The area of quadrilateral with vertices (2, 4), (0, 4), (0, - 4), (2, - 4) is equal to (sq. units)
(a)8
(b)12
(c)16
(d)32

Options

A. 8
B. 12
C. 16
D. 32

Answer

To find the area of the quadrilateral with vertices at (2, 4), (0, 4), (0, -4), and (2, -4), we can use the formula for the area of a polygon based on its vertices. The formula is:

\[ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| \]

### Step 1: Assign the vertices
Let:
- \( (x_1, y_1) = (2, 4) \)
- \( (x_2, y_2) = (0, 4) \)
- \( (x_3, y_3) = (0, -4) \)
- \( (x_4, y_4) = (2, -4) \)

### Step 2: Substitute the coordinates into the formula
Now we substitute the coordinates into the area formula:

\[ \text{Area} = \frac{1}{2} \left| 2 \cdot 4 + 0 \cdot (-4) + 0 \cdot (-4) + 2 \cdot 4 - (4 \cdot 0 + 4 \cdot 0 + (-4) \cdot 2 + (-4) \cdot 2) \right| \]

### Step 3: Calculate the terms
Calculating the terms inside the absolute value:

- Left to right:
- \( 2 \cdot 4 = 8 \)
- \( 0 \cdot (-4) = 0 \)
- \( 0 \cdot (-4) = 0 \)
- \( 2 \cdot 4 = 8 \)

So, the sum of the left to right products is:
\[ 8 + 0 + 0 + 8 = 16 \]

- Right to left:
- \( 4 \cdot 0 = 0 \)
- \( 4 \cdot 0 = 0 \)
- \( -4 \cdot 2 = -8 \)
- \( -4 \cdot 2 = -8 \)

So, the sum of the right to left products is:
\[ 0 + 0 - 8 - 8 = -16 \]

### Step 4: Calculate the area
Now substituting these sums back into the area formula:

\[ \text{Area} = \frac{1}{2} \left| 16 - (-16) \right| = \frac{1}{2} \left| 16 + 16 \right| = \frac{1}{2} \left| 32 \right| = \frac{32}{2} = 16 \]

Thus, the area of the quadrilateral is \( 16 \) square units.

### Conclusion
The area of the quadrilateral is \( \boxed{16} \).