Question

Let S1 be the plane figure consisting of the points (x,y) given by the inequalities |x-1|<=2 and |y+2|<=3. Let S2 be the plane figure given by the inequalities x-y>=-2, y>=1, and x<=3. Let S be the union of S1 and S2. The area of S is

Accepted Answer

The correct answer is (C) 32. S1 is a rectangle: x in [-1,3] (width 4), y in [-5,1] (height 6), area = 24. S2 is bounded by y>=1, x<=3, and y<=x+2, forming a right triangle with vertices (-1,1), (3,1), (3,5): base 4, height 4, area = 8. The two regions only touch along the boundary line y=1 (S1 requires y<=1, S2 requires y>=1), a zero-area overlap. Union area = 24 + 8 - 0 = 32.

Explanation