Welcome dear students! Today we are going to learn about Areas Related To Circles from Class 10 Maths. Let us begin with section 11.1, which covers the areas of sector and segment of a circle. You have already come across the terms sector and segment of a circle in your earlier classes. Recall that the portion or part of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle, and the portion enclosed between a chord and the corresponding arc is called a segment of the circle. Thus, in Figure 11.1, the shaded region OAPB is a sector of the circle with centre O. ∠AOB is called the angle of the sector. Note that in this figure, the unshaded region OAQB is also a sector of the circle. For obvious reasons, OAPB is called the minor sector and OAQB is called the major sector. You can also see that the angle of the major sector is 360° – ∠AOB.
[CHECKPOINT]
Now, look at Figure 11.2, in which AB is a chord of the circle with centre O. The shaded region APB is a segment of the circle. You can also note that the unshaded region AQB is another segment of the circle formed by the chord AB. For obvious reasons, APB is called the minor segment and AQB is called the major segment. Please remember this important remark: When we write ‘segment’ and ‘sector’, we will mean the ‘minor segment’ and the ‘minor sector’ respectively, unless stated otherwise. Now with this knowledge, let us try to find some relations or formulae to calculate their areas. Let OAPB be a sector of a circle with centre O and radius r, as shown in Figure 11.3. Let the degree measure of ∠AOB be θ. You know that the area of a circle is πr². We can consider this circular region to be a sector forming an angle of 360° at the centre O.
[CHECKPOINT]
By applying the Unitary Method, we can arrive at the area of the sector OAPB as follows. When the degree measure of the angle at the centre is 360, the area of the sector = πr². So, when the degree measure is 1, the area of the sector = πr²/360. Therefore, when the degree measure is θ, the area of the sector = (θ × πr²)/360. Thus, we obtain the formula: Area of the sector of angle θ = (θ/360) × πr², where r is the radius and θ is the angle in degrees. Now, a natural question arises: Can we find the length of the arc APB corresponding to this sector? Yes. Again, by applying the Unitary Method and taking the whole circumference of the circle as 2πr, we obtain the required length of the arc APB as (θ × 2πr)/360. So, length of an arc of a sector of angle θ = (θ/360) × 2πr.
[CHECKPOINT]
Now let us consider the area of the segment APB of a circle with centre O and radius r, as shown in Figure 11.4. You can see that: Area of the segment APB = Area of the sector OAPB – Area of ∆OAB. This equals (θ/360) × πr² – area of ∆OAB. Note that from Figure 11.3 and Figure 11.4, you can observe that: Area of the major sector OAQB = πr² – Area of the minor sector OAPB, and Area of the major segment AQB = πr² – Area of the minor segment APB. Let us now take some examples to understand these concepts. Example 1: Find the area of the sector of a circle with radius 4 cm and angle 30°. Also, find the area of the corresponding major sector. Use π = 3.14.
[CHECKPOINT]
Given sector is OAPB, as shown in Figure 11.5. Area of the sector = (θ/360) × πr². Substituting the values, we get (30/360) × 3.14 × 4² cm². This simplifies to (1/12) × 3.14 × 16 cm² = 50.24/12 cm² = 4.1866... cm², which is approximately 4.19 cm². Area of the corresponding major sector = πr² – area of sector OAPB. This equals (3.14 × 16 – 4.19) cm² = 46.05 cm², or approximately 46.1 cm². Alternatively, area of the major sector = ((360 – θ)/360) × πr². Substituting gives ((360 – 30)/360) × 3.14 × 16 cm² = (330/360) × 50.24 cm² = 46.05 cm², or approximately 46.1 cm².
[CHECKPOINT]
Example 2: Find the area of the segment AYB shown in Figure 11.6, if the radius of the circle is 21 cm and ∠AOB = 120°. Use π = 22/7. Area of the segment AYB = Area of sector OAYB – Area of ∆OAB. Let us call this equation (1). Now, area of the sector OAYB = (120/360) × (22/7) × 21² cm² = (1/3) × (22/7) × 441 cm² = 462 cm². Let us call this equation (2). For finding the area of ∆OAB, draw OM ⊥ AB as shown in Figure 11.7. Note that OA = OB. Therefore, by RHS congruence, ∆AMO ≅ ∆BMO. So, M is the mid-point of AB and ∠AOM = ∠BOM = (1/2) × 120° = 60°. Let OM = x cm. So, from ∆OMA, OM/OA = cos 60°. Or, x/21 = 1/2. Therefore, x = 21/2. So, OM = 21/2 cm.
[CHECKPOINT]
Also, AM/OA = sin 60° = √3/2. So, AM = (21√3)/2 cm. Therefore, AB = 2 × AM = 2 × (21√3)/2 cm = 21√3 cm. So, area of ∆OAB = (1/2) × AB × OM = (1/2) × 21√3 × (21/2) cm² = (441√3)/4 cm². Let us call this equation (3). Therefore, area of the segment AYB = 462 – (441√3)/4 cm², from equations (1), (2), and (3). This simplifies to (21/4)(88 – 21√3) cm². Now let us proceed to Exercise 11.1. Unless stated otherwise, use π = 22/7. We will solve each question step by step.
[CHECKPOINT]
Question 1: Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. Solution: Area = (60/360) × π × 6² = (1/6) × (22/7) × 36 = (6 × 22)/7 = 132/7 cm². Question 2: Find the area of a quadrant of a circle whose circumference is 22 cm. Solution: Circumference = 2πr = 22. So, 2 × (22/7) × r = 22, which gives r = 7/2 = 3.5 cm. A quadrant has an angle of 90°. Area = (90/360) × π × r² = (1/4) × (22/7) × (3.5)² = (1/4) × (22/7) × 12.25 = 9.625 cm². Question 3: 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 hand sweeps 360°. In 5 minutes, it sweeps (5/60) × 360° = 30°. Area = (30/360) × π × 14² = (1/12) × (22/7) × 196 = 154/3 cm² ≈ 51.33 cm².
[CHECKPOINT]
Question 4: 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. Use π = 3.14. Solution: (i) Area of minor sector = (90/360) × 3.14 × 10² = 78.5 cm². Area of ∆OAB = (1/2) × 10 × 10 = 50 cm². Area of minor segment = 78.5 – 50 = 28.5 cm². (ii) Area of major sector = ((360 – 90)/360) × 3.14 × 10² = (270/360) × 314 = 0.75 × 314 = 235.5 cm². Question 5: 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. Solution: (i) Arc length = (60/360) × 2πr = (1/6) × 2 × (22/7) × 21 = 22 cm. (ii) Sector area = (60/360) × πr² = (1/6) × (22/7) × 441 = 231 cm². (iii) Since the angle is 60° and OA = OB, ∆OAB is equilateral. Area of ∆OAB = (√3/4) × 21² = (441√3)/4 cm². Area of segment = 231 – (441√3)/4 cm².
[CHECKPOINT]
Question 6: 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. Use π = 3.14 and √3 = 1.73. Solution: Sector area = (60/360) × 3.14 × 15² = 117.75 cm². ∆OAB is equilateral. Area of ∆OAB = (√3/4) × 15² = (1.73/4) × 225 = 97.3125 cm². Area of minor segment = 117.75 – 97.3125 = 20.4375 cm². Area of major segment = πr² – minor segment = (3.14 × 225) – 20.4375 = 706.5 – 20.4375 = 686.0625 cm². Question 7: 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. Use π = 3.14 and √3 = 1.73. Solution: Sector area = (120/360) × 3.14 × 12² = 150.72 cm². Draw OM ⊥ AB. In ∆OMA, ∠AOM = 60°. OM = 12 cos 60° = 6 cm. AM = 12 sin 60° = 12 × (√3/2) = 6√3 cm. AB = 12√3 cm. Area of ∆OAB = (1/2) × 12√3 × 6 = 36√3 = 36 × 1.73 = 62.28 cm². Area of segment = 150.72 – 62.28 = 88.44 cm².
[CHECKPOINT]
Question 8: 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. Use π = 3.14. Solution: (i) The horse grazes a quadrant of radius 5 m. Area = (90/360) × 3.14 × 5² = 19.625 m². (ii) If rope is 10 m, area = (1/4) × 3.14 × 10² = 78.5 m². Increase = 78.5 – 19.625 = 58.875 m². Question 9: 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 as shown in Figure 11.9. Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch. Solution: (i) Circumference = πd = (22/7) × 35 = 110 mm. Length of 5 diameters = 5 × 35 = 175 mm. Total length = 110 + 175 = 285 mm. (ii) Each sector angle = 360°/10 = 36°. Area = (36/360) × π × (17.5)² = (1/10) × (22/7) × 306.25 = 96.25 mm².
[CHECKPOINT]
Question 10: An umbrella has 8 ribs which are equally spaced as shown in Figure 11.10. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. Solution: Angle between ribs = 360°/8 = 45°. Area = (45/360) × π × 45² = (1/8) × (22/7) × 2025 = 22275/28 cm² ≈ 795.54 cm². Question 11: 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. Solution: Area cleaned by one wiper = (115/360) × π × 25². Total area for two wipers = 2 × (115/360) × (22/7) × 625 = (230/360) × (13750/7) = 316250/252 ≈ 1254.96 cm². Question 12: 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. Use π = 3.14. Solution: Area = (80/360) × 3.14 × (16.5)² = (2/9) × 3.14 × 272.25 = 189.97 km².
[CHECKPOINT]
Question 13: A round table cover has six equal designs as shown in Figure 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. Solution: Each design is a segment subtending 60° at the centre. Sector area = (60/360) × (22/7) × 28² = 410.67 cm². Area of equilateral triangle = (√3/4) × 28² = (1.7/4) × 784 = 333.2 cm². Area of one segment = 410.67 – 333.2 = 77.47 cm². Total area for 6 designs = 6 × 77.47 = 464.82 cm². Cost = 464.82 × 0.35 = ₹162.69. Question 14: Tick the correct answer in the following: 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. Solution: The correct formula is (p/360) × πR². Hence, option (D) is correct.
[CHECKPOINT]
Let us now review the summary of this chapter in section 11.2. In this chapter, you have studied the following points. First, the length of an arc of a sector of a circle with radius r and angle with degree measure θ is (θ/360) × 2πr. Second, the area of a sector of a circle with radius r and angle with degree measure θ is (θ/360) × πr². Third, the area of a segment of a circle equals the area of the corresponding sector minus the area of the corresponding triangle. These are the key results you must remember for your examinations. Practice these formulae regularly and apply them carefully to different types of problems. Always check whether you are asked for a minor or major sector or segment, and use the appropriate formula. These concepts frequently appear in board questions. Make sure you are comfortable with the unitary method derivation and the geometric construction for finding triangle areas in segment problems. Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]