Two medians drawn from acute angles of a right angled triangles intersect at an angle `pi//6`. If the length of the hypotenuse of the triangle is 3 units, then area of the traingle (in sq. units) is

Options

A. `sqrt(3)`
B. 3
C. `sqrt(2)`
D. 9

Explanation

To solve the problem, we need to find the area of a right-angled triangle given that two medians drawn from the acute angles intersect at an angle of π/6 (30 degrees) and the length of the hypotenuse is 3 units.

### Step-by-Step Solution:

1. **Identify the Right-Angled Triangle**:
Let the triangle be denoted as \( \triangle ABC \) where \( C \) is the right angle. Let \( AB \) be the hypotenuse with length \( c = 3 \) units.

2. **Use the Pythagorean Theorem**:
According to the Pythagorean theorem, we have:
\[ a^2 + b^2 = c^2 \]
where \( a \) and \( b \) are the lengths of the other two sides. Substituting the value of \( c \):
\[ a^2 + b^2 = 3^2 = 9 \quad \text{(Equation 1)} \]

3. **Calculate the Centroid**:
The centroid \( G \) of triangle \( ABC \) can be calculated using the formula:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
For our triangle, we can place \( A(0, b) \), \( B(a, 0) \), and \( C(0, 0) \). Thus:
\[ G\left(\frac{0 + a + 0}{3}, \frac{b + 0 + 0}{3}\right) = \left(\frac{a}{3}, \frac{b}{3}\right) \]

4. **Find the Slopes of the Medians**:
- The slope of median \( AG \) from \( A \) to \( G \):
\[ \text{slope of } AG = \frac{\frac{b}{3} - b}{\frac{a}{3} - 0} = \frac{-\frac{2b}{3}}{\frac{a}{3}} = -\frac{2b}{a} \]
- The slope of median \( BG \) from \( B \) to \( G \):
\[ \text{slope of } BG = \frac{\frac{b}{3} - 0}{\frac{a}{3} - a} = \frac{\frac{b}{3}}{-\frac{2a}{3}} = -\frac{b}{2a} \]

5. **Use the Angle Between the Medians**:
The angle \( \theta \) between the two medians is given as \( \frac{\pi}{6} \). The tangent of the angle between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting the slopes:
\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} = \left| \frac{-\frac{2b}{a} + \frac{b}{2a}}{1 + \left(-\frac{2b}{a}\right)\left(-\frac{b}{2a}\right)} \right| \]

6. **Simplifying the Expression**:
After substituting and simplifying, we can find a relationship between \( a \) and \( b \).

7. **Calculate the Area**:
The area \( A \) of triangle \( ABC \) can be calculated using:
\[ A = \frac{1}{2} \times a \times b \]
We can express \( b \) in terms of \( a \) using the relationships derived from the slopes and the angle condition.

8. **Final Calculation**:
After finding \( a \) and \( b \), substitute back into the area formula to find the area of the triangle.

### Area Result:
The area of triangle \( ABC \) is \( \sqrt{3} \) square units.