Introduction
This chapter extends your knowledge of circles to finding the areas and perimeters of sectors and segments of circles. You will learn to calculate the area of a sector (the region between two radii and the arc), the area of a segment (the region between a chord and the arc), and solve problems involving combinations of plane figures such as circles, triangles, and rectangles.
Perimeter and Area of a Circle
The circumference (perimeter) of a circle with radius r is 2.pi.r and the area is pi.r^2, where pi is approximately 22/7 or 3.14159. These formulas form the basis for calculating areas and perimeters of sectors and segments.
Key Points
- •Circumference = 2.pi.r = pi.d (where d is diameter)
- •Area = pi.r^2
- •pi = 22/7 or 3.14159 (use 22/7 unless stated otherwise)
- •Diameter = 2 x radius
- •Area in terms of diameter: pi.d^2/4
Worked Example
Find the area and circumference of a circle with radius 7 cm. Area = pi.r^2 = 22/7 x 7 x 7 = 154 sq cm Circumference = 2.pi.r = 2 x 22/7 x 7 = 44 cm
Area of Sector and Length of Arc
A sector is the region enclosed by two radii and the arc between them. The area of a sector with central angle theta (in degrees) and radius r is (theta/360) x pi.r^2. The length of the arc of the sector is (theta/360) x 2.pi.r. A semicircle is a sector with theta = 180 degrees, and a quadrant is a sector with theta = 90 degrees.
Key Points
- •Area of sector = (theta/360) x pi.r^2
- •Length of arc = (theta/360) x 2.pi.r
- •Perimeter of sector = 2r + arc length
- •Semicircle area = pi.r^2/2, quadrant area = pi.r^2/4
- •theta is the central angle of the sector in degrees
Worked Example
Find the area of a sector of angle 60 degrees in a circle of radius 21 cm. Area = (60/360) x (22/7) x 21 x 21 = (1/6) x (22/7) x 441 = (1/6) x 22 x 63 = 231 sq cm
Watch Out
When using pi = 22/7, choose r values that are multiples of 7 to simplify calculations. In exams, the values are usually designed for clean answers.
Area of Segment
A segment of a circle is the region between a chord and its corresponding arc. The area of a minor segment = Area of sector - Area of triangle formed by the two radii and the chord. The area of the major segment = Area of circle - Area of minor segment.
Key Points
- •Area of minor segment = Area of sector - Area of triangle
- •Area of major segment = Area of circle - Area of minor segment
- •For a 60-degree sector: triangle is equilateral (side = radius)
- •For a 90-degree sector: triangle is right isosceles (legs = radius)
- •Area of triangle with angle theta between sides r: (1/2).r^2.sin(theta)
Worked Example
Find the area of the minor segment of a circle of radius 14 cm, cut off by a central angle of 90 degrees. Area of sector = (90/360) x 22/7 x 14 x 14 = 154 sq cm Area of triangle = 1/2 x 14 x 14 = 98 sq cm Area of segment = 154 - 98 = 56 sq cm
Combination of Plane Figures
Many problems involve finding the area of shaded regions formed by combinations of circles, triangles, rectangles, and other shapes. The general approach is to identify the component shapes, calculate their individual areas, and then add or subtract as appropriate. Common combinations include a circle inscribed in a square, a triangle inscribed in a circle, or overlapping circular regions.
Key Points
- •Shaded area = Larger area - Smaller area (for region between two shapes)
- •For overlapping regions, identify which areas to subtract
- •Circle inscribed in square: diameter = side of square
- •Circle circumscribing a square: diameter = diagonal of square
- •Always draw or visualise the figure clearly before computing
Worked Example
A circle is inscribed in a square of side 14 cm. Find the area of the remaining portion (square - circle). Radius of inscribed circle = 14/2 = 7 cm Area of square = 14 x 14 = 196 sq cm Area of circle = 22/7 x 7 x 7 = 154 sq cm Remaining area = 196 - 154 = 42 sq cm
Quick Summary
- ✓Circumference = 2.pi.r, Area = pi.r^2
- ✓Area of sector = (theta/360) x pi.r^2
- ✓Arc length = (theta/360) x 2.pi.r
- ✓Area of segment = Area of sector - Area of triangle
- ✓For combination figures: add/subtract component areas
- ✓Use pi = 22/7 unless stated otherwise
Key Formulas
Circumference = 2.pi.r
Area of circle = pi.r^2
Area of sector = (theta/360) x pi.r^2
Arc length = (theta/360) x 2.pi.r
Perimeter of sector = 2r + arc length
Area of segment = Area of sector - Area of triangle
Area of equilateral triangle = (sqrt(3)/4) x side^2
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