The area of a pentagon whose vertices are (4,1) (3,6) , (-5,1) , (-3,-3) and (-3,0) , is

Options

A. 30 sq. units
B. 60 sq. units
C. 9 sq. units
D. none of these

Answer

To find the area of the pentagon with vertices at (4,1), (3,6), (-5,1), (-3,-3), and (-3,0), we can use the formula for the area of a polygon given its vertices.

### Step-by-step solution:

1. **List the vertices**:
Let the vertices of the pentagon be:
- A(4, 1) → (x1, y1)
- B(3, 6) → (x2, y2)
- C(-5, 1) → (x3, y3)
- D(-3, -3) → (x4, y4)
- E(-3, 0) → (x5, y5)

2. **Apply the area formula**:
The area \( A \) of a polygon with vertices \((x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)\) is given by the formula:
\[ A = \frac{1}{2} \left| \sum_{i=1}^{n} (x_i y_{i+1} - y_i x_{i+1}) \right| \]
where \( (x_{n+1}, y_{n+1}) \) is the same as \( (x_1, y_1) \).

3. **Set up the coordinates**:
We will write the coordinates in a cyclic manner:
- A(4, 1)
- B(3, 6)
- C(-5, 1)
- D(-3, -3)
- E(-3, 0)
- A(4, 1) (to close the polygon)

4. **Calculate the sums**:
- Calculate \( \sum (x_i y_{i+1}) \):
\[ x_1y_2 + x_2y_3 + x_3y_4 + x_4y_5 + x_5y_1 = 4 \cdot 6 + 3 \cdot 1 + (-5) \cdot (-3) + (-3) \cdot 0 + (-3) \cdot 1 \]
\[ = 24 + 3 + 15 + 0 - 3 = 39 \]

- Calculate \( \sum (y_i x_{i+1}) \):
\[ y_1x_2 + y_2x_3 + y_3x_4 + y_4x_5 + y_5x_1 = 1 \cdot 3 + 6 \cdot (-5) + 1 \cdot (-3) + (-3) \cdot (-3) + 0 \cdot 4 \]
\[ = 3 - 30 - 3 + 9 + 0 = -21 \]

5. **Substitute into the area formula**:
\[ A = \frac{1}{2} \left| 39 - (-21) \right| = \frac{1}{2} \left| 39 + 21 \right| = \frac{1}{2} \left| 60 \right| = 30 \]

6. **Conclusion**:
The area of the pentagon is \( 30 \) square units.