The perimeter of a triangle whose sides are in A.P. is $$21\,cm$$  and the product of lengths of the shortest side and the longest side exceeds the length of the other side by $$6\,cm.$$  The longest side of the triangle is

Solution :
Let the sides of the triangle be $$a - d, a, a + d$$    then
Perimeter $$ = \left( {a + d} \right) + a + \left( {a - d} \right) = 21$$
∴ $$a = 7$$
Again, $$\left( {a - d} \right)\left( {a + d} \right) = a + 6$$
$$\eqalign{ & \Rightarrow {a^2} - {d^2} = a + 6 \cr & \Rightarrow 49 - {d^2} = 13 \cr} $$
$$\therefore d = \pm 6.$$   Hence, the sides of the triangle are $$1\,cm, 7\,cm, 13\,cm.$$