JEE Main 2017 · 9 Apr · Shift 1, Mathematics question 89

The triangle on sides 2 and 5 has area $\frac{5\sqrt3}2$, and its third side (the diagonal) is $\sqrt{4+25-10}=\sqrt{19}$. The opposite angle is 120°, so the other sides $c,d$ satisfy $\frac{\sqrt3}4cd=4\sqrt3-\frac{5\sqrt3}2$, i.e. $cd=6$, and $c^2+d^2+cd=19$. Then $(c+d)^2=25$ and the perimeter is $2+5+5=12$.