This MCQ module is based on: Areas of Sectors and Segments
TOPIC 27 OF 35
Areas of Sectors and Segments
🎓 Class 10
Mathematics
CBSE
Theory
Ch 11 — Areas Related to Circles
⏱ ~30 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Areas of Sectors and Segments
Targeting Class 10 level in Mensuration, with Intermediate difficulty.
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11.1 Areas of Sector and Segment of a Circle
You have already met circles in earlier classes and learned two useful formulas: a circle of radius \(r\) has circumference \(2\pi r\) and area \(\pi r^2\). The present chapter asks: what if we want only a slice of the circle, not the entire disc?
Look at Fig. 11.1 below. A straight line OP joins the centre O to a point P on the circle. A second radius OQ makes an angle with OP. The two radii together with the arc PQ bound a pizza-slice-shaped region. Such a region is called a sectori.
Fig. 11.1 — Minor sector OPQ (shaded) and the rest, the major sector
Sector
The portion (or part) of a circular region enclosed by two radii and the corresponding arc is a sector. The smaller of the two is the minor sector, the larger the major sector. When not stated, "sector" means the minor sector.Next, draw a chord PQ — the straight segment joining the endpoints of the arc. The chord divides the circular region into two parts called segmentsi.
Fig. 11.2 — Minor segment (below chord PQ) and major segment (above)
Segment
The region cut off from a circular disc by a chord is a segment. The chord divides the disc into a minor segment (smaller area) and a major segment (larger area).Formulas — Length of arc and Area of sector
A complete circle corresponds to an angle of 360° at the centre and has circumference \(2\pi r\). A sector whose radii make an angle θ degrees at the centre is a fraction \(\theta/360\) of the whole. So:
Formulas
Length of arc = \(\dfrac{\theta}{360}\times 2\pi r\)Area of sector of angle θ = \(\dfrac{\theta}{360}\times \pi r^2\)
These follow directly from the unitary method: if 360° at the centre corresponds to the whole circle, 1° corresponds to 1/360 of the whole, and θ° to θ/360 of it.
Area of a Segment
The area of the segment cut off by chord PQ from a circle centre O equals the area of the sector OPQ minus the area of the triangle OPQ:
\[\text{Area of segment}=\text{Area of sector}-\text{Area of }\triangle OPQ.\]Fig. 11.5 — Segment (orange) = sector (orange+blue) − triangle OAB (blue)
Activity — Pizza-slice fractions
Cut out a paper circle of radius 6 cm. Fold it carefully to mark angles of 60°, 90° and 120° at the centre. Compute the sector areas using the formula \(\theta\cdot\pi r^2/360\). Cut along one radius and verify by placing the slices side-by-side that the three sector areas are in the ratio 60 : 90 : 120 = 2 : 3 : 4.
60°: (60/360)(22/7)(36) = 132/7 ≈ 18.86 cm². 90°: (90/360)(22/7)(36) = 198/7 ≈ 28.29 cm². 120°: (120/360)(22/7)(36) = 264/7 ≈ 37.71 cm². Ratios 18.86 : 28.29 : 37.71 ≈ 2 : 3 : 4 ✓.
Example 1 — Area of a sector
Find the area of a sector of a circle of radius 4 cm and central angle 30°. (Take π = 3.14.)
Solution.
\[\text{Area}=\frac{30}{360}\times 3.14\times 4^2=\frac{1}{12}\times 50.24=\frac{50.24}{12}\approx 4.19\ \text{cm}^2.\]Area of the corresponding major sector = πr² − 4.19 = 3.14 × 16 − 4.19 ≈ 46.05 cm². (Alternatively = (330/360) × πr².)
Example 2 — Area of a segment
Find the area of the segment AYB of a circle of radius 21 cm with central angle ∠AOB = 120°. (Use π = 22/7.)
Solution. Area of sector OAYB = (120/360) × (22/7) × 21² = (1/3) × (22/7) × 441 = 462 cm².
To find area of △OAB, drop perpendicular OM from O to AB. Then M is mid-point of AB and ∠AOM = 60°.
\[\cos 60°=\frac{OM}{OA}\Rightarrow OM=21\times \tfrac12=10.5\ \text{cm}.\] \[\sin 60°=\frac{AM}{OA}\Rightarrow AM=21\times \tfrac{\sqrt3}{2}=\tfrac{21\sqrt3}{2}\ \text{cm};\ AB=2\,AM=21\sqrt3.\] \[\text{Area of }\triangle OAB=\tfrac12\times AB\times OM=\tfrac12\times 21\sqrt3\times 10.5=\tfrac{441\sqrt3}{4}\ \text{cm}^2.\] \[\text{Area of segment}=462-\tfrac{441\sqrt3}{4}=\tfrac{21}{4}(88-21\sqrt3)\ \text{cm}^2.\]Exercise 11.1
(Unless stated otherwise, take π = 22/7.)
Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Area = (60/360) × (22/7) × 36 = (1/6)(792/7) = 132/7 ≈ 18.86 cm².
Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.
2πr = 22 ⇒ r = 7/2 cm. Quadrant (90°) = (1/4)πr² = (1/4)(22/7)(49/4) = 1078/112 = 77/8 cm² ≈ 9.625 cm².
Q3. The length of minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
In 5 minutes, angle swept = (5/60) × 360° = 30°. Area = (30/360) × (22/7) × 196 = (1/12)(4312/7) = 4312/84 ≈ 51.33 cm².
Q4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the areas of the corresponding minor segment and major sector. (Use π = 3.14.)
Sector (90°) = (1/4)(3.14)(100) = 78.5 cm². Triangle (right, legs 10 each) = (1/2)(10)(10) = 50 cm². Minor segment = 78.5 − 50 = 28.5 cm². Major sector = (270/360)(3.14)(100) = 235.5 cm².
Q5. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments. (Use π = 3.14, √3 = 1.73.)
Sector = (60/360)(3.14)(225) = 117.75 cm². Equilateral triangle side 15: area = (√3/4)(225) = 56.25 × 1.73 / 1 ≈ 97.3125 cm². Wait: (225 × 1.73)/4 = 97.3125 cm². Minor segment = 117.75 − 97.3125 = 20.4375 cm². Major segment = π·r² − minor = 706.5 − 20.4375 = 686.0625 cm².
Q6. A chord of a circle of radius 12 cm subtends 120° at centre. Find the area of the corresponding segment. (Use π = 3.14, √3 = 1.73.)
Sector = (120/360)(3.14)(144) = 150.72 cm². Triangle: OM = 12·cos 60° = 6, AM = 12·sin 60° = 6√3, so AB = 12√3, area = (1/2)(12√3)(6) = 36√3 ≈ 62.28 cm². Segment = 150.72 − 62.28 = 88.44 cm².
Q7. A horse is tied to a peg at one corner of a square grass field of side 15 m by means of a 5 m rope. Find (i) the area of the field the horse can graze; (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14.)
The horse grazes a quadrant. (i) (1/4)(3.14)(25) = 19.625 m². (ii) New area = (1/4)(3.14)(100) = 78.5 m². Increase = 58.875 m².
Q8. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find (i) the total length of the silver wire required; (ii) the area of each sector of the brooch.
r = 17.5 mm. (i) Circumference = 2πr = 110 mm. 5 diameters = 5 × 35 = 175 mm. Total = 285 mm. (ii) Each sector angle = 36°. Area = (36/360)(22/7)(17.5)² = (1/10)(22/7)(306.25) = 96.25 mm².
Competency-Based Questions
Q1. A ceiling fan has 3 blades, each 50 cm long, rotating at 120 rpm. Apply the sector-area formula to compute the total area swept per second.
L3 Apply120 rpm = 2 rps, so in 1 s all three blades together sweep 2 full circles. But one full circle already = π(50)² = 2500π cm². So area per second = 2 × 2500π = 5000π ≈ 15 708 cm². (The three blades cover the same circle three times, not add extra area — only the circle area matters.)
Q2. Analyse why the minor-segment area is always less than a quarter of the whole circle when the central angle is 90°.
L4 AnalyseMinor segment (90°) = sector(90°) − triangle OAB = (πr²/4) − (r²/2). Since π/4 ≈ 0.785 and 1/2 = 0.5, segment = r²(π/4 − 1/2) ≈ 0.285 r², which is less than quarter (0.25πr² ≈ 0.785 r²) because the triangle area is subtracted.
Q3. Evaluate: if the central angle doubles (say from 30° to 60°), does the segment area double?
L5 EvaluateNo. The sector area doubles linearly, but the triangle area does NOT double (it depends on sin θ). Segment = sector − triangle, so the relationship is nonlinear. For example, at θ = 30°: sector ∝ 30/360, triangle = r²/2·sin 30° = r²/4. At 60°: sector doubles but triangle area = r²·√3/4 ≈ 0.433 r², not 0.5 r². Hence segment ratio is not 2:1.
Q4. Design a problem where a goat is tied to the corner of a rectangular garden 10 m × 6 m with a 7 m rope. Compute the grazed area taking into account that rope can bend around the two nearest corners.
L6 CreateQuarter circle of radius 7 inside: (1/4)π(49) = 49π/4. When rope reaches a corner (at distance 6 along one side), remaining rope = 1 m sweeps a quarter on adjacent side = (1/4)π(1)² = π/4. Similarly for the other side (at distance 10) — but 10 > 7, so no wrap on that side. Total = 49π/4 + π/4 = 50π/4 = 25π/2 ≈ 39.27 m².
Assertion & Reason
A: A sector of angle 180° is a semicircle.
R: The fraction θ/360 becomes 1/2 when θ = 180°.
R: The fraction θ/360 becomes 1/2 when θ = 180°.
Answer: A. Half of a circle's area ↔ semicircle.
A: The area of the minor segment of a circle is always less than the area of the minor sector.
R: Segment = sector − triangle, and the triangle has positive area whenever θ ≠ 0° and θ ≠ 360°.
R: Segment = sector − triangle, and the triangle has positive area whenever θ ≠ 0° and θ ≠ 360°.
Answer: A. Subtracting a positive triangle area yields a smaller number.
A: Length of arc with angle 60° in a circle of radius 21 cm is 22 cm.
R: Arc = (θ/360) × 2πr, so (60/360)(2)(22/7)(21) = 22.
R: Arc = (θ/360) × 2πr, so (60/360)(2)(22/7)(21) = 22.
Answer: A. Computation: (1/6) × (44/7) × 21 = 44 × 21 /(6×7) = 924/42 = 22 cm ✓.
Frequently Asked Questions
What is a sector of a circle?
A sector is the region bounded by two radii of a circle and the arc between them. A pizza slice is an everyday example of a sector.
What is a segment of a circle?
A segment is the region bounded by a chord and the arc it cuts off. A circle is divided by any chord into a minor segment and a major segment.
What is the formula for arc length?
Arc length = (theta/360) x 2 pi r, where theta is the central angle in degrees and r is the radius of the circle.
How do you find the area of a minor segment?
Area of minor segment = area of corresponding sector - area of triangle formed by the two radii and the chord.
What value of pi is used in NCERT Class 10?
NCERT Class 10 uses pi = 22/7 unless stated otherwise; occasionally pi = 3.14 is specified in the question.
What is the area of a semicircle of radius r?
Area of a semicircle = (1/2) pi r squared, which follows from the sector formula with angle 180 degrees.
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