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

Solution:

Given  :
Base = 80 cm

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

Area of an isosceles triangle $=\left(\frac{1}{4} b \sqrt{4 a^{2}-b^{2}}\right)$

$\Rightarrow 360=\frac{1}{4} \times 80 \sqrt{4 a^{2}-80^{2}}$

$\Rightarrow 360=20 \sqrt{4 a^{2}-6400}$

$\Rightarrow 18=2 \sqrt{a^{2}-1600}$

 

$\Rightarrow 9=\sqrt{a^{2}-1600}$

Squaring both the sides, we get:

$\Rightarrow 81=a^{2}-1600$

$\Rightarrow a^{2}=1681$

$\Rightarrow a=41 \mathrm{~cm}$

Perimeter $=(2 a+b)$

$=[2(41)+80]=82+80=162 \mathrm{~cm}$

So, the perimeter of the triangle is 162 cm.