If the area of the quadrilateral ABCD whose vertices are A(1, 1), B(7, -3), C(12, 2) and D(7, 21) is `Delta`. Find the sum of the digits of `Delta`.

Answer

To find the area of the quadrilateral ABCD with vertices A(1, 1), B(7, -3), C(12, 2), and D(7, 21), we can break it down into two triangles: ABD and BCD. We can use the formula for the area of a triangle given by its vertices.

### Step-by-Step Solution:

1. **Identify the vertices**:
- A(1, 1)
- B(7, -3)
- C(12, 2)
- D(7, 21)

2. **Calculate the area of triangle ABD**:
- The formula for the area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
- For triangle ABD:
- A(1, 1), B(7, -3), D(7, 21)
- Plugging in the coordinates:
\[ \text{Area}_{ABD} = \frac{1}{2} \left| 1(-3 - 21) + 7(21 - 1) + 7(1 + 3) \right| \]
\[ = \frac{1}{2} \left| 1(-24) + 7(20) + 7(4) \right| \]
\[ = \frac{1}{2} \left| -24 + 140 + 28 \right| \]
\[ = \frac{1}{2} \left| 144 \right| = 72 \]

3. **Calculate the area of triangle BCD**:
- For triangle BCD:
- B(7, -3), C(12, 2), D(7, 21)
- Plugging in the coordinates:
\[ \text{Area}_{BCD} = \frac{1}{2} \left| 7(2 - 21) + 12(21 + 3) + 7(-3 - 2) \right| \]
\[ = \frac{1}{2} \left| 7(-19) + 12(24) + 7(-5) \right| \]
\[ = \frac{1}{2} \left| -133 + 288 - 35 \right| \]
\[ = \frac{1}{2} \left| 120 \right| = 60 \]

4. **Total area of quadrilateral ABCD**:
- Now, add the areas of triangles ABD and BCD:
\[ \text{Area}_{ABCD} = \text{Area}_{ABD} + \text{Area}_{BCD} = 72 + 60 = 132 \]

5. **Find the sum of the digits of Delta**:
- The area \( \Delta = 132 \)
- Sum of the digits: \( 1 + 3 + 2 = 6 \)

### Final Answer:
The sum of the digits of the area \( \Delta \) is **6**.