Area (in sq. units) of the region outside

$\frac{|x|}{2}+\frac{|y|}{3}=1$

$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$

Area of Ellipse $=\pi a b=6 \pi$

Required area,

$=\pi \times 2 \times 3-($ Area of quadrilateral $)$

$=6 \pi-\frac{1}{2} 6 \times 4$

$=6 \pi-12$

$=6(\pi-2)$