Exercise 11.1 - Areas Related to Circles

📝 Important Notes & Formulas

📐 Area of Sector

When angle θ is in degrees:

A = (θ/360) × πr²

📏 Arc Length

For central angle θ (in degrees):

L = (θ/360) × 2πr

🔺 Area of Triangle (Two Radii)

Triangle formed by two radii & chord:

A = ½ r² sin(θ)

🥧 Area of Segment

Segment = Sector − Triangle

A = (θ/360)πr² − ½r²sin(θ)

📐 Quadrant

One-fourth of a circle (θ = 90°):

A = ¼ πr²

⏱️ Clock Angles

Minute hand: 360° in 60 min

Angle per min = 6°

❓ MCQ Solutions — Click an option to reveal step-by-step answer!

1Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Area of Sector = (θ/360) × πr²

Step 1: Given: r = 6 cm, θ = 60°, π = 22/7

Step 2: Area = (60/360) × (22/7) × 6² = (1/6) × (22/7) × 36

Step 3: Area = (22 × 36) / (7 × 6) = 792/42 = 132/7 cm²

2Find the area of a quadrant of a circle whose circumference is 22 cm.

Circumference = 2πr → Quadrant Area = ¼ πr²

Step 1: 2πr = 22 → 2 × (22/7) × r = 22 → r = 22 × 7 / 44 = 7/2 cm

Step 2: Area of quadrant = ¼ × (22/7) × (7/2)² = ¼ × (22/7) × (49/4)

Step 3: = (22 × 49) / (7 × 16) = 1078/112 = 77/8 cm²

3The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

In 60 min, minute hand sweeps 360° → In 1 min = 6° → In 5 min = 30°

Step 1: Angle swept in 5 min = 5 × 6° = 30°

Step 2: Area = (30/360) × (22/7) × 14² = (1/12) × (22/7) × 196

Step 3: = (22 × 196) / (7 × 12) = 4312/84 = 154/3 cm²

4A 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. (Use π = 3.14)

Segment = Sector − Triangle | Major Sector = πr² − Minor Sector

Step 1: Minor Sector Area = (90/360) × 3.14 × 10² = ¼ × 3.14 × 100 = 78.5 cm²

Step 2: Triangle Area = ½ × 10 × 10 = 50 cm² (right angle at centre)

Step 3: Minor Segment = 78.5 − 50 = 28.5 cm²

Step 4: Major Sector = 3.14 × 100 − 78.5 = 314 − 78.5 = 235.5 cm²

5In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) length of arc (ii) area of sector (iii) area of segment.

Arc = (θ/360)×2πr | Sector = (θ/360)×πr² | Segment = Sector − ½r²sinθ

(i) Arc length = (60/360) × 2 × (22/7) × 21 = (1/6) × 2 × 22 × 3 = 22 cm

(ii) Sector area = (60/360) × (22/7) × 21² = (1/6) × (22/7) × 441 = 231 cm²

(iii) Triangle area = ½ × 21² × sin(60°) = ½ × 441 × (√3/2) = 441√3/4 cm²

Ans Segment = 231 − 441√3/4 = 231 − 190.96 ≈ 40.04 cm²

6A 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 and √3 = 1.73)

Sector = (θ/360)×πr² | Triangle = ½r²sinθ | Segment = Sector − Triangle

Step 1: Sector area = (60/360) × 3.14 × 15² = (1/6) × 3.14 × 225 = 117.75 cm²

Step 2: Triangle area = ½ × 15² × sin(60°) = ½ × 225 × 1.73/2 = 97.3125 cm²

Step 3: Minor Segment = 117.75 − 97.3125 = 20.4375 ≈ 20.43 cm²

Step 4: Circle area = 3.14 × 225 = 706.5 cm²

Step 5: Major Segment = 706.5 − 20.4375 = 686.0625 ≈ 686.07 cm²

7A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment. (Use π = 3.14 and √3 = 1.73)

Segment = (θ/360)×πr² − ½r²sinθ

Step 1: Sector = (120/360) × 3.14 × 12² = (1/3) × 3.14 × 144 = 150.72 cm²

Step 2: Triangle = ½ × 12² × sin(120°) = ½ × 144 × (√3/2) = 72 × 1.73/2... wait, sin(120°)=sin(60°)=√3/2

Step 3: Triangle = ½ × 144 × 1.73/2 = 36 × 1.73 = 62.28 cm²

Step 4: Segment = 150.72 − 62.28 = 88.44 cm²

8A 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. (Use π = 3.14)

At corner of square → Grazing area = Quadrant = ¼ πr²

(i) Grazing area = ¼ × 3.14 × 5² = ¼ × 3.14 × 25 = 19.625 m²

(ii) New area = ¼ × 3.14 × 10² = ¼ × 3.14 × 100 = 78.5 m²

Ans Increase = 78.5 − 19.625 = 58.875 m²

9A 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.

r = 35/2 = 17.5 mm | 10 sectors → each angle = 36°

(i) Circumference = 2 × (22/7) × 17.5 = 110 mm

5 diameters = 5 × 35 = 175 mm

Total wire = 110 + 175 = 285 mm

(ii) Each sector = (36/360) × (22/7) × 17.5² = (1/10) × (22/7) × 306.25

= (22 × 306.25) / 70 = 6737.5/70 = 385/4 mm² = 96.25 mm²

10An 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.

8 ribs → 8 equal sectors → each angle = 360°/8 = 45°

Step 1: Area between ribs = Area of one sector = (45/360) × (22/7) × 45²

Step 2: = (1/8) × (22/7) × 2025 = (22 × 2025) / 56 = 44550/56

Step 3: = 795.5357... ≈ 795.53 cm²

11A 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.

Area of one wiper sweep = Sector area = (θ/360) × πr²

Step 1: One wiper = (115/360) × 3.14 × 25² = (115/360) × 3.14 × 625

Step 2: = (115 × 1962.5) / 360 = 225687.5 / 360 ≈ 627.465 cm²

Step 3: Two wipers = 2 × 627.465 = 1254.93 cm² ≈ 1254.9 cm²

12To 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. (Use π = 3.14)

Area = (θ/360) × πr²

Step 1: Area = (80/360) × 3.14 × (16.5)² = (2/9) × 3.14 × 272.25

Step 2: = (2 × 3.14 × 272.25) / 9 = 1709.73 / 9

Step 3: = 189.97 km²

13A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹0.35 per cm². (Use √3 = 1.7)

6 designs → 6 segments | Each central angle = 360°/6 = 60°

Step 1: Area of one sector = (60/360) × (22/7) × 28² = (1/6) × (22/7) × 784 = 410.67 cm²

Step 2: Area of equilateral Δ = (√3/4) × 28² = (1.7/4) × 784 = 333.2 cm²

Step 3: One segment = 410.67 − 333.2 = 77.47 cm²

Step 4: 6 segments = 6 × 77.47 = 464.8 cm²

Step 5: Cost = 464.8 × 0.35 = ₹162.68... wait, let me recalculate

Recalc: Sector = (1/6)×22×112 = 2464/6 = 410.67 | Δ = 1.7×196 = 333.2 | Segment = 77.47 | 6×77.47=464.82 | Cost=162.69

Correct: Using exact: Sector=1232/3, Δ=333.2, Segment=77.47, Total=464.82, Cost=₹162.69

14Tick the correct answer: Area of a sector of angle p (in degrees) of a circle with radius R is:

Area of Sector = (θ/360) × πr² → Here θ = p, r = R

Step 1: The standard formula for sector area with angle p° and radius R is:

Step 2: Area = (p/360) × πR²

Step 3: This matches Option B

Note: Option A gives arc length (wrong multiplier), C gives arc length formula, D is incorrect.

📐 Exercise 11.1

📝 Important Notes & Formulas

📐 Area of Sector

📏 Arc Length

🔺 Area of Triangle (Two Radii)

🥧 Area of Segment

📐 Quadrant

⏱️ Clock Angles

🧮 Quick Calculators (With Units)

Sector Area Calculator

Arc Length Calculator

Segment Area Calculator

🔬 Interactive Geometry Visualizer

❓ MCQ Solutions — Click an option to reveal step-by-step answer!