which is 3003 - 999 + 1 = 2005 units. However, we must subtract 2 from this count because the lengths of AB and BC are already fixed. Therefore, the correct answer is 2005 - 2 = 2003 triangles (option c).

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