CBSE Class 10 Mathematics Areas Related to Circles Notes Set 11

Circle: Circumference and Area

For a circle of radius r: circumference = 2πr, area = πr², and diameter d = 2r. Unless the question says otherwise, π is taken as 22/7 (when the radius is a multiple of 7) or 3.14.
A circle of larger radius encloses more area for the same wire, and doubling the radius multiplies the area by 4 but the circumference only by 2.

Ring (Path Around a Circle)

If R is the outer radius and r the inner radius, the area of the ring is π(R² − r²) = π(R + r)(R − r). The width of the path is R − r, and the difference of the two circumferences is 2π(R − r).

Sector and Arc

A sector is the region between two radii and the arc they cut. If the angle of the sector is θ degrees:
Length of arc = \( \frac{θ}{360} × 2πr \)
Area of sector = \( \frac{θ}{360} × πr^2 \) = ½ × arc length × r
Perimeter of sector = arc length + 2r
Minor sector has θ < 180°; the remaining part of the circle is the major sector.

Segment

A segment is the region between a chord and its arc.
Area of minor segment = area of sector − area of the triangle formed by the two radii and the chord = \( \frac{θ}{360} πr^2 − ½ r^2 sin θ \).
Area of major segment = πr² − area of minor segment.
For θ = 90°, minor segment = r²(π/4 − ½); for θ = 60°, it is r²(π/6 − √3/4).

Formulas for Common Figure Combinations

1. Largest circle in a square of side a has radius a/2; the area left over is a² − πa²/4.
2. Largest square in a circle of radius r has diagonal 2r and area 2r².
3. Inradius of an equilateral triangle of side a is a/(2√3) and its circumradius is a/√3.
4. A wheel of radius r covers a distance 2πr in one revolution, so the number of revolutions = distance ÷ 2πr.
5. The minute hand of length l sweeps a sector of 6° each minute; the hour hand sweeps 0.5° each minute.
6. Area of a quadrant = πr²/4, area of a semicircle = πr²/2 and perimeter of a semicircle = πr + 2r.

Worked Examples

Example 1. The circumference of a circle exceeds its diameter by 16.5 cm. Find the radius (π = 22/7).
Solution: πd − d = 16.5, so d(15/7) = 16.5 and d = 7.7 cm. Radius = 3.85 cm.

Example 2. Find the area of a sector of angle 60° in a circle of radius 6 cm (π = 22/7).
Solution: Area = (60/360) × (22/7) × 36 = 132/7 ≈ 18.86 cm².

Example 3. Find the area of the minor segment of a circle of radius 14 cm if the chord subtends 90° at the centre (π = 22/7).
Solution: Sector = ¼ × (22/7) × 196 = 154 cm², triangle = ½ × 14 × 14 = 98 cm². Segment = 154 − 98 = 56 cm².

Example 4. A horse is tied to a corner of a 50 m × 50 m field by a rope of 14 m. Find the area it can graze (π = 22/7).
Solution: The grazed region is a quadrant of radius 14 m: ¼ × (22/7) × 196 = 154 m².

Example 5. A wire in the shape of a circle of radius 21 cm is bent into a square. Find the area of the square.
Solution: Wire length = 2 × (22/7) × 21 = 132 cm. Side = 33 cm, area = 1089 cm².

Example 6. A circular park of radius 20 m has a path 5 m wide around it. Find the area of the path (π = 3.14).
Solution: Outer radius = 25 m. Area = 3.14 × (625 − 400) = 706.5 m².