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: ABC and ACD. We will use the formula for the area of a triangle given its vertices.

### Step 1: Calculate the area of triangle ABC

The formula for the area of a triangle with vertices at (x1, y1), (x2, y2), and (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 ABC:
- A(1, 1) → (x1, y1)
- B(7, -3) → (x2, y2)
- C(12, 2) → (x3, y3)

Substituting the coordinates into the formula:

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

Calculating each term:

\[ = \frac{1}{2} \left| 1 \cdot (-5) + 7 \cdot 1 + 12 \cdot 4 \right| \]
\[ = \frac{1}{2} \left| -5 + 7 + 48 \right| \]
\[ = \frac{1}{2} \left| 50 \right| = \frac{50}{2} = 25 \]

### Step 2: Calculate the area of triangle ACD

For triangle ACD:
- A(1, 1) → (x1, y1)
- C(12, 2) → (x2, y2)
- D(7, 21) → (x3, y3)

Substituting the coordinates into the formula:

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

Calculating each term:

\[ = \frac{1}{2} \left| 1 \cdot (-19) + 12 \cdot 20 + 7 \cdot (-1) \right| \]
\[ = \frac{1}{2} \left| -19 + 240 - 7 \right| \]
\[ = \frac{1}{2} \left| 214 \right| = \frac{214}{2} = 107 \]

### Step 3: 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:

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

### Step 4: Find the sum of the digits of Delta

Now, we need to find the sum of the digits of Delta (the area of the quadrilateral):

\[ \Delta = 132 \]

The sum of the digits is:

\[ 1 + 3 + 2 = 6 \]

### Final Answer

The sum of the digits of Delta is **6**.

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