NIMCET 2011 — Mathematics PYQ

1. Identify the four lines:

• Quadrant I ($x\ge 0,y\ge 0$): $x+y=1$

• Quadrant II (x < 0, y \ge 0): $-x+y=1$

• Quadrant III (x < 0, y < 0): $-x-y=1$

• Quadrant IV (x \ge 0, y < 0): $x-y=1$

2. Visualize the shape:

These four lines form a square (often called a diamond shape) with vertices at:

3. Calculate the Area:

There are two easy ways to find the area:

• Method A (Area of 4 Triangles):

  The square is composed of 4 identical right-angled triangles (one in each quadrant). Each triangle has a base of 1 unit and a height of 1 unit.

  $$\text{Area of one triangle}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 1\times 1=\frac{1}{2}$$
  $$\text{Total Area}=4\times \left(\frac{1}{2}\right)=2$$

• Method B (Formula for $|x|+|y|=k$):

  For any equation of the form $|x|+|y|=k$, the enclosed area is always:

  $$\text{Area}=2k^2$$

  Here $k=1$, so:

  $$\text{Area}=2(1{)}^2=2$$

The correct option is (b) 2.