Find the area of quadrilateral whose vertices are `(-4,5) , (0,7), (5,-5) and (-4-2)

Options

A. ` 54 (1)/(2) ` sq. units
B. ` 56 (1)/(2)` sq. units
C. ` 58 (1)/(2)` sq. units
D. ` 60 (1)/(2) ` sq. units

Answer

To find the area of the quadrilateral with vertices at \((-4, 5)\), \((0, 7)\), \((5, -5)\), and \((-4, -2)\), we can divide the quadrilateral into two triangles and then calculate the area of each triangle separately.

### Step-by-Step Solution:

1. **Identify the vertices of the quadrilateral:**
- Let \( A = (-4, 5) \)
- Let \( B = (0, 7) \)
- Let \( C = (5, -5) \)
- Let \( D = (-4, -2) \)

2. **Divide the quadrilateral into two triangles:**
- We can divide the quadrilateral into triangles \( ABC \) and \( ACD \).

3. **Calculate the area of triangle \( ABC \):**
- The formula for the area of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) is:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
- Substituting the coordinates of points \( A \), \( B \), and \( C \):
\[ \text{Area}_{ABC} = \frac{1}{2} \left| (-4)(7 - (-5)) + (0)(-5 - 5) + (5)(5 - 7) \right| \]
- Simplifying:
\[ = \frac{1}{2} \left| (-4)(12) + 0 + 5(-2) \right| \]
\[ = \frac{1}{2} \left| -48 - 10 \right| \]
\[ = \frac{1}{2} \left| -58 \right| = \frac{58}{2} = 29 \]

4. **Calculate the area of triangle \( ACD \):**
- Using the same area formula:
\[ \text{Area}_{ACD} = \frac{1}{2} \left| (-4)(-2 - (-5)) + (5)(-5 - 5) + (-4)(5 - (-2)) \right| \]
- Simplifying:
\[ = \frac{1}{2} \left| (-4)(3) + 5(-10) + (-4)(7) \right| \]
\[ = \frac{1}{2} \left| -12 - 50 - 28 \right| \]
\[ = \frac{1}{2} \left| -90 \right| = \frac{90}{2} = 45 \]

5. **Calculate the total area of the quadrilateral:**
- The total area is the sum of the areas of triangles \( ABC \) and \( ACD \):
\[ \text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD} = 29 + 45 = 74 \]

### Final Answer:
The area of the quadrilateral is \( 74 \) square units.