Ch 11: Areas Related to Circles (Part 1)
Master the core formulas for Area of Sectors, Length of Arcs, and Area of Segments. (Based on New NCERT Syllabus).
Exercise 11.1 (Questions 1 to 7)
💡 Formula: Area of Sector = (θ/360) × πr² | Length of Arc = (θ/360) × 2πr
Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
Here, r = 6 cm and θ = 60°.
Area of Sector = (θ / 360) × πr²
= (60 / 360) × (22/7) × (6)²
= (1/6) × (22/7) × 36
= (22/7) × 6 = 132 / 7 = 18.86 cm² (approx).
Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Circumference = 2πr = 22
2 × (22/7) × r = 22 ⇒ r = 7/2 cm.
A quadrant has an angle θ = 90°.
Area of quadrant = (1/4) × πr²
= (1/4) × (22/7) × (7/2) × (7/2)
= 77 / 8 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.
Solution:
In 60 minutes, the minute hand sweeps 360°.
In 5 minutes, it sweeps: θ = (360 / 60) × 5 = 30°.
Radius (r) = 14 cm.
Area swept = Area of sector = (30 / 360) × πr²
= (1/12) × (22/7) × 14 × 14
= (1/12) × 22 × 2 × 14 = 154 / 3 cm².
Q4 to Q7. Area of corresponding Minor and Major Segments.
Core Concept for Segments (Q4, Q5, Q6, Q7):
Area of Minor Segment = Area of Sector - Area of Triangle formed by radii and chord.
If θ = 90° (Right Triangle): Area = ½ × Base × Height = ½ × r × r.
If θ = 60° (Equilateral Triangle): Area = (√3 / 4) × r².
If θ = 120°: Draw a perpendicular from center to chord, split into two 60° right triangles, and use trigonometry (sin/cos) to find base and height.
Area of Major Segment = Area of Circle (πr²) - Area of Minor Segment.
Exercise 11.1 (Questions 8 to 14)
Q8. A horse is tied to a peg at one corner of a square shaped grass field of side 15m by means of a 5m long rope. Find (i) area of that part of the field in which the horse can graze. (ii) the increase in grazing area if rope were 10m.
Solution:
The corner of a square has an angle θ = 90° (Quadrant).
(i) For r = 5m:
Area = (90/360) × πr² = (1/4) × 3.14 × 5² = (1/4) × 3.14 × 25 = 19.625 m².
(ii) For r = 10m:
New Area = (1/4) × 3.14 × 10² = (1/4) × 314 = 78.5 m².
Increase in area = 78.5 - 19.625 = 58.875 m².
Q13. A 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.
Solution:
The 6 designs are 6 equal segments. Angle subtended by each segment at center θ = 360° / 6 = 60°.
Since θ = 60°, the triangle formed by the radii and the chord is an Equilateral Triangle.
Area of 1 sector = (60/360) × (22/7) × (28)² = (1/6) × (22/7) × 28 × 28 = 410.66 cm².
Area of 1 equilateral triangle = (√3 / 4) × r² = (1.7 / 4) × (28)² = 1.7 × 7 × 28 = 333.2 cm².
Area of 1 design (segment) = Area of Sector - Area of Triangle
= 410.66 - 333.2 = 77.46 cm².
Area of 6 designs = 6 × 77.46 = 464.76 cm².
Cost = 464.76 × 0.35 = ₹162.68 (Approx).
Q14. Tick the correct answer: Area of a sector of angle p (in degrees) of a circle with radius R is
Formula is: (p / 360) × πR².
Multiply numerator and denominator by 2:
(p / 720) × 2πR².
Option (D) is correct: p/720 × 2πR².