Discussion
Let S1 be the plane figure consisting of the points (x, y) given by the inequalities \(|x - 1| \le 2\) and \(|y + 2| \le 3\). Let S2 be the plane figure given by the inequalities \(x - y \ge -2\), \(y \ge 1\), and \(x \le 3\). Let S be the union of S1 and S2. The area of S is
Answer: C
The area of S is 32.
S1 is the rectangle \(-1 \le x \le 3\), \(-5 \le y \le 1\), with area 4 × 6 = 24.
S2 is bounded by \(y \ge 1\), \(x \le 3\) and \(y \le x + 2\): a triangle with corners (-1, 1), (3, 1) and (3, 5), with area \(\tfrac{1}{2} \times 4 \times 4 = 8\).
The two regions meet only along the segment y = 1, which has no area, so the union has area 24 + 8 = 32. 26 and 28 undercount the triangle, and 34 counts something twice.
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