The area of the triangle with vertices `(a, 0), (a cos theta, b sin theta), (a cos theta, -b sin theta)` is

Options

A. `(3 sqrt3ab)/(4)`
B. `3sqrtab`
C. `(sqrt (3 ab ))/(4)`
D. `3sqrt (3ab)`

Answer

To find the area of the triangle with vertices at the points \((a, 0)\), \((a \cos \theta, b \sin \theta)\), and \((a \cos \theta, -b \sin \theta)\), we can use the formula for the area of a triangle given by three vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

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

### Step 1: Assign the coordinates
Let:
- \( (x_1, y_1) = (a, 0) \)
- \( (x_2, y_2) = (a \cos \theta, b \sin \theta) \)
- \( (x_3, y_3) = (a \cos \theta, -b \sin \theta) \)

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

\[ \text{Area} = \frac{1}{2} \left| a(b \sin \theta - (-b \sin \theta)) + a \cos \theta(-b \sin \theta - 0) + a \cos \theta(0 - b \sin \theta) \right| \]

### Step 3: Simplify the expression
Calculating each term:

1. The first term:
\[ a(b \sin \theta - (-b \sin \theta)) = a(b \sin \theta + b \sin \theta) = a(2b \sin \theta) \]

2. The second term:
\[ a \cos \theta(-b \sin \theta - 0) = -a b \sin \theta \cos \theta \]

3. The third term:
\[ a \cos \theta(0 - b \sin \theta) = -a b \sin \theta \cos \theta \]

Putting it all together:

\[ \text{Area} = \frac{1}{2} \left| a(2b \sin \theta) - ab \sin \theta \cos \theta - ab \sin \theta \cos \theta \right| \]

This simplifies to:

\[ \text{Area} = \frac{1}{2} \left| 2ab \sin \theta - 2ab \sin \theta \cos \theta \right| \]

### Step 4: Factor out common terms
Factoring out \(2ab\):

\[ \text{Area} = \frac{1}{2} \left| 2ab \sin \theta (1 - \cos \theta) \right| \]

### Step 5: Final simplification
This simplifies to:

\[ \text{Area} = |ab \sin \theta (1 - \cos \theta)| \]

Thus, the area of the triangle is:

\[ \text{Area} = |ab \sin \theta (1 - \cos \theta)| \]