Areas Related to Circles

A circle is the set of all points in a plane that are at a fixed distance (the radius, r) from a fixed point called the centre. Two basic measurements describe a circle: how far it is around, and how much region it encloses.

The circumference is the distance once around the circle: C = 2πr = πd, where d = 2r is the diameter. The region enclosed has area A = πr². Here π (pi) is the constant ratio of a circle's circumference to its diameter; in this chapter we use π = 22/7 or π = 3.14 as convenient.

Worked example: A circular garden has radius 7 m. Taking π = 22/7, circumference = 2 × 22/7 × 7 = 44 m and area = 22/7 × 7² = 22/7 × 49 = 154 m².

If only the circumference is known, first find r. For example, if C = 44 cm then 2πr = 44, so r = (44 × 7)/(2 × 22) = 7 cm, and hence A = 22/7 × 49 = 154 cm².

A useful idea for combination problems: if a wire of a given length is bent into a circle, its length equals the circumference; the same wire re-bent into another shape keeps the same total length (perimeter).

A part of the circumference lying between two points on the circle is an arc. When the two radii to the ends of an arc are drawn, the enclosed region is a sector, and the angle between those radii at the centre is the central angle θ.

The complete circle corresponds to a central angle of 360° and its length (circumference) is 2πr. An arc of central angle θ is simply the fraction θ/360° of the whole circle. Therefore the length of the arc is l = (θ/360°) × 2πr.

Worked example: Find the length of an arc of a circle of radius 21 cm that subtends 60° at the centre (π = 22/7). l = 60/360 × 2 × 22/7 × 21 = 1/6 × 132 = 22 cm.

The perimeter of a sector is the arc length together with the two bounding radii: P = l + 2r. For the sector above, P = 22 + 2 × 21 = 64 cm.

The idea of a fraction of the whole also applies to clocks: as the minute hand turns, its tip traces an arc, and in 60 minutes it turns through the full 360°, i.e. 6° every minute.

A sector is the region enclosed by two radii and the arc between them. Just as with arc length, the area of a sector is the fraction θ/360° of the area of the whole circle: A = (θ/360°) × πr².

Because the arc length is l = (θ/360°) × 2πr, the sector area can also be written as A = ½ × l × r. This form is handy when the arc length is already known.

A quadrant is a sector of angle 90° (a quarter of the circle) and a semicircle is a sector of angle 180° (half the circle).

Worked example: A sector has radius 14 cm and central angle 90° (π = 22/7). A = 90/360 × 22/7 × 14² = 1/4 × 22/7 × 196 = 1/4 × 616 = 154 cm².

The smaller of the two sectors formed (with θ < 180°) is the minor sector; the remaining part of the circle is the major sector, whose central angle is 360° − θ. The two areas together add up to the area of the whole circle, πr².

A chord divides a circle into two parts called segments. The region lying between a chord and its corresponding arc is a segment; the smaller part is the minor segment and the larger part is the major segment.

To find a segment's area, take the sector cut off by the two radii and remove the triangle formed by those two radii and the chord: Area of segment = Area of sector − Area of triangle.

The triangle has two sides equal to r with the central angle θ included between them, so its area is ½ r² sin θ. Hence the area of the minor segment = (θ/360°) πr² − ½ r² sin θ.

Worked example: A chord of a circle of radius 10 cm subtends 90° at the centre (π = 3.14). Sector = 90/360 × 3.14 × 10² = 1/4 × 314 = 78.5 cm². The triangle is right-angled at the centre, so its area = ½ × 10 × 10 = 50 cm². Minor segment = 78.5 − 50 = 28.5 cm².

The major segment equals the whole circle minus the minor segment: area of circle = 3.14 × 100 = 314 cm², so major segment = 314 − 28.5 = 285.5 cm².

Many real designs — flower beds, floor tiles, brooches, running tracks — are built by combining circles, sectors and segments with squares, rectangles and triangles. The area of such a shaded region is found by adding or subtracting the areas of the simple figures that make it up.

The method is always the same: break the figure into standard shapes, compute each area separately, and then combine them correctly (add the parts you keep, subtract the parts that are removed). Keep every measurement in the same units.

Worked example 1: A square of side 14 cm has a circle inscribed in it, touching all four sides. The circle's diameter equals the side, so r = 7 cm. Shaded area (square minus circle) = 14² − 22/7 × 7² = 196 − 154 = 42 cm².

Worked example 2: From each of the four corners of a square of side 14 cm, a quadrant of radius 7 cm is cut off. The four quadrants together form exactly one full circle of radius 7 cm (area 154 cm²). Remaining area = 196 − 154 = 42 cm².

When several equal sectors appear at the corners of a polygon, add their central angles: at the four corners of a square they total 4 × 90° = 360°, i.e. one whole circle; at the three corners of a triangle they total 180°, i.e. a semicircle.