ABC is an equilateral triangle of side 2 cm. With A, B, C as centre and radius 1 cm three arcs are drawn. The area of the region within the triangle bounded by the three arcs is

Options

A. (a) `(3sqrt(3) - (pi)/(2)) cm^(2)`
B. (b) `(sqrt(3) - (3pi)/(2)) cm^(2)`
C. (c) `(sqrt(3) - (pi)/(2))cm^(2)`
D. (d) `((pi)/(2) - sqrt(3)) cm^(2)`

Answer

To solve the problem, we need to find the area of the region within the equilateral triangle ABC that is bounded by the three arcs drawn with centers A, B, and C and a radius of 1 cm.

### Step-by-Step Solution:

1. **Calculate the Area of the Equilateral Triangle:**
- The formula for the area \( A \) of an equilateral triangle with side length \( s \) is given by:
\[ A = \frac{\sqrt{3}}{4} s^2 \]
- Here, the side length \( s = 2 \) cm. Therefore, the area of triangle ABC is:
\[ A = \frac{\sqrt{3}}{4} \times (2)^2 = \frac{\sqrt{3}}{4} \times 4 = \sqrt{3} \text{ cm}^2 \]

2. **Calculate the Area of One Sector:**
- Each arc corresponds to a sector of a circle with a radius of 1 cm and a central angle of 60 degrees (since the triangle is equilateral).
- The area \( A_s \) of a sector is given by:
\[ A_s = \frac{\theta}{360} \times \pi r^2 \]
- For our case, \( \theta = 60 \) degrees and \( r = 1 \) cm:
\[ A_s = \frac{60}{360} \times \pi \times (1)^2 = \frac{1}{6} \pi \text{ cm}^2 \]

3. **Calculate the Total Area of the Three Sectors:**
- Since there are three sectors (one at each vertex of the triangle), the total area of the sectors is:
\[ A_{total\_sectors} = 3 \times A_s = 3 \times \frac{1}{6} \pi = \frac{1}{2} \pi \text{ cm}^2 \]

4. **Calculate the Required Area:**
- The required area \( A_{required} \) is the area of the triangle minus the total area of the sectors:
\[ A_{required} = A_{triangle} - A_{total\_sectors} = \sqrt{3} - \frac{1}{2} \pi \text{ cm}^2 \]

### Final Answer:
The area of the region within the triangle bounded by the three arcs is:
\[ \sqrt{3} - \frac{1}{2} \pi \text{ cm}^2 \]