Sector, segment, arc length, and areas of combinations of figures
Circumference (Perimeter) = 2πr
Area = πr²
where r = radius of circle and π = 22/7 or 3.14 (approximately)
A sector is the region bounded by two radii of a circle and the arc between them.
• Minor sector: the smaller region (angle < 180°)
• Major sector: the larger region (angle > 180°)
• Semicircle: when angle = 180°
Length of arc = (θ/360°) × 2πr
Area of sector = (θ/360°) × πr²
where θ is the central angle in degrees
A sector is like a pizza slice! If the full pizza is 360°, a slice with angle θ is (θ/360)th of the total pizza. So its area is (θ/360) × total area. The curved edge of the slice is the arc length!
Q: A circle has radius 6 cm. Find the area and arc length of a sector with angle 60°.
Arc length = (60/360) × 2π × 6 = (1/6) × 12π = 2π cm ≈ 6.28 cm
Area of sector = (60/360) × π × 6² = (1/6) × 36π = 6π cm² ≈ 18.86 cm²
A segment is the region between a chord and the arc it cuts off.
• Minor segment: smaller region (corresponding to minor arc)
• Major segment: larger region (corresponding to major arc)
Area of Minor Segment = Area of Sector – Area of Triangle
Area of Major Segment = Area of Circle – Area of Minor Segment
Q: Find area of minor segment of a circle of radius 14 cm, if the central angle is 60°.
Area of sector AOB = (60/360) × π × 14² = (1/6) × 22/7 × 196 = (1/6) × 616 = 102.67 cm²
Since ∠AOB = 60° and OA = OB = 14 cm (radii), triangle AOB is equilateral.
Area of equilateral △AOB = (√3/4) × 14² = (√3/4) × 196 = 49√3 ≈ 84.87 cm²
Area of minor segment = 102.67 – 84.87 = ≈ 17.8 cm²
Many problems involve finding areas of regions that are combinations of circles, triangles, rectangles, etc.
Q: A garden is in the shape of a rectangle 40 m × 20 m with a semicircle on one of the shorter sides. Find total area.
Area of rectangle = 40 × 20 = 800 m²
Radius of semicircle = 20/2 = 10 m
Area of semicircle = (1/2) × π × 10² = 50π ≈ 157 m²
Total area = 800 + 157 = 957 m²
Q: A circle of radius 7 cm is inscribed in a square. Find area of the shaded region (four corners).
Side of square = diameter = 2 × 7 = 14 cm
Area of square = 14² = 196 cm²
Area of circle = π × 7² = 22/7 × 49 = 154 cm²
Shaded area (four corners) = 196 – 154 = 42 cm²
Q: Four equal circles each of radius 5 cm are drawn inside a square of side 10 cm. Find shaded area (central region).
Area of square = 100 cm²
Each circle has area = π × 5² = 25π, and 4 circles but they overlap inside the square... (advanced combination)
Area of 4 quarter-circles from corners = 4 × (1/4)πr² = π × 5² = 25π ≈ 78.5 cm²
Shaded region = 100 – 25π ≈ 21.5 cm²
Circumference = 2πr
Area = πr²
Arc = (θ/360) × 2πr
Area = (θ/360) × πr²
Area = Area(sector) – Area(triangle)
Equilateral (60°): (√3/4)a²
Right (90°): (1/2) × base × height