Answer
By area addition postulate
Area(RUS) = Area(R) + Area(S)
Area(RUS), Area(R) and Area(S) are all positive numbers so let assume Area(S) as p
Area(RUS) = Area(R) + p
So by definition of inequality Area(RUS) > Area(R)
Work Step by Step
Given region R U S
Let R and S be enclosed regions that do not overlap
As we know area addition postulate -- eq1
Area(RUS) = Area(R) + Area(S)
Area(RUS), Area(R) and Area(S) are all positive numbers so let assume Area(S) as p
So put the value of Area(S) in eq1
Area(RUS) = Area(R) + p
So by definition of inequality Area(RUS) > Area(R)
