Areas Related to Circles

Sector of a circle

The part of a circular region enclosed by two radii and the arc between them. The angle between the radii at the centre is called the angle of the sector.

Minor and major sectors

Two radii split a circle into two sectors: the smaller is the minor sector and the larger the major sector. If the minor sector's angle is θ, the major sector's is 360° − θ. 'Sector' alone means the minor one.

Segment of a circle

The part of a circular region enclosed between a chord and the arc it cuts off.

Minor and major segments

A chord splits a circle into two segments: the smaller is the minor segment and the larger the major segment. 'Segment' alone means the minor one unless stated otherwise.

Area of a sector

(θ/360) × πr², where r is the radius and θ the angle of the sector in degrees.

Why is the area of a sector of angle θ equal to (θ/360) × πr²?

A whole circle is a sector of angle 360° with area πr². By the unitary method, 1° of angle corresponds to πr²/360 of area, so θ degrees correspond to θ times that.

Length of an arc

(θ/360) × 2πr for the arc of a sector of angle θ degrees in a circle of radius r. It is the same fraction of the circumference as the sector is of the circle.

Area of a segment

Area of the segment = area of the corresponding sector − area of the triangle formed by the two radii and the chord, i.e. (θ/360) × πr² − area of △OAB.

Area of a major sector or major segment

Subtract the minor one from the whole circle: major sector = πr² − minor sector, and major segment = πr² − minor segment. The major sector's area is also ((360 − θ)/360) × πr².

How do you find the area of a 30° sector of a circle of radius 4 cm, and of the corresponding major sector (π = 3.14)?

Sector = (30/360) × 3.14 × 16 ≈ 4.19 cm². The major sector is the rest of the circle, 3.14 × 16 − 4.19 ≈ 46.1 cm². Using (330/360) × 3.14 × 16 gives the same result.

In a segment problem, how do you find the area of triangle OAB when the chord AB subtends angle θ at the centre?

Draw OM perpendicular to AB. RHS congruence shows OM bisects both the chord and the angle AOB, so ∠AOM = θ/2. Then OM = r cos(θ/2) and AM = r sin(θ/2), and the triangle's area is ½ × AB × OM = AM × OM.

How do you find the area of the segment cut off by a chord subtending 120° in a circle of radius 21 cm?

Sector area = (120/360) × (22/7) × 21² = 462 cm². With OM ⊥ AB, ∠AOM = 60°, so OM = 21/2 and AM = 21√3/2, giving △OAB = 441√3/4 cm². The segment is 462 − 441√3/4 = (21/4)(88 − 21√3) cm².

How do you find the area swept by a clock's minute hand in a given number of minutes?

The minute hand turns 360° in 60 minutes, which is 6° per minute, so in 5 minutes it turns 30°. The area swept is a sector with that angle and radius equal to the length of the hand.

How do you find the central angle of a sector when its area and the radius are known?

Set the given area equal to (θ/360) × πr² and solve for θ: θ = (area × 360) ÷ (πr²). Check that the answer is less than 360°.

How are the area of a sector and the length of its arc related?

Area = (θ/360) × πr² and arc length = (θ/360) × 2πr, so the area is ½ × arc length × r. Knowing the arc and the radius is enough to find the area without finding θ.

What makes up the perimeter of a sector, and how is it found?

Its boundary is two radii plus the arc, so the perimeter is 2r + (θ/360) × 2πr. A common slip is to give only the arc length.

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