Three mutually touching circles (radius 3.5 cm) – Area of the central curvilinear region:\nThree equal circles of radius 3.5 cm touch each other pairwise. Find the area enclosed between the three circles (the central curved triangle).

Introduction / Context:
When three congruent circles touch pairwise, their centres form an equilateral triangle of side 2r. The curvilinear central region’s area equals the area of that equilateral triangle minus the areas of three 60° circular sectors.

Given Data / Assumptions:

• Radius r = 3.5 cm.
• Centres form equilateral side = 2r = 7 cm.
• Use π ≈ 22/7 for clean cancellation.

Concept / Approach:
Area required = Area(∆) − 3 * Area(sector, 60°). For an equilateral triangle of side a: area = (√3/4)a^2. Each 60° sector area = (60/360) * π r^2 = (1/6)π r^2.

Step-by-Step Solution:

Verification / Alternative check:
Direct numeric with π = 3.1416 yields a very close value to 1.87 sq cm.

Why Other Options Are Wrong:
1.967, 1.767, 1.567 deviate from the equilateral-minus-sectors computation.

Common Pitfalls:
Using side = r instead of 2r; subtracting three semicircles by mistake; degree-to-sector fraction errors.

Final Answer:
1.867