Find the area of a quadrilateral ABCD having vertices at A(l, 2), B(1, 0), C(4,0) and D(4,4).

Answer

To find the area of the quadrilateral ABCD with vertices at A(1, 2), B(1, 0), C(4, 0), and D(4, 4), we can divide the quadrilateral into two triangles: triangle ACD and triangle ABC. We will calculate the area of each triangle using the formula for the area of a triangle given by its vertices.

### Step-by-Step Solution:

1. **Identify the vertices**:
- A(1, 2)
- B(1, 0)
- C(4, 0)
- D(4, 4)

2. **Calculate the area of triangle ACD**:
- Use the formula for the area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
- For triangle ACD:
- A(1, 2) → (x1, y1)
- C(4, 0) → (x2, y2)
- D(4, 4) → (x3, y3)
- Substitute the coordinates into the formula:
\[ \text{Area}_{ACD} = \frac{1}{2} \left| 1(0 - 4) + 4(4 - 2) + 4(2 - 0) \right| \]
- Calculate:
\[ = \frac{1}{2} \left| 1(-4) + 4(2) + 4(2) \right| \]
\[ = \frac{1}{2} \left| -4 + 8 + 8 \right| \]
\[ = \frac{1}{2} \left| 12 \right| = 6 \]

3. **Calculate the area of triangle ABC**:
- For triangle ABC:
- A(1, 2) → (x1, y1)
- B(1, 0) → (x2, y2)
- C(4, 0) → (x3, y3)
- Substitute the coordinates into the formula:
\[ \text{Area}_{ABC} = \frac{1}{2} \left| 1(0 - 0) + 1(0 - 2) + 4(2 - 0) \right| \]
- Calculate:
\[ = \frac{1}{2} \left| 0 + 1(-2) + 4(2) \right| \]
\[ = \frac{1}{2} \left| -2 + 8 \right| = \frac{1}{2} \left| 6 \right| = 3 \]

4. **Total area of quadrilateral ABCD**:
- Add the areas of triangles ACD and ABC:
\[ \text{Area}_{ABCD} = \text{Area}_{ACD} + \text{Area}_{ABC} = 6 + 3 = 9 \]

### Final Answer:
The area of quadrilateral ABCD is **9 square units**.