Let one of the sides of a triangle be  $17cm$  and the sum of all the sides of the triangle be $40cm$. If the sum of two adjacent sides is  $35cm$, then the area (in sq. cms) of the triangle is

According to given informations,
Let  $a=17$,   $a+b+c=40$  and   $a+b=35$
So,  $b=18$  and  $c=5$
 $\because$   Area of required triangle is
 $\sqrt{s(s-a)(s-b)(s-c)}$
   $[where,s=\frac{a+b+c}{2}=20]$
 $=\sqrt{20\left(3\right)\left(2\right)\left(15\right)}=30\sqrt{2}$
Hence, option (3) is correct