A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length `3,4 and 5` units. Then area of the triangleis equal to:

Options

A. `(9sqrt3(1+sqrt3))/(pi^(2))` unit
B. `(3 sqrt3(sqrt3-1))/(pi^(2))` sq unit
C. `(9 sqrt3(1+sqrt3))/(2pi^(2))`sq unit
D. `(9 sqrt3(sqrt3-1))/(2pi^(2))` sq unit

Answer

To find the area of the triangle inscribed in a circle with arc lengths of 3, 4, and 5 units, we can follow these steps:

### Step 1: Calculate the Circumference of the Circle
The total circumference \( C \) of the circle can be calculated by adding the lengths of the arcs:
\[ C = 3 + 4 + 5 = 12 \text{ units} \]

### Step 2: Find the Radius of the Circle
Using the formula for the circumference of a circle \( C = 2\pi R \), we can solve for the radius \( R \):
\[ 2\pi R = 12 \implies R = \frac{12}{2\pi} = \frac{6}{\pi} \]

### Step 3: Determine the Angles of the Triangle
The angles at the center of the circle corresponding to each arc can be calculated using the formula \( \theta = \frac{s}{R} \), where \( s \) is the arc length.

- For arc of length 3:
\[ \theta_A = \frac{3}{R} = \frac{3}{\frac{6}{\pi}} = \frac{3\pi}{6} = \frac{\pi}{2} \text{ radians} \]

- For arc of length 4:
\[ \theta_B = \frac{4}{R} = \frac{4}{\frac{6}{\pi}} = \frac{4\pi}{6} = \frac{2\pi}{3} \text{ radians} \]

- For arc of length 5:
\[ \theta_C = \frac{5}{R} = \frac{5}{\frac{6}{\pi}} = \frac{5\pi}{6} \text{ radians} \]

### Step 4: Convert Angles to Degrees
To find the sine values, we convert the angles from radians to degrees:
- \( \theta_A = 90^\circ \)
- \( \theta_B = 120^\circ \)
- \( \theta_C = 150^\circ \)

### Step 5: Calculate the Area of the Triangle
The area \( \Delta \) of the triangle can be calculated using the formula:
\[ \Delta = \frac{1}{2} R^2 (\sin A + \sin B + \sin C) \]
Where:
- \( A = 90^\circ \)
- \( B = 120^\circ \)
- \( C = 150^\circ \)

Calculating the sine values:
- \( \sin(90^\circ) = 1 \)
- \( \sin(120^\circ) = \frac{\sqrt{3}}{2} \)
- \( \sin(150^\circ) = \frac{1}{2} \)

Substituting these values into the area formula:
\[ \Delta = \frac{1}{2} \left(\frac{6}{\pi}\right)^2 \left(1 + \frac{\sqrt{3}}{2} + \frac{1}{2}\right) \]
\[ = \frac{1}{2} \left(\frac{36}{\pi^2}\right) \left(1 + \frac{\sqrt{3}}{2} + \frac{1}{2}\right) \]
\[ = \frac{18}{\pi^2} \left(2 + \sqrt{3}\right) \]

### Final Answer
Thus, the area of the triangle is:
\[ \Delta = \frac{18(2 + \sqrt{3})}{\pi^2} \]