The base of an isosceles triangle measures 80 cm and its area is 360 cm2.

Let $\triangle P Q R$ be an isosceles triangle and $P X \perp Q R$.

Now,

Area of triangle $=360 \mathrm{~cm}^{2}$

$\Rightarrow \frac{1}{2} \times Q R \times P X=360$

$\Rightarrow h=\frac{720}{80}=9 \mathrm{~cm}$

Now,

$Q X=\frac{1}{2} \times 80=40 \mathrm{~cm}$ and $P X=9 \mathrm{~cm}$

Also,

$P Q=\sqrt{Q X^{2}+P X^{2}}$

$a=\sqrt{40^{2}+9^{2}}=\sqrt{1600+81}=\sqrt{1681}=41 \mathrm{~cm}$

∴ Perimeter = 80 + 41 + 41  = 162 cm