Question

All the three sides of a $\Delta ABC$ 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

The correct option is C $2003$

The length of each side is less than the sum of the lengths of the other two sides and greater than the difference between these lengths.
So, $AC<AB+BCorAC<3003$
$\Longrightarrow \quad AC\quad \le \quad 3002$Similarly,$AC > AB - BC or AC > 999\Longrightarrow \quad AC\quad \ge \quad 1000$ThepossiblenumberoftrianglesthatcanbeformedinclusiveofbothvaluesofACare$(3002 - 1000) + 1 = 2003$