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Areas Related to Circles

All 6 textbook questions with direct answer previews. Open any question for simple explanations and exam-ready answers.

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6
Q1

Define the terms (i) sector of a circle, (ii) segment of a circle. Distinguish between minor and major sectors and segments.

(i) Sector: The region of a circle enclosed by two radii and the corresponding arc is called a sector. (ii) Segment: The region of a circle enclosed between a chord and the corresponding arc is called a segment. The minor sector is the smaller region bounded by the two radii and the minor arc. The major sector is the larger region bounded by the two radii and the major arc. Similarly, the minor segment is the smaller region between a chord and the minor arc, and the major segment is the larger region between a chord and the major arc. Unless stated otherwise, 'sector' and 'segment' refer to the minor sector and minor segment.
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Q2

Derive the formulas for (i) the area of a sector of angle θ\theta and radius rr, and (ii) the length of the arc of a sector of angle θ\theta and radius rr.

Using the unitary method: For a full circle (angle 360∘360^\circ): area =πr2= \pi r^2, circumference =2πr= 2\pi r. For angle 1∘1^\circ: area of sector =πr2360= \frac{\pi r^2}{360}, arc length =2πr360= \frac{2\pi r}{360}. For angle θ\theta: (i) Area of sector=θ360×πr2\text{Area of sector} = \frac{\theta}{360} \times \pi r^2, (ii) Arc length=θ360×2πr\text{Arc length} = \frac{\theta}{360} \times 2\pi r.
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Q3

Derive the formula for the area of a segment of a circle. Write the expression for the area of the minor segment APBAPB of a circle with centre OO, radius rr, and sector angle θ\theta.

The area of the minor segment is the area of the corresponding minor sector minus the area of the triangle formed by the two radii and the chord. Area of minor segment APB=Area of sector OAPB−Area of △OAB=θ360×πr2−Area of △OAB\text{Area of minor segment } APB = \text{Area of sector } OAPB - \text{Area of } \triangle OAB = \frac{\theta}{360} \times \pi r^2 - \text{Area of } \triangle OAB.
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Q4

Find the area of the sector of a circle with radius 4 cm4\text{ cm} and of angle 30∘30^\circ. Also, find the area of the corresponding major sector. (Use π=3.14\pi = 3.14)

Area of minor sector =30360×3.14×4×4=112×3.14×16=50.2412≈4.19 cm2= \frac{30}{360} \times 3.14 \times 4 \times 4 = \frac{1}{12} \times 3.14 \times 16 = \frac{50.24}{12} \approx 4.19\text{ cm}^2. Area of major sector =πr2−minor sector=3.14×16−4.19=50.24−4.19=46.05 cm2= \pi r^2 - \text{minor sector} = 3.14 \times 16 - 4.19 = 50.24 - 4.19 = 46.05\text{ cm}^2 (approx 46.1 cm246.1\text{ cm}^2).
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Q5

Find the area of the segment AYBAYB of a circle with radius 21 cm21\text{ cm} if ∠AOB=120∘\angle AOB = 120^\circ. (Use π=227\pi = \frac{22}{7})

Sector area: Area of sector OAYB=120360×227×21×21=13×227×441=462 cm2\text{Area of sector } OAYB = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{3} \times \frac{22}{7} \times 441 = 462\text{ cm}^2. Triangle area: Draw OM⊥ABOM \perp AB. Since OA=OBOA = OB, MM bisects ABAB and ∠AOM=∠BOM=60∘\angle AOM = \angle BOM = 60^\circ. In right △OMA\triangle OMA, cos⁡60∘=OM21\cos 60^\circ = \frac{OM}{21}, so OM=212 cmOM = \frac{21}{2}\text{ cm}. sin⁡60∘=AM21\sin 60^\circ = \frac{AM}{21}, so AM=2132 cmAM = \frac{21\sqrt{3}}{2}\text{ cm} and AB=213 cmAB = 21\sqrt{3}\text{ cm}. Area of △OAB=12×213×212=44134 cm2\triangle OAB = \frac{1}{2} \times 21\sqrt{3} \times \frac{21}{2} = \frac{441\sqrt{3}}{4}\text{ cm}^2. Segment area: Area of segment AYB=462−44134=214(88−213) cm2\text{Area of segment } AYB = 462 - \frac{441\sqrt{3}}{4} = \frac{21}{4}(88 - 21\sqrt{3})\text{ cm}^2.
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Q6

How would you find the area of the major sector and the major segment of a circle? Write the formulas.

For a circle of radius rr with minor sector angle θ\theta: (i) Area of major sector=πr2−Area of minor sector=360−θ360×πr2\text{Area of major sector} = \pi r^2 - \text{Area of minor sector} = \frac{360 - \theta}{360} \times \pi r^2. (ii) Area of major segment=πr2−Area of minor segment=πr2−(θ360×πr2−Area of △OAB)\text{Area of major segment} = \pi r^2 - \text{Area of minor segment} = \pi r^2 - \left(\frac{\theta}{360} \times \pi r^2 - \text{Area of } \triangle OAB\right).
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