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 divide the quadrilateral into two triangles and calculate their areas separately.

### Step 1: Divide the quadrilateral into two triangles
We can divide quadrilateral ABCD into triangles ABC and ACD.

### Step 2: Calculate the area of triangle ABC
The vertices of triangle ABC are A(1, 1), B(7, -3), and C(12, 2). We can use the formula for the area of a triangle given by vertices (x1, y1), (x2, y2), (x3, y3):

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates of A, B, and C:

\[ \text{Area}_{ABC} = \frac{1}{2} \left| 1(-3 - 2) + 7(2 - 1) + 12(1 + 3) \right| \]

Calculating each term:

- \(1(-3 - 2) = 1 \times -5 = -5\)
- \(7(2 - 1) = 7 \times 1 = 7\)
- \(12(1 + 3) = 12 \times 4 = 48\)

Now, summing these values:

\[ \text{Area}_{ABC} = \frac{1}{2} \left| -5 + 7 + 48 \right| = \frac{1}{2} \left| 50 \right| = \frac{50}{2} = 25 \]

### Step 3: Calculate the area of triangle ACD
The vertices of triangle ACD are A(1, 1), C(12, 2), and D(7, 21). Using the same area formula:

\[ \text{Area}_{ACD} = \frac{1}{2} \left| 1(2 - 21) + 12(21 - 1) + 7(1 - 2) \right| \]

Calculating each term:

- \(1(2 - 21) = 1 \times -19 = -19\)
- \(12(21 - 1) = 12 \times 20 = 240\)
- \(7(1 - 2) = 7 \times -1 = -7\)

Now, summing these values:

\[ \text{Area}_{ACD} = \frac{1}{2} \left| -19 + 240 - 7 \right| = \frac{1}{2} \left| 214 \right| = \frac{214}{2} = 107 \]

### Step 4: Calculate the total area of quadrilateral ABCD
Now, we can find the total area of quadrilateral ABCD by adding the areas of triangles ABC and ACD:

\[ \Delta = \text{Area}_{ABC} + \text{Area}_{ACD} = 25 + 107 = 132 \]

### Step 5: Find the sum of the digits of Delta
Now, we need to find the sum of the digits of 132:

- The digits are 1, 3, and 2.
- Summing these digits: \(1 + 3 + 2 = 6\)

Thus, the final answer is:

\[ \text{Sum of the digits of } \Delta = 6 \]