Areas Related to Circles Formulas for CBSE Class 10 With Solved Examples
Areas related to circles allow students to calculate the exact space enclosed by circular shapes and their segments. Using radius, diameter, chord length, and arc measures, these formulas help determine the area of a full circle, sectors, segments, and rings. They also connect to real-world applications in geometry, construction, design, and physics, where precise measurements of circular regions are required. These concepts also support real-world applications in geometry, construction, design, and physics and are commonly practised through structured Class 10 online CBSE maths tuition. In this blog, you will get to know the key formulas for areas related to circles in Class 10 Maths and learn how to apply each one effectively. Areas Related to Circles – Class 10 Maths Formulas Understanding areas related to circles helps calculate spaces, lengths, and segments in circular shapes accurately. These formulas are essential in CBSE Class 10 Maths for solving questions on full circles, sectors, segments, rings, and combinations of circular regions. They are widely applied in construction, design, sports fields, and real-life measurement problems where precision is required. Core Formulas for Areas and Circumference of a Circle Students can use the following key formulas related to circles in Class 10 for quick reference and practice: Circumference of a circle = 2πr Area enclosed by a circle = πr² Length of an arc of a sector with radius r and central angle θ = (θ / 360) × 2πr Area of a sector with radius r and angle θ = (θ / 360) × πr² Area of a segment of a circle = Area of the sector − Area of the triangle Read more: Complete Guide to Class 10 Maths Formulas Chapter by Chapter Practical Applications of Circle Area Formulas Read more: Quadratic Equations Formulas CBSE Class 10 Maths Tips to Master Areas Related to Circle Formulas Read more: Coordinate Geometry Class 10 Formulas with Solved Examples Area Related to Circle Class 10 – Solved Examples Example 1 A circular park has a boundary length of 220 m. The grass inside the park costs ₹5 per m² to maintain. Find the total maintenance cost of the park. Solution: Circumference of the park = 220 m Let the radius be r m 2πr = 220 r = 220 ÷ (2 × 3.14) r = 35 m Area of the park = πr²= 3.14 × 35²= 3846.5 m² Cost per m² = ₹5 Total maintenance cost = 3846.5 × 5= ₹19,232.50 Example 2 Two circular fields have radii 14 m and 21 m. Find the radius of a circle whose circumference is equal to the difference of the circumferences of the two fields. Solution: Circumference of first field= 2 × 3.14 × 14= 87.92 m Circumference of second field= 2 × 3.14 × 21= 131.88 m Difference of circumferences= 131.88 − 87.92= 43.96 m Let the required radius be R 2πR = 43.96 R = 43.96 ÷ (2 × 3.14) R = 7 m







