The area of an isosceles \( \Large \triangle ABC \) with AB = AC and altitude AD = 3 cm is 12 sq cm. What is its perimeter?

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