CBSE Class 10 · Maths · board exam 2027
Chapter 11: Areas Related to Circles — notes
The chapter’s 6 parts in the book’s order — the first idea of each and the points Medha Ma’am writes on the board — and the summary at the end. The full lesson and a check question for each part are in the app.
1. Perimeter and area of a circle
First, let us go back a little, to what you studied in earlier classes.
- • Centre = O, radius = r
- • Diameter = 2r
- • Circumference = the distance once round the circle
- • circumference/diameter = π
- • Circumference = π × diameter
- • Circumference = π × 2r = 2πr
- • π is an irrational number
- • Aryabhata: π ≈ 3.1416
- • We take: π = 22/7 or 3.14
- • Length of the rectangle = 1/2 × 2πr = πr
- • Breadth of the rectangle = r
- • Area = length × breadth = πr × r
- • Area of a circle = πr²
- • Example: r = 14 cm, circumference = ?
- • Circumference = 2πr = 2 × 22/7 × 14
- • 14/7 = 2
- • Circumference = 2 × 22 × 2 = 88
- • Answer: circumference = 88 cm
- • Area = πr² = 22/7 × 14 × 14
- • = 22 × 2 × 14 = 44 × 14
- • 44 × 10 = 440; 44 × 4 = 176
- • 440 + 176 = 616
- • Answer: area = 616 cm²
- • Example: fencing cost ₹5280 at ₹24 per m
- • Ploughing at ₹0.50 per m², cost of ploughing = ?
- • Length of the fence = circumference of the field
- • Length of the fence = 5280/24
- • 24 × 200 = 4800; 5280 − 4800 = 480
- • 480/24 = 20
- • 5280/24 = 200 + 20 = 220
- • Circumference = 220 m
- • 2πr = 220
- • 2 × 22/7 × r = 220 ⇒ 44/7 × r = 220
- • r = 220 × 7/44
- • 220/44 = 5 ⇒ r = 5 × 7 = 35
- • Radius r = 35 m
- • Area = πr² = 22/7 × 35 × 35
- • 35/7 = 5
- • Area = 22 × 5 × 35 = 110 × 35 = 3850
- • Area = 3850 m²
- • 0.50 = 1/2
- • Cost of ploughing = 3850 × 1/2 = 1925
- • Answer: cost of ploughing = ₹1925
2. Perimeter of a semicircle and area of a ring
The semicircle.
- • Arc = 1/2 × 2πr = πr
- • Straight part = diameter = 2r
- • Perimeter of a semicircle = πr + 2r = (π + 2)r
- • π + 2 = 22/7 + 2 = 22/7 + 14/7 = 36/7
- • Perimeter = 36/7 × r
- • Example: r = 7 cm
- • Perimeter = 36/7 × 7 = 36
- • Check: arc = 22/7 × 7 = 22, diameter = 2 × 7 = 14
- • 22 + 14 = 36
- • Answer: perimeter = 36 cm
- • Ring = outer circle − inner circle
- • Area of the ring = πR² − πr² = π(R² − r²)
- • Example: R = 14 cm, r = 7 cm
- • 14² = 196, 7² = 49
- • 196 − 49 = 147
- • Area = 22/7 × 147
- • 147/7 = 21 ⇒ area = 22 × 21 = 462
- • Answer: 462 cm²
3. Sector and segment of a circle
Two new names of this chapter.
- • Sector = two radii + the arc between them
- • Sector OAPB: radii OA, OB and arc APB
- • ∠AOB = the angle of the sector
- • Minor sector = OAPB, major sector = OAQB
- • Angle of the major sector = 360° − ∠AOB
- • Segment = chord + arc
- • Segment APB: chord AB and arc APB
- • A sector contains the centre O, a segment does not
- • Minor segment = APB, major segment = AQB
- • Only 'sector' or 'segment' written: it means the minor one
4. Area of a sector and length of an arc
We will make the formula for the area of a sector ourselves, by the unitary method.
- • Angle 360° → area = πr²
- • Angle 1° → area = πr²/360
- • Angle θ → area = πr²/360 × θ
- • Area of the sector of angle θ = θ/360 × πr²
- • Angle 360° → arc = 2πr
- • Length of the arc l = θ/360 × 2πr
- • Perimeter of a sector = r + l + r = 2r + l
- • Example: r = 4 cm, θ = 30°, π = 3.14
- • Area = 30/360 × 3.14 × 4 × 4
- • 30/360 = 1/12
- • 4 × 4 = 16
- • 3.14 × 16 = 48 + 2.24 = 50.24
- • Area = 50.24/12 ≈ 4.19
- • Answer: about 4.19 cm²
- • Whole circle = 3.14 × 16 = 50.24
- • Major sector = 50.24 − 4.19 = 46.05
- • Answer: about 46.1 cm²
- • Angle of the major sector = 360° − 30° = 330°
- • 330/360 = 11/12
- • Area = 11/12 × 50.24 ≈ 46.05
- • Example: r = 21 cm, θ = 60°, l = ?
- • l = 60/360 × 2 × 22/7 × 21
- • 60/360 = 1/6; 21/7 = 3
- • l = 1/6 × 2 × 22 × 3 = 132/6 = 22
- • Answer: length of the arc = 22 cm
- • Minute hand, in 1 minute: 360°/60 = 6°
- • Hour hand, in 1 hour: 360°/12 = 30°
- • Length of the hand = r, angle turned = θ
5. Area of a segment
The area of a segment.
- • Segment APB = sector OAPB − △OAB
- • Area of the segment = θ/360 × πr² − area of △OAB
- • Major segment = πr² − minor segment
- • If θ = 90°: area of △OAB = 1/2 × r × r
- • If θ = 60°: △OAB is equilateral, side = r
- • Height² = a² − (a/2)² = a² − a²/4 = 3a²/4
- • Height = √3/2 a
- • Equilateral triangle = 1/2 × a × √3/2 a = √3/4 a²
- • Example: r = 21 cm, ∠AOB = 120°
- • Sector = 120/360 × 22/7 × 21 × 21
- • 120/360 = 1/3; 21/7 = 3
- • Sector = 1/3 × 22 × 3 × 21 = 22 × 21 = 462
- • Sector OAPB = 462 cm²
- • OM ⟂ AB; ∠AOM = 120°/2 = 60°
- • OM/OA = cos 60° = 1/2
- • OM = 21/2
- • AM/OA = sin 60° = √3/2
- • AM = 21√3/2
- • AB = 2 × AM = 21√3
- • △OAB = 1/2 × AB × OM
- • △OAB = 1/2 × 21√3 × 21/2 = 441√3/4
- • 462 = 21/4 × 88; 441 = 21 × 21
- • Segment APB = 462 − 441√3/4 = 21/4 × (88 − 21√3)
- • Answer: 21/4(88 − 21√3) cm²
6. Areas of combinations of plane figures
The last part.
- • Shaded part = big figure − the part removed
- • or: shaded part = sum of the areas of the shapes
- • Example: square of side 14 cm, 4 equal circles
- • Diameter = 14/2 = 7
- • Radius r = 7/2 cm
- • One circle = 22/7 × 7/2 × 7/2
- • = 22/2 × 7/2 = 11 × 7/2 = 77/2
- • Four circles = 4 × 77/2 = 2 × 77 = 154
- • Square = 14 × 14 = 196
- • Shaded part = 196 − 154 = 42
- • Answer: 42 cm²
- • Example: square ABCD, side = 10 cm, a semicircle on each side, π = 3.14
- • Unshaded regions: I, II, III, IV
- • Radius r = 10/2 = 5
- • Two semicircles = 2 × 1/2 × 3.14 × 5 × 5 = 3.14 × 25 = 78.5
- • Regions I + III = 100 − 78.5 = 21.5
- • Regions II + IV = 21.5
- • All four regions = 2 × 21.5 = 43
- • Shaded design = 100 − 43 = 57
- • Answer: 57 cm²
Summary
Very good! You worked very hard today. First we recalled the circumference, 2 π r, and the area, π r². Then we saw the perimeter of a semicircle and the area of a ring. Then we learnt to tell a sector from a segment, and we made the formulas for the area of a sector and the length of an arc ourselves, by the unitary method. For a segment, we took the triangle away from the sector. And at the end, we found the areas of designs by adding and subtracting. Write all of today's formulas once in your notebook without looking. And remember, when 22 upon 7 comes in, cancel the 7 first. Today's homework is on your screen. In the next class we will study Surface Areas and Volumes. If anything is not clear, press the Ask button and ask me. Done, you did very good work today!
AskMedha · Medha Ma’am — CBSE Class 10 Maths, Chapter 11. Written from the textbook, checked twice.
Questions parents and students ask
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