Geometric Setup

When three circles of equal radius ($r$) are placed such that each touches the other two, their centers form an equilateral triangle. The side length ($s$) of this triangle is equal to the sum of the radii of two touching circles, which is $2r$. The area we need to find is the region in the center bounded by the arcs of the three circles.

Calculations

• Given radius, $r = 3.5 \text{ cm}$.
• Side length of the equilateral triangle formed by the centers: $s = 2r = 2 \times 3.5 = 7 \text{ cm}$.
• Area of the equilateral triangle ($A_{\triangle}$): Using the formula $A_{\triangle} = \frac{\sqrt{3}}{4} s^2$. Given $\sqrt{3} \approx 1.732$. $A_{\triangle} = \frac{1.732}{4} \times (7 \text{ cm})^2 = \frac{1.732}{4} \times 49 \text{ cm}^2 = 0.433 \times 49 \text{ cm}^2 = 21.217 \text{ cm}^2$.
• Area of one sector ($A_{\text{sector}}$): Each angle of the equilateral triangle is $60^\circ$. The portion of each circle inside the triangle is a sector with a central angle of $60^\circ$. Using the formula $A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2$. $A_{\text{sector}} = \frac{60^\circ}{360^\circ} \times \pi (3.5 \text{ cm})^2 = \frac{1}{6} \times \pi \times 12.25 \text{ cm}^2$. Using $\pi \approx \frac{22}{7}$: $A_{\text{sector}} = \frac{1}{6} \times \frac{22}{7} \times 12.25 \text{ cm}^2 = \frac{1}{6} \times 22 \times 1.75 \text{ cm}^2 = \frac{38.5}{6} \text{ cm}^2 \approx 6.4167 \text{ cm}^2$.
• Total area of the three sectors inside the triangle: Total Sector Area $= 3 \times A_{\text{sector}} = 3 \times \frac{38.5}{6} \text{ cm}^2 = \frac{38.5}{2} \text{ cm}^2 = 19.25 \text{ cm}^2$.
• Area of the enclosed portion ($A_{\text{enclosed}}$): This is the area of the equilateral triangle minus the total area of the three sectors. $A_{\text{enclosed}} = A_{\triangle} - \text{Total Sector Area}$ $A_{\text{enclosed}} = 21.217 \text{ cm}^2 - 19.25 \text{ cm}^2 = 1.967 \text{ cm}^2$.

Conclusion

The area of the portion enclosed by the three circles is approximately $1.967 \text{ cm}^2$. This matches Option B.