Area Of Circles Introduction Chapter 12 Class 10 Maths NCERT

Areas Related to Circles: Sectors, Segments, and Combined Figures

This chapter in CBSE Class 10 Mathematics applies the concepts of circles — circumference, area, arcs, sectors, and segments — to solve practical problems involving the area of regions bounded by circular and linear boundaries. Students learn to calculate the area and perimeter of sector-shaped and segment-shaped regions, and to find areas of combined figures involving circles, triangles, squares, and other polygons. These skills have direct applications in engineering, architecture, design, and everyday problem-solving.

The circumference of a circle with radius r is given by C = 2πr, and its area is A = πr². The value of π (pi) is approximately 3.14159, or 22/7 for most calculations. An arc is a portion of the circumference of a circle. The length of an arc subtending an angle θ (in degrees) at the centre is: Arc length = (θ/360) × 2πr. This formula makes sense because a full circle (θ = 360°) gives the entire circumference. A sector is the region enclosed by two radii and an arc — it looks like a "slice of pie." The area of a sector with central angle θ is: Area of sector = (θ/360) × πr². The perimeter of a sector includes the two radii plus the arc: Perimeter = 2r + arc length = 2r + (θ/360) × 2πr. A segment is the region between a chord and the corresponding arc. The area of a minor segment is found by subtracting the area of the triangle formed by the two radii and the chord from the area of the sector: Area of segment = Area of sector − Area of triangle. The triangle can be calculated using the formula Area = ½r²sin θ (when θ is in degrees), or by constructing an altitude from the centre to the chord and using the Pythagorean theorem.

Board examination problems frequently involve combined figures. Common problem types include: (1) Finding the area of a ring (annulus) between two concentric circles: Area = π(R² − r²), where R and r are the outer and inner radii. (2) Calculating the area of the path around a circular garden or track. (3) Finding the area swept by the minute hand or hour hand of a clock in a given time — the tip of the hand traces an arc, and the region swept is a sector. For example, the minute hand of a clock (length 14 cm) sweeps a sector of 90° in 15 minutes, with area = (90/360) × π × 14² = 154 cm². (4) Finding the area of a shaded region in a diagram that combines circles with squares, equilateral triangles, or other regular polygons. For example, a circle inscribed in a square — the circle's diameter equals the side of the square. (5) Finding the area of segments when a chord subtends a specific angle at the centre. (6) Problems involving a sector cut from a circular sheet of paper and rolled into a cone — the arc length of the sector becomes the circumference of the base of the cone, and the radius of the sector becomes the slant height of the cone. Students should always draw a clear diagram, identify the geometric shapes involved, apply the appropriate formula, and express answers in terms of π unless a numerical approximation is specifically requested.

• Circumference = 2πr, Area = πr²; arc length = (θ/360) × 2πr for central angle θ degrees.
• Area of sector = (θ/360) × πr²; perimeter of sector = 2r + arc length.
• Area of segment = area of sector − area of triangle formed by radii and chord (use ½r²sin θ for the triangle).
• Annulus (ring) area = π(R² − r²); clock hand problems use sector formulas with angles from time.
• For combined figures: draw diagram, identify shapes, apply formulas separately, then add or subtract areas as needed.