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 \( A(4,1) \), \( B(3,6) \), \( C(-5,1) \), \( D(-3,-3) \), and \( E(-3,0) \), we can use the formula for the area of a polygon given its vertices:

\[ \text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n} (x_i y_{i+1} - y_i x_{i+1}) \right| \]

where \( (x_{n+1}, y_{n+1}) = (x_1, y_1) \) to close the polygon.

### Step 1: List the coordinates
The coordinates of the vertices are:
- \( A(4,1) \) → \( (x_1, y_1) = (4, 1) \)
- \( B(3,6) \) → \( (x_2, y_2) = (3, 6) \)
- \( C(-5,1) \) → \( (x_3, y_3) = (-5, 1) \)
- \( D(-3,-3) \) → \( (x_4, y_4) = (-3, -3) \)
- \( E(-3,0) \) → \( (x_5, y_5) = (-3, 0) \)

### Step 2: Set up the formula
We will calculate:
\[ \text{Area} = \frac{1}{2} \left| (x_1y_2 + x_2y_3 + x_3y_4 + x_4y_5 + x_5y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_5 + y_5x_1) \right| \]

### Step 3: Substitute the coordinates into the formula
Substituting the coordinates into the formula, we have:
\[ \text{Area} = \frac{1}{2} \left| (4 \cdot 6 + 3 \cdot 1 + (-5) \cdot (-3) + (-3) \cdot 0 + (-3) \cdot 1) - (1 \cdot 3 + 6 \cdot (-5) + 1 \cdot (-3) + (-3) \cdot (-3) + 0 \cdot 4) \right| \]

### Step 4: Calculate each term
Calculating the first part:
- \( 4 \cdot 6 = 24 \)
- \( 3 \cdot 1 = 3 \)
- \( -5 \cdot -3 = 15 \)
- \( -3 \cdot 0 = 0 \)
- \( -3 \cdot 1 = -3 \)

So, the sum is:
\[ 24 + 3 + 15 + 0 - 3 = 39 \]

Now, calculating the second part:
- \( 1 \cdot 3 = 3 \)
- \( 6 \cdot -5 = -30 \)
- \( 1 \cdot -3 = -3 \)
- \( -3 \cdot -3 = 9 \)
- \( 0 \cdot 4 = 0 \)

So, the sum is:
\[ 3 - 30 - 3 + 9 + 0 = -21 \]

### Step 5: Combine the results
Now substituting back into the area formula:
\[ \text{Area} = \frac{1}{2} \left| 39 - (-21) \right| = \frac{1}{2} \left| 39 + 21 \right| = \frac{1}{2} \left| 60 \right| = \frac{60}{2} = 30 \]

### Final Answer
The area of the pentagon is \( 30 \) square units.

---