Triangle ABC has integral sides measuring 2001 units and 1002 units repectively. The no. of such triangles is?? Share with your friends Share 5 Varun.Rawat answered this We have ABC as the given ∆.Let AB = 2001 units ; BC = 1002 unitsWe know that sum of any 2 sides of a triangle is always greater than third side.So, AB + BC > CA⇒2001 + 1002 > CA⇒CA < 3003 unitsAlso, we know that, difference of any 2 sides of a triangle is always less than the third side.So, AB - BC < CA⇒2001 - 1002 < CA⇒CA > 999Hence, 999 < CA < 3003Lowest integral value that CA can take is 1000 and highest integral value that CA can take is 3002.Total number of possible values of CA = 2003hence, possible number of such triangles = 2003 16 View Full Answer