A circle of radius 'a' is divided into 6 equal sectors. An equilateral triangle is drawn on the chord of each sector to lie outside the circle. Area of the resulting figure is :

Options

A. `3a^2(pi + sqrt3)`
B. `3sqrt3 a^2`
C. `3(a^2sqrt3 + pi)`
D. `(3sqrt3 pia^2)/2

Answer

To find the area of the resulting figure formed by a circle of radius 'a' divided into 6 equal sectors, with an equilateral triangle drawn on the chord of each sector, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circle and Sectors**:
- A circle of radius 'a' is divided into 6 equal sectors. Each sector corresponds to a central angle of \(60^\circ\) (since \(360^\circ / 6 = 60^\circ\)).

2. **Identifying the Chords**:
- The chord of each sector is the line segment connecting the two points on the circumference of the circle that define the sector.

3. **Drawing the Equilateral Triangle**:
- An equilateral triangle is drawn on each chord, lying outside the circle. Since the triangle is equilateral, all its sides are equal.

4. **Finding the Length of the Chord**:
- The length of the chord \(c\) for a sector with a central angle of \(60^\circ\) in a circle of radius \(a\) can be calculated using the formula:
\[ c = 2a \sin\left(\frac{60^\circ}{2}\right) = 2a \sin(30^\circ) = 2a \cdot \frac{1}{2} = a \]

5. **Area of One Equilateral Triangle**:
- The area \(A_t\) of an equilateral triangle with side length \(s\) is given by:
\[ A_t = \frac{\sqrt{3}}{4} s^2 \]
- Here, the side length \(s\) of each equilateral triangle is equal to the length of the chord, which we found to be \(a\). Therefore:
\[ A_t = \frac{\sqrt{3}}{4} a^2 \]

6. **Total Area of All Triangles**:
- Since there are 6 sectors, and thus 6 equilateral triangles, the total area \(A\) of the triangles is:
\[ A = 6 \cdot A_t = 6 \cdot \frac{\sqrt{3}}{4} a^2 = \frac{3\sqrt{3}}{2} a^2 \]

7. **Final Area of the Resulting Figure**:
- The resulting figure consists of the area of the circle and the area of the 6 equilateral triangles. However, since the triangles are drawn outside the circle, we only need the area of the triangles:
\[ \text{Total Area} = 6 \cdot A_t = 6 \cdot \frac{\sqrt{3}}{4} a^2 = \frac{3\sqrt{3}}{2} a^2 \]

### Final Answer:
The area of the resulting figure is:
\[ \frac{3\sqrt{3}}{2} a^2 \]