FInd the area of the figure given on the coordinate plane (XOY) as a system of inequalities:
x(x + y - sqrt2) < 0
x^2 + y^2 < 2

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Where's the figure? Your question states "figure given" but you forgot to upload the picture of the figure.

x^2 + y^2 < 2

The above figure consists of all points within the circle x^2 + y^2 = 2 i.e. circle whose centre is the Origin and whose radius is sqrt 2

Regarding second condition: x(x + y - sqrt2) < 0

This is possible when

1. x is negative but x + y - sqrt2 is positive.

Note: x + y - sqrt2 = 0 is a straight line as follows:

Suppose the circle cuts positive side of X axis at A and positive side of Y axis at B
then the line is AB

So x + y - sqrt2 is positive only for points lying above AB

None of the points in the circle meet the above TWO conditions

2. x is positive but x + y - sqrt2 is negative.

All points within the circle, on right of Y-axis, but below the straight line AB
meet the above 2 conditions.

The figure consists of:

Triangle OAB + a quarter of the circle

AREA = 1/2(OA times OB) + 1/4 pi r^2

OA = OB = sqrt 2

r = sqrt 2

AREA = 1/2 2 + 1/4 pi 2

= 1 + 1/2 pi

= Approx 1 + 1/2 3.14

= 2.57