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Delta ABC has integeral sides AB, BC mea...

`Delta ABC` has integeral sides AB, BC measuring 2001 units 1002 units respechively the number of such triangle is

A

2001

B

202

C

2003

D

2004

Text Solution

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The correct Answer is:
To find the number of triangles that can be formed with sides AB = 2001 units, BC = 1002 units, and AC as an unknown integer side, we will use the triangle inequality theorem. The theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ### Step-by-Step Solution: 1. **Identify the sides of the triangle**: - Let AB = 2001 units - Let BC = 1002 units - Let AC = x (unknown side) 2. **Apply the triangle inequality**: We need to satisfy the following conditions: - Condition 1: AB + BC > AC - Condition 2: AB + AC > BC - Condition 3: AC + BC > AB 3. **Substituting the known values into the inequalities**: - From Condition 1: \[ 2001 + 1002 > x \implies 3003 > x \implies x < 3003 \] - From Condition 2: \[ 2001 + x > 1002 \implies x > 1002 - 2001 \implies x > -999 \text{ (which is always true since x is positive)} \] - From Condition 3: \[ x + 1002 > 2001 \implies x > 2001 - 1002 \implies x > 999 \] 4. **Combine the inequalities**: From the inequalities derived, we have: \[ 999 < x < 3003 \] 5. **Determine the integer values for x**: The integer values for x can range from 1000 to 3002 inclusive. To find the total number of integer solutions: - The smallest integer value for x is 1000. - The largest integer value for x is 3002. 6. **Calculate the number of possible integer values**: The number of integers from 1000 to 3002 can be calculated as: \[ 3002 - 1000 + 1 = 2003 \] ### Conclusion: The total number of triangles that can be formed with the given sides is **2003**. ---
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