(3, 2), (-3, 2) and `(0,2sqrt(3))` are, the vertices of ……………...triangle of area ………. .

Options

A. isosceles, 81 sq. units
B. scalene, `9sqrt(3)` sq. units
C. equilateral, `9sqrt(3)` sq. units
D. right angled, 81 sq. units

Answer

To find the area of the triangle with vertices at (3, 2), (-3, 2), and (0, 2√3), we will follow these steps:

### Step 1: Identify the vertices
The vertices of the triangle are given as:
- A(3, 2)
- B(-3, 2)
- C(0, 2√3)

### Step 2: Plot the points on the coordinate plane
We can sketch the coordinate plane and plot the points:
- Point A(3, 2) is located 3 units to the right of the origin and 2 units up.
- Point B(-3, 2) is located 3 units to the left of the origin and 2 units up.
- Point C(0, 2√3) is located at the origin's vertical line and 2√3 units up.

### Step 3: Determine the base and height of the triangle
The base of the triangle can be determined by the distance between points A and B. The height can be determined by the vertical distance from point C to the line formed by points A and B.

- **Base (AB)**: The distance between A and B can be calculated as:
\[ \text{Base} = |x_A - x_B| = |3 - (-3)| = 3 + 3 = 6 \]

- **Height (from C to line AB)**: The height is the vertical distance from point C to the line y = 2 (the y-coordinate of points A and B).
\[ \text{Height} = y_C - y_{AB} = 2\sqrt{3} - 2 \]

### Step 4: Calculate the area of the triangle
The area \(A\) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Substituting the values we found:
\[ A = \frac{1}{2} \times 6 \times (2\sqrt{3} - 2) \]
\[ A = 3 \times (2\sqrt{3} - 2) \]
\[ A = 6\sqrt{3} - 6 \]

### Step 5: Conclusion
The area of the triangle formed by the vertices (3, 2), (-3, 2), and (0, 2√3) is \(6\sqrt{3} - 6\).

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