Let AB = CA = a cm and base = b cm
Now, area of the \( \Large \triangle ABC \)= \( \Large \frac{1}{2}\times b\times h \)
=> 12= \( \Large \frac{1}{2}\times b\times 3 \)

b= \( \Large \frac{12\times 2}{3} \)=8cm
Here, BD=CD=\( \Large \frac{b}{2}=\frac{8}{2} \)=4cm
In right angled \( \Large \triangle ABD \) , by Pythagoras theorem,
AB=\( \Large \sqrt{BD^{2}+AD^{2}} \) => a=\( \Large \sqrt{4^{2}+3^{2}} \)
=\( \Large \sqrt{16+9}=\sqrt{25} \)=5cm
Now, perimeter of an isosceles triangle
= 2a + b = \( \Large 2\times 5+8 \)
= 10 + 8 = 18 cm