Question

The area enclosed within the curve \( |x| + |y| = 2 \) is

• A. 16 sq.unit
• B. 24 sq.unit
• C. 32 sq.unit
• D. 8 sq.unit

Hint:

The equation \( |x| + |y| = a \) always represents a square centered at the origin, with vertices at \( (a, 0), (-a, 0), (0, a), \) and \( (0, -a) \). The length of its diagonals is \( 2a \). The area of such a square is easily calculated as \( \frac{1}{2} \times (2a) \times (2a) = 2a^2 \). In this case, \( a=2 \), so the area is \( 2 \times 2^2 = 8 \).

Answer 1

The Correct Option is D

The curve \( |x| + |y| = 2 \) describes a square rotated by 45 degrees. We can see this by considering the four quadrants: \[\begin{array}{rl} \bullet & \text{Quadrant I (\( x \ge 0, y \ge 0 \)): \( x + y = 2 \)} \\ \bullet & \text{Quadrant II (\( x < 0, y \ge 0 \)): \( -x + y = 2 \)} \\ \bullet & \text{Quadrant III (\( x < 0, y < 0 \)): \( -x - y = 2 \implies x + y = -2 \)} \\ \bullet & \text{Quadrant IV (\( x \ge 0, y < 0 \)): \( x - y = 2 \)} \\ \end{array}\] These four lines form the sides of a geometric figure. Let's find the intercepts with the axes: \[\begin{array}{rl} \bullet & \text{x-intercepts (set y=0): \( |x| = 2 \implies x = \pm 2 \). The vertices are (2, 0) and (-2, 0).} \\ \bullet & \text{y-intercepts (set x=0): \( |y| = 2 \implies y = \pm 2 \). The vertices are (0, 2) and (0, -2).} \\ \end{array}\] The figure is a square with vertices at (2, 0), (0, 2), (-2, 0), and (0, -2).
The area of a square can be calculated in several ways. One way is to treat it as a rhombus. The diagonals of this square lie along the x and y axes.
Length of diagonal 1 (along x-axis) = \( 2 - (-2) = 4 \).
Length of diagonal 2 (along y-axis) = \( 2 - (-2) = 4 \).
Area of a rhombus = \( \frac{1}{2} \times d_1 \times d_2 \).
Area = \( \frac{1}{2} \times 4 \times 4 = 8 \) sq. units.