In `triangleABC` (not shown), `angleABC=60^(@) and ACbotBC`. If `AB=x`, then what is the area of `triangleABC`, in terms of x?

Options

A. `(x^(2)sqrt(3))/(8)`
B. `(x^(2)sqrt(3))/(4)`
C. `(x^(2)sqrt(3))/(2)`
D. `x^(2)sqrt(3)`

Answer

To find the area of triangle ABC in terms of \( x \), we can follow these steps:

### Step 1: Identify the triangle properties
Given:
- \( \angle ABC = 60^\circ \)
- \( AC \perp BC \) (which means \( AC \) is the height and \( BC \) is the base of the triangle)
- \( AB = x \)

### Step 2: Determine the angles of the triangle
Since the sum of angles in a triangle is \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
Given \( \angle B = 60^\circ \) and \( \angle C = 90^\circ \) (because \( AC \perp BC \)):
\[ \angle A + 60^\circ + 90^\circ = 180^\circ \]
\[ \angle A = 30^\circ \]

### Step 3: Use trigonometric ratios to find \( AC \) and \( BC \)
Using the sine and cosine functions:
1. For \( AC \):
\[ \sin(60^\circ) = \frac{AC}{AB} \implies \frac{\sqrt{3}}{2} = \frac{AC}{x} \]
Rearranging gives:
\[ AC = x \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}x}{2} \]

2. For \( BC \):
\[ \cos(60^\circ) = \frac{BC}{AB} \implies \frac{1}{2} = \frac{BC}{x} \]
Rearranging gives:
\[ BC = x \cdot \frac{1}{2} = \frac{x}{2} \]

### Step 4: Calculate the area of triangle ABC
The area \( A \) of triangle ABC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, \( BC \) is the base and \( AC \) is the height:
\[ \text{Area} = \frac{1}{2} \times BC \times AC = \frac{1}{2} \times \frac{x}{2} \times \frac{\sqrt{3}x}{2} \]
Simplifying this:
\[ \text{Area} = \frac{1}{2} \times \frac{x}{2} \times \frac{\sqrt{3}x}{2} = \frac{\sqrt{3}x^2}{8} \]

### Final Answer
Thus, the area of triangle ABC in terms of \( x \) is:
\[ \text{Area} = \frac{\sqrt{3}x^2}{8} \]

---