Area (in sq.

Question

Area (in sq. units) of the region outside $\dfrac{|x|}2+\dfrac{|y|}3=1$ and inside the ellipse $\dfrac{x^2}4+\dfrac{y^2}9=1$ is:

Accepted Answer

Area of Ellipse = $$\pi $$ab = 6$$\pi $$

$$ \therefore $$ Required area = Area of ellipse
-- 4 (Area of triangle OAB)

= 6$$\pi $$ - $$4\left( {{1 \over 2} \times 2 \times 3} \right)$$

= 6$$\pi $$ - 12

= $$6\left( {\pi  - 2} \right)$$ sq.units

Solutions · 1