What is the area of the triangle whose vertices are` (3, 0), (0, 4) and (3, 4)` ?

Options

A. 6 sq. unit
B. 7.5 sq. unit
C. 9 sq. unit
D. 12 sq. unit

Explanation

To find the area of the triangle with vertices at \( A(3, 0) \), \( B(0, 4) \), and \( C(3, 4) \), we can use the formula for the area of a triangle given by the coordinates of its vertices:

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

### Step 1: Identify the coordinates
Let:
- \( A(3, 0) \) → \( (x_1, y_1) \)
- \( B(0, 4) \) → \( (x_2, y_2) \)
- \( C(3, 4) \) → \( (x_3, y_3) \)

Thus, we have:
- \( x_1 = 3, y_1 = 0 \)
- \( x_2 = 0, y_2 = 4 \)
- \( x_3 = 3, y_3 = 4 \)

### Step 2: Substitute the coordinates into the formula
Substituting the values into the area formula:

\[ \text{Area} = \frac{1}{2} \left| 3(4 - 4) + 0(4 - 0) + 3(0 - 4) \right| \]

### Step 3: Simplify the expression
Calculating each term:
- The first term: \( 3(4 - 4) = 3 \times 0 = 0 \)
- The second term: \( 0(4 - 0) = 0 \)
- The third term: \( 3(0 - 4) = 3 \times -4 = -12 \)

So, we have:

\[ \text{Area} = \frac{1}{2} \left| 0 + 0 - 12 \right| = \frac{1}{2} \left| -12 \right| = \frac{1}{2} \times 12 = 6 \]

### Step 4: Conclusion
Thus, the area of the triangle is:

\[ \text{Area} = 6 \text{ square units} \]

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