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The given equation of closed figure is $x^2+y^2-|x|\cdot y=1$

For this equation if we remove the mod on x, the area won't change.

So the equation is $x^2+y^2-x\cdot y=1$ $\longrightarrow 1$

First we need to eliminate $x\cdot y$ for easing the problems. This can be done by rotating the axis.(Rotating the axis will not change the area).

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For rotating axis:-

If X and Y are coordinates after rotation and if $\theta$ is the angle of rotation, then-

$x=Xcos\theta -Ysin\theta$ $\longrightarrow 2$

$y=Xsin\theta +Ycos\theta$ $\longrightarrow 3$

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Substituting 2&3 in 1 , we get,

$X^2+Y^2-(X^2 sin\theta cos\theta +XYcos^2\theta-XYsin^2\theta-Y^2 sin\theta cos\theta)=1$

For eliminating XY , $\theta$ must be $\frac{\pi}{4}$ ,

$\implies \dfrac{X^2}{(\sqrt{2})^2}+\dfrac{Y^2}{(\sqrt{\frac{2}{3}})^2}=1$

This is in standard form of ellipse equation $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

Area= $\pi ab$

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Here $a=\sqrt{2}$ and $b=\sqrt{\dfrac{2}{3}}$

$Area=\pi \cdot \sqrt{2} \cdot \sqrt{\frac{2}{3}}$

$\boxed{Area=\dfrac{2}{\sqrt{3}} \Pi}$