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 use the formula for the area of a polygon given its vertices. The quadrilateral can be divided into two triangles by drawing a diagonal, and we can calculate the area of each triangle separately.

### 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**:
We can divide quadrilateral ABCD into triangles ABC and ACD.

3. **Calculate the area of triangle ABC**:
The formula for the area of a triangle given 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 ABC:
- \(x_1 = -4, y_1 = 8\)
- \(x_2 = -3, y_2 = -4\)
- \(x_3 = 0, y_3 = -5\)

Plugging in the values:
\[ \text{Area}_{ABC} = \frac{1}{2} \left| -4(-4 - (-5)) + (-3)(-5 - 8) + 0(8 - (-4)) \right| \]
\[ = \frac{1}{2} \left| -4(1) + (-3)(-13) + 0 \right| \]
\[ = \frac{1}{2} \left| -4 + 39 \right| \]
\[ = \frac{1}{2} \left| 35 \right| = \frac{35}{2} \]

4. **Calculate the area of triangle ACD**:
For triangle ACD:
- \(x_1 = -4, y_1 = 8\)
- \(x_2 = 0, y_2 = -5\)
- \(x_3 = 5, y_3 = 6\)

Plugging in the values:
\[ \text{Area}_{ACD} = \frac{1}{2} \left| -4(-5 - 6) + 0(6 - 8) + 5(8 - (-5)) \right| \]
\[ = \frac{1}{2} \left| -4(-11) + 0 + 5(13) \right| \]
\[ = \frac{1}{2} \left| 44 + 65 \right| \]
\[ = \frac{1}{2} \left| 109 \right| = \frac{109}{2} \]

5. **Calculate the total area of quadrilateral ABCD**:
\[ \text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD} \]
\[ = \frac{35}{2} + \frac{109}{2} = \frac{144}{2} = 72 \]

### Final Answer:
The area of quadrilateral ABCD is \(72\) square units.