Chapter 12: Areas Related to Circles (NCERT Solutions)

Exercise 12.1: Perimeter and Area of Circle

Q1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

2πR = 2π(19) + 2π(9).
R = 19 + 9 = 28 cm.

Q2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

πR² = π(8)² + π(6)².
R² = 64 + 36 = 100.
R = 10 cm.

Q3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Gold radius = 10.5 cm. Area = π(10.5)² = 110.25π = 346.5 cm².
Red: π(21)² - π(10.5)² = 441π - 110.25π = 330.75π = 1039.5 cm².
Blue: π(31.5)² - π(21)² = 992.25π - 441π = 551.25π = 1732.5 cm².
Black: π(42)² - π(31.5)² = 1764π - 992.25π = 771.75π = 2425.5 cm².
White: π(52.5)² - π(42)² = 2756.25π - 1764π = 992.25π = 3118.5 cm².

Q4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Circumference = πd = π(80) cm = 0.8π m.
Distance in 10 min = 66 × (10/60) = 11 km = 11000 m.
Revolutions = 11000/(0.8π) = 11000/(2.513) ≈ 4375.

Q5. Tick the correct answer. The area of a sector of a circle with radius 6 cm if angle of the sector is 60° is: (A) 132/7 cm² (B) 6π cm² (C) 66/7 cm² (D) None

Area = (θ/360°) × πr².
= (60°/360°) × π(6)².
= (1/6) × 36π = 6π cm².
Answer: (B) 6π cm².

Exercise 12.2: Sectors and Segments

Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Area = (60/360) × π(6)² = (1/6) × 36π = 6π cm² = 18.86 cm².

Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.

2πr = 22 ⇒ r = 22/(2π) = 11/π = 3.5 cm.
Quadrant area = (1/4)πr² = (1/4)π(3.5)² = (1/4)π(12.25) = 9.625 cm².

Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

In 5 min, angle = (5/60) × 360° = 30°.
Area = (30/360) × π(14)² = (1/12) × 196π = 49π/3 = 51.33 cm².

Q4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector.

(i) Minor segment:
Sector area = (90/360) × π(10)² = 25π cm².
Triangle area = (1/2) × 10 × 10 = 50 cm².
Segment = 25π - 50 = 78.5 - 50 = 28.5 cm².

(ii) Major sector:
= (270/360) × π(10)² = 75π = 235.5 cm².

Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord.

(i) Arc length: = (θ/360°) × 2πr = (60/360) × 2π(21) = 22 cm = 22 cm.

(ii) Sector area: = (60/360) × π(21)² = 231 cm² = 231 cm².

(iii) Segment area:
Triangle: Area = (1/2)r²sinθ = (1/2)(21)²sin60° = (441/2)(√3/2) = 441√3/4 = 190.9 cm².
Segment = 231 - 190.9 = 40.1 cm².

Q6. 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 of the circle.

Minor sector = (60/360) × π(15)² = 37.5π = 117.75 cm².
Triangle = (1/2)(15)²sin60° = (225/2)(√3/2) = 97.3 cm².
Minor segment = 117.75 - 97.3 = 20.45 cm².
Major segment = π(15)² - 20.45 = 706.5 - 20.45 = 686.05 cm².

Q7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.

Sector = (120/360) × π(12)² = 48π = 150.72 cm².
Triangle = (1/2)(12)²sin120° = 72(√3/2) = 62.28 cm².
Segment = 150.72 - 62.28 = 88.44 cm².

Q8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find: (i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m.

(i) Grazing area = (1/4)π(5)² = 6.25π = 19.625 m².

(ii) New area = (1/4)π(10)² = 25π = 78.5 m².
Increase = 78.5 - 19.625 = 58.875 m².

Q9. 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.

(i) Circumference + 5 diameters = π(35) + 5(35) = 110 + 175 = 285 mm.

(ii) Each sector = (1/10) × π(17.5)² = 30.625π = 96.25 mm².

Q10. An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Each sector = (1/8) × π(45)² = (2025π/8) = 795.94 cm².

Q11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Each wiper area = (115/360) × π(25)² = 625.98 cm².
Total = 2 × 625.98 = 1251.96 cm².

Q12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.

Area = (80/360) × π(16.5)² = (2/9) × 272.25π = 190.06 km².

Q13. A round table cover has six equal designs. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm².

Each design = (1/6) × π(28)² = 410.67 cm².
Total area = 6 × 410.67 = 2464 cm².
Cost = 2464 × 0.35 = Rs 862.40.

Q14. Tick the correct answer. Area of a sector of angle p (in degrees) of a circle with radius R is: (A) p/180 × 2πR (B) p/180 × πR² (C) p/360 × 2πR (D) p/360 × πR²

Area of sector = (θ/360°) × πR².
Answer: (D) p/360 × πR².

Exercise 12.3: Combination of Plane Figures

Q1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

RQ = √(24² + 7²) = √625 = 25 cm (hypotenuse, radius).
Semicircle area = (1/2)π(12.5)² = 245.31 cm².
Triangle area = (1/2) × 24 × 7 = 84 cm².
Shaded = 245.31 - 84 = 161.31 cm².

Q2. Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Shaded area = (40/360)[π(14)² - π(7)²].
= (1/9)[196π - 49π] = (1/9)(147π) = 51.33 cm².

Q3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Square area = 14² = 196 cm².
Each semicircle radius = 7 cm, area = (1/2)π(7)² = 77 cm².
Shaded = 196 - 2(77) = 196 - 154 = 42 cm².

Q4. Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Equilateral triangle area = (√3/4)(12)² = 36√3 = 62.28 cm².
Sector area (60°) = (60/360)π(6)² = 6π = 18.84 cm².
Shaded = 62.28 - 18.84 = 43.44 cm².

Q5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut. Find the area of the remaining portion of the square.

Square area = 16 cm².
4 quadrants = 1 full circle area = π(1)² = π cm².
Circle area = π(1)² = π cm².
Remaining = 16 - π - π = 16 - 2π = 9.72 cm².

Q6-Q16. [Additional complex problems involving combinations of circles, triangles, squares, rectangles, and their areas]

These problems follow similar approaches using:
• Area of basic shapes (circle, triangle, square, rectangle)
• Sector and segment formulas
• Pythagorean theorem for missing dimensions
• Addition/subtraction of areas for shaded regions

Areas Related to Circles - RD Sharma Important Questions

Q1. A race track is in the form of a ring whose inner and outer circumferences are 352 m and 396 m respectively. Find the width of the track and its area.

2πr₁ = 352 ⇒ r₁ = 56 m.
2πr₂ = 396 ⇒ r₂ = 63 m.
Width = 63 - 56 = 7 m.
Area = π(63² - 56²) = π(3969 - 3136) = 833π = 2618 m².

Q2. A wire when bent in the form of a square encloses an area of 484 cm². The same wire is now bent to form a circle. Find the area enclosed by the circle.

Side of square = √484 = 22 cm.
Perimeter = 4 × 22 = 88 cm.
This becomes circumference: 2πr = 88 ⇒ r = 44/π = 14 cm.
Circle area = π(14)² = 616 cm².

Q3. The area of a circular playground is 22176 m². Find the cost of fencing this ground at the rate of Rs 50 per metre.

πr² = 22176 ⇒ r² = 7056 ⇒ r = 84 m.
Circumference = 2π(84) = 528 m.
Cost = 528 × 50 = Rs 26,400.

Q4. The diameter of a wheel of a car is 63 cm. Find the distance travelled by it in 500 revolutions.

Circumference = πd = π(63) = 198 cm.
Distance = 500 × 198 = 99000 cm = 990 m.

Q5. Find the area of the sector of a circle of radius 5 cm, if the angle of the sector is 150°.

Area = (150/360) × π(5)² = (5/12) × 25π = 32.72 cm².

Q6. Four equal circles are described about the four corners of a square so that each touches two of the others. Find the area of the space enclosed between the circumferences of the circles, each side of the square measuring 14 cm.

Radius of each circle = 7 cm.
Square area = 196 cm².
4 quadrants = 1 circle area = π(7)² = 154 cm².
Space = 196 - 154 = 42 cm².

Q7. A piece of wire 20 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle.

Arc length = (θ/360°) × 2πr.
20 = (60/360) × 2πr.
20 = (πr/3).
r = 60/π = 19.1 cm.

Q8. The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Find the area of the sector.

Perimeter = 2r + arc.
16.4 = 2(5.2) + arc.
arc = 6 cm.
Arc = (θ/360) × 2πr.
6 = (θ/360) × 2π(5.2).
θ = 65.97°.
Area = (65.97/360) × π(5.2)² = 15.6 cm².

Q9. AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Large circle area = π(7)² = 154 cm².
Small circle radius = 3.5 cm, area = π(3.5)² = 38.5 cm².
Shaded = 154/2 - 38.5 = 77 - 38.5 = 38.5 cm².

Q10. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc and area of the minor segment.

Arc = (60/360) × 2π(21) = 22 cm.
Sector = 231 cm², Triangle = 190.9 cm².
Segment = 40.1 cm².

Q11-Q20. [Additional problems on circles with inscribed and circumscribed figures, combination problems, and application-based questions]

Key concepts:
• Sector area = (θ/360) × πr²
• Arc length = (θ/360) × 2πr
• Segment area = Sector area - Triangle area
• Ring area = π(R² - r²)

Areas Related to Circles - Formulas & PYQs

Key Formulas

1. Circle

Circumference = 2πr = πd

Area = πr²

2. Sector of Circle

Arc length = (θ/360°) × 2πr

Area of sector = (θ/360°) × πr²

Perimeter of sector = 2r + arc length

3. Segment of Circle

Area of segment = Area of sector - Area of triangle

For minor segment with angle θ:
Area = (1/2)r²[(πθ/180) - sin θ]

4. Annulus (Ring)

Area = π(R² - r²)

where R = outer radius, r = inner radius

5. Special Cases

Semicircle: Area = πr²/2, Perimeter = πr + 2r

Quadrant: Area = πr²/4, Perimeter = πr/2 + 2r

Previous Year Questions (CBSE/JKBOSE)

Q1. Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector. (CBSE 2020)

Minor sector = (30/360) × π(4)² = (4π/3) cm².
Major sector = π(4)² - (4π/3) = 16π - 4π/3 = (44π/3) cm².

Q2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles. (CBSE 2019)

NCERT Ex 12.1 Q2. Radius = 10 cm.

Q3. In Fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region. (CBSE 2018)

Quadrant area = (1/4)π(3.5)² = 9.625 cm².
Triangle area = (1/2) × 3.5 × 2 = 3.5 cm².
Shaded = 9.625 - 3.5 = 6.125 cm².

Q4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour? (CBSE 2017)

NCERT Ex 12.1 Q4. Revolutions = 4375.

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