The vertices of a triangle are A(4, 4), B(3, -2) and C(-3, 16). The area of the triangle is

Options

A. 30 sq units
B. 36 sq units
C. 27 sq units
D. 40 sq units

Answer

To find the area of the triangle with vertices A(4, 4), B(3, -2), and C(-3, 16), we can use the formula for the area of a triangle given its vertices in coordinate geometry:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step-by-Step Solution:

1. **Identify the coordinates**:
- Let \( A(4, 4) \) be \( (x_1, y_1) \)
- Let \( B(3, -2) \) be \( (x_2, y_2) \)
- Let \( C(-3, 16) \) be \( (x_3, y_3) \)

Thus, we have:
- \( x_1 = 4, y_1 = 4 \)
- \( x_2 = 3, y_2 = -2 \)
- \( x_3 = -3, y_3 = 16 \)

2. **Substitute the coordinates into the area formula**:
\[ \text{Area} = \frac{1}{2} \left| 4(-2 - 16) + 3(16 - 4) + (-3)(4 - (-2)) \right| \]

3. **Calculate each term**:
- First term:
\[ 4(-2 - 16) = 4(-18) = -72 \]
- Second term:
\[ 3(16 - 4) = 3(12) = 36 \]
- Third term:
\[ -3(4 - (-2)) = -3(4 + 2) = -3(6) = -18 \]

4. **Combine the terms**:
\[ \text{Area} = \frac{1}{2} \left| -72 + 36 - 18 \right| \]
\[ = \frac{1}{2} \left| -72 + 36 - 18 \right| = \frac{1}{2} \left| -54 \right| = \frac{1}{2} \times 54 \]

5. **Final calculation**:
\[ \text{Area} = 27 \]

Thus, the area of the triangle is \( 27 \) square units.