Areas Related to Circles: NCERT Class 10 Maths Ex 12.2 Solutions

Key Terms: Sector and Segment

Sector of a Circle

The region of a circle enclosed by two radii and the corresponding arc. The smaller area is called the 'minor sector' and the larger area is the 'major sector'. Think of it as a slice of pizza.

Segment of a Circle

The region of a circle enclosed by a chord and the corresponding arc. The smaller area is the 'minor segment' and the larger area is the 'major segment'. Think of the crust portion of a pizza slice after you've eaten the pointy part.

Angle of the Sector (θ)

The angle formed by the two radii at the center of the circle. This angle is crucial for calculating the area of the sector and is measured in degrees.

Understanding the Formulas for Sector and Segment

To solve problems from Exercise 12.2, we need two key formulas. Let's understand how they are derived.

1. Area of a Sector:
The total area of a circle is πr², which corresponds to a full angle of 360° at the center. A sector is just a fraction of this total area. The fraction is determined by its central angle, θ. So, the area of a sector is simply the fraction (θ/360) of the total area.

Formula: Area of a Sector = (θ / 360°) × πr²

Where:

• θ is the angle of the sector in degrees.
• r is the radius of the circle.

2. Area of a Segment:
There's no direct, single formula for a segment. Instead, we find it by subtracting. If you look at a minor segment, you can see it's part of a minor sector. Specifically, the segment is what's left after you remove the triangle formed by the two radii and the chord.

Formula: Area of a Segment = Area of the corresponding Sector – Area of the corresponding Triangle

To find the area of the triangle (usually named ΔOAB, where O is the center), you can use:

• If θ = 90°: The triangle is a right-angled triangle. Area = (1/2) × base × height = (1/2) × r × r = (1/2)r².
• If θ = 60°: The triangle is equilateral. Area = (√3/4) × side² = (√3/4)r².
• For any θ: You can use trigonometry. Area = (1/2)r²sin(θ).

Worked Example: Finding Area of Sector and Segment

• Problem: In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find: (i) the area of the sector formed by the arc and (ii) the area of the segment formed by the corresponding chord. (Use π = 22/7). Solution: Given: Radius (r) = 21 cm Angle (θ) = 60° (i) Area of the Sector Step 1: Write down the formula for the area of a sector. Area of Sector = (θ / 360°) × πr² Step 2: Substitute the given values into the formula. Area = (60° / 360°) × (22/7) × (21)² Step 3: Simplify the expression. Area = (1/6) × (22/7) × 21 × 21 Area = (1/6) × 22 × 3 × 21 Area = 11 × 21 Step 4: Calculate the final area. Area = 231 cm² (ii) Area of the Segment Step 1: Write down the formula for the area of a segment. Area of Segment = Area of Sector – Area of Triangle (ΔOAB) Step 2: Find the area of the triangle. Since θ = 60° and the two sides OA and OB are radii, the triangle ΔOAB is an equilateral triangle (all angles are 60°). The side of the triangle is equal to the radius, r = 21 cm. Area of equilateral ΔOAB = (√3 / 4) × (side)² Area = (√3 / 4) × (21)² Area = (√3 / 4) × 441 = (441√3) / 4 cm² * Step 3: Substitute the values to find the segment's area. Area of Segment = 231 - (441√3 / 4) cm² Final Answer: The area of the sector is 231 cm² and the area of the segment is (231 - 441√3 / 4) cm².

Important Tips for Your Board Exams

Students often make small mistakes in this chapter that can cost them marks. Keep these points in mind:

1. Major vs. Minor: Read the question carefully! If it asks for the area of the major sector, the angle to use is (360° - θ), not θ. Similarly, the area of the major segment is Area of Circle - Area of Minor Segment.
2. Value of π: Always check if a specific value for π (like 3.14 or 22/7) is given in the question. Using the wrong one can lead to a different answer and a loss of marks.
3. Area of the Triangle: Be very careful when calculating the area of the triangle for the segment formula. Identify the type of triangle correctly (right-angled, equilateral) based on the central angle θ. If it's not a special angle, you may need to use trigonometry.
4. Units: Always write the correct units in your final answer. For area, the unit will be square units (e.g., cm², m²).

Practice Questions with Solutions

• Q: Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°. (Use π=22/7) A: Step 1: Given radius (r) = 6 cm, angle (θ) = 60°. Formula for area of sector = (θ/360°) πr². Step 2: Area = (60/360) (22/7) 6 6 = (1/6) (22/7) 36 = (22/7) * 6 = 132/7 cm². Final answer: 132/7 cm² (approx 18.86 cm²).
• Q: A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use π=3.14) A: Step 1: Given r=10 cm, θ=90°. Area of sector = (θ/360°) πr² = (90/360) 3.14 10² = (1/4) 3.14 100 = 78.5 cm². Step 2: Area of Δ formed by chord and radii = (1/2) r r sin(90°) = (1/2) 10 10 * 1 = 50 cm². Step 3: Area of minor segment = Area of sector - Area of Δ = 78.5 - 50 = 28.5 cm². Final answer: 28.5 cm².
• Q: Two concentric circles have radii 7 cm and 14 cm. If the angle of the sector is 45°, find the area of the shaded region between the two sectors. (Use π=22/7) A: Step 1: Outer radius (R)=14 cm, inner radius (r)=7 cm, angle (θ)=45°. Step 2: Area of outer sector = (45/360) πR² = (1/8) (22/7) 14 14 = 77 cm². Step 3: Area of inner sector = (45/360) πr² = (1/8) (22/7) 7 7 = 38.5 cm². Step 4: Shaded area = Area of outer sector - Area of inner sector = 77 - 38.5 = 38.5 cm². Final answer: 38.5 cm².
• Q: A square ABCD has side 14 cm. With A as centre, a quadrant of a circle is drawn with radius 14 cm. Find the area of the shaded region (square - quadrant). (Use π=22/7) A: Step 1: Side of square = 14 cm. Area of square = side² = 14 14 = 196 cm². Step 2: Radius of quadrant (r) = 14 cm, angle (θ) = 90°. Area of quadrant = (θ/360°) πr² = (90/360) (22/7) 14 14 = (1/4) 22 2 14 = 154 cm². Step 3: Shaded area = Area of square - Area of quadrant = 196 - 154 = 42 cm². Final answer: 42 cm².

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