Find the area of quadrilateral ABCD, whose vertices are `A(-4,8),B(-3,-4),C(0,-5)and D(5,6).`

Answer

To find the area of quadrilateral ABCD with vertices A(-4, 8), B(-3, -4), C(0, -5), and D(5, 6), we can divide the quadrilateral into two triangles, ABD and BCD, and then calculate the area of each triangle using the formula for the area of a triangle given its vertices.

### Step-by-Step Solution

1. **Identify the vertices:**
- A = (-4, 8)
- B = (-3, -4)
- C = (0, -5)
- D = (5, 6)

2. **Divide the quadrilateral into two triangles:**
- Triangle ABD
- Triangle BCD

3. **Calculate the area of triangle ABD:**
- Use the formula for the area of a triangle given 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 ABD:
- \(x_1 = -4, y_1 = 8\)
- \(x_2 = -3, y_2 = -4\)
- \(x_3 = 5, y_3 = 6\)
- Substitute the values into the formula:
\[ \text{Area}_{ABD} = \frac{1}{2} \left| -4(-4 - 6) + -3(6 - 8) + 5(8 - (-4)) \right| \]
\[ = \frac{1}{2} \left| -4(-10) + -3(-2) + 5(12) \right| \]
\[ = \frac{1}{2} \left| 40 + 6 + 60 \right| \]
\[ = \frac{1}{2} \left| 106 \right| = 53 \text{ square units} \]

4. **Calculate the area of triangle BCD:**
- For triangle BCD:
- \(x_1 = -3, y_1 = -4\)
- \(x_2 = 0, y_2 = -5\)
- \(x_3 = 5, y_3 = 6\)
- Substitute the values into the formula:
\[ \text{Area}_{BCD} = \frac{1}{2} \left| -3(-5 - 6) + 0(6 - (-4)) + 5(-4 - (-5)) \right| \]
\[ = \frac{1}{2} \left| -3(-11) + 0(10) + 5(1) \right| \]
\[ = \frac{1}{2} \left| 33 + 0 + 5 \right| \]
\[ = \frac{1}{2} \left| 38 \right| = 19 \text{ square units} \]

5. **Calculate the total area of quadrilateral ABCD:**
\[ \text{Area}_{ABCD} = \text{Area}_{ABD} + \text{Area}_{BCD} = 53 + 19 = 72 \text{ square units} \]

### Final Answer
The area of quadrilateral ABCD is **72 square units**.