{"slug":"cbse-class-10-maths-areas-related-to-circles","title":"Areas Related to Circles","description":"Sectors, segments, arc length and their areas, weighted towards the sector-angle, arc-length and minute-hand questions set across the 2026 papers.","board":"CBSE","grade":"Class 10","subject":"Maths","chapter":"Areas Related to Circles","url":"https://www.flipnlearn.app/deck/cbse-class-10-maths-areas-related-to-circles","studyUrl":"https://www.flipnlearn.app/deck/cbse-class-10-maths-areas-related-to-circles/study","cards":[{"id":"sector-of-a-circle","kind":"term-definition","front":"Sector of a circle","back":"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."},{"id":"minor-and-major-sectors","kind":"term-definition","front":"Minor and major sectors","back":"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."},{"id":"segment-of-a-circle","kind":"term-definition","front":"Segment of a circle","back":"The part of a circular region enclosed between a chord and the arc it cuts off."},{"id":"minor-and-major-segments","kind":"term-definition","front":"Minor and major segments","back":"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."},{"id":"area-of-a-sector","kind":"formula","front":"Area of a sector","back":"(θ/360) × πr², where r is the radius and θ the angle of the sector in degrees."},{"id":"why-the-sector-area-formula-works","kind":"concept-question","front":"Why is the area of a sector of angle θ equal to (θ/360) × πr²?","back":"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."},{"id":"length-of-an-arc","kind":"formula","front":"Length of an arc","back":"(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."},{"id":"area-of-a-segment","kind":"formula","front":"Area of a segment","back":"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."},{"id":"area-of-a-major-sector-or-segment","kind":"formula","front":"Area of a major sector or major segment","back":"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."},{"id":"how-to-find-a-sector-and-its-major-sector","kind":"worked-step","front":"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)?","back":"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."},{"id":"how-to-find-the-area-of-triangle-oab","kind":"worked-step","front":"In a segment problem, how do you find the area of triangle OAB when the chord AB subtends angle θ at the centre?","back":"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."},{"id":"how-to-find-the-area-of-a-segment","kind":"worked-step","front":"How do you find the area of the segment cut off by a chord subtending 120° in a circle of radius 21 cm?","back":"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."},{"id":"how-to-find-the-area-swept-by-a-minute-hand","kind":"worked-step","front":"How do you find the area swept by a clock's minute hand in a given number of minutes?","back":"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."},{"id":"how-to-find-the-angle-of-a-sector","kind":"worked-step","front":"How do you find the central angle of a sector when its area and the radius are known?","back":"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°."},{"id":"sector-area-and-arc-length","kind":"concept-question","front":"How are the area of a sector and the length of its arc related?","back":"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."},{"id":"perimeter-of-a-sector","kind":"concept-question","front":"What makes up the perimeter of a sector, and how is it found?","back":"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."}]}