Areas Related to Circles Class 10: Notes and Questions

This guide covers Areas Related to Circles Class 10, chapter 11 of NCERT Class 10 Mathematics. You will learn how to find the area of a sector, the length of an arc and the area of a segment, and how to handle major regions and real-life problems like clock hands and wipers. Important questions with full working and common mistakes are at the end.

Quick recap of the circle

• Circumference = 2πr (a length, in cm or m).
• Area = πr² (in square units, such as cm² or m²).
• Take π = 22/7 unless told otherwise. Sometimes 3.14 is used.

Example: if the circumference is 22 cm, then 22 = 2 × (22/7) × r = (44/7)r, so r = (22 × 7)/44 = 3.5 cm.

Sectors and segments

• A sector is the region between two radii and the arc joining them, like a slice of pizza. The angle between the radii at the centre is the sector angle θ.
• A segment is the region between a chord and its arc.

Each comes in a minor (smaller) and major (larger) version. If a question just says "sector" or "segment", it means the minor one. The major sector angle is 360° − θ; for example, if the minor sector angle is 60°, the major sector angle is 300°.

Area of a sector and length of an arc

A full circle is a sector of 360° with area πr². So a 1° sector has area πr²/360, and by the unitary method:

Area of sector = (θ/360) × πr²

In the same way, the arc is that fraction of the circumference:

Length of arc l = (θ/360) × 2πr

Example (sector): r = 4 cm, θ = 30°, π = 3.14. Area = (30/360) × 3.14 × 16 = (1/12) × 50.24 ≈ 4.19 cm².

Example (arc): r = 21 cm, θ = 60°. l = (1/6) × 2 × (22/7) × 21 = (1/6) × 132 = 22 cm.

Area of a segment

A sector contains the segment and the triangle formed by the two radii and the chord. So:

Minor segment = Minor sector − Triangle OAB

To find the triangle's area, drop a perpendicular from the centre to the chord. It splits the isosceles triangle into two right triangles, so you can find half the chord and the height using sin and cos of θ/2. Then use (1/2) × base × height.

Do not subtract the triangle from a major sector to get a major segment. Always use circle area minus minor segment.

Worked example: a 120° segment

Radius 21 cm, central angle 120°.

1. Sector area = (120/360) × (22/7) × 21 × 21 = 462 cm².
2. Draw OM ⊥ AB. Then ∠AOM = 60°.
3. Height OM = 21 × cos 60° = 21 × 1/2 = 10.5 cm.
4. Half-chord AM = 21 × sin 60° = 21√3/2, so chord AB = 21√3 cm.
5. Triangle area = (1/2) × 21√3 × 10.5 = 441√3/4 cm².
6. Segment area = 462 − 441√3/4 cm².

In general, for θ = 120° the chord is r√3, the height is r/2, the triangle is r²√3/4 and the sector is πr²/3. So

Minor segment (120°) = r²(π/3 − √3/4)

For r = 6 cm: sector = 12π cm², triangle = (1/2)(6√3)(3) = 9√3 cm², so the segment is 12π − 9√3 cm², about 22.11 cm².

Real-life uses

• Clock hands: the minute hand turns 360° in 60 minutes, which is 6° per minute. In 5 minutes it sweeps 30°, so the area swept is a 30° sector with radius equal to the hand's length.
• Wipers and lighthouse beams: the area covered is a sector. A lighthouse beam with range 16.5 km sweeping 80° covers (80/360) × π × (16.5)².
• Designs: for a pattern made of several equal sectors or segments, find the area of one piece and multiply by the number of pieces.

Example: a minute hand 14 cm long in 5 minutes sweeps (30/360) × (22/7) × 14 × 14 = (1/12) × 616 ≈ 51.33 cm².

Remember this

• Sector area = (θ/360) × πr². Arc length = (θ/360) × 2πr.
• Minor segment = sector − triangle.
• Major sector: use 360° − θ, or subtract from πr².
• Major segment = πr² − minor segment.
• Minute hand: 6° per minute.
• Areas are in square units; arcs and radii are in linear units.

Important questions with answers

1. Find the area of a sector of radius 6 cm and angle 60°. (60/360) × (22/7) × 36 = (1/6) × (22/7) × 36 = 132/7 ≈ 18.86 cm².

2. The radius is 7 cm and an arc is 11 cm long. Find the angle of the sector. 11 = (θ/360) × 2 × (22/7) × 7 = (θ/360) × 44. So θ/360 = 1/4 and θ = 90°.

3. A circle has radius 7 cm. Find the area of the major sector when the minor sector angle is 60°. Circle area = (22/7) × 49 = 154 cm². Major sector = (300/360) × 154 = (5/6) × 154 ≈ 128.33 cm².

4. Find the area of the minor segment of a circle of radius 14 cm when the angle at the centre is 90°. Sector = (90/360) × (22/7) × 196 = 154 cm². The two radii are perpendicular, so the triangle has base 14 cm and height 14 cm: area = (1/2) × 14 × 14 = 98 cm². Segment = 154 − 98 = 56 cm².

5. Find the area of the segment of a circle of radius 21 cm with central angle 120°. 462 − 441√3/4 cm², as worked out above.

6. The area of a circle is 100 cm² and a minor sector is 20 cm². Find the major sector. 100 − 20 = 80 cm².

7. How many degrees does a minute hand sweep in 10 minutes? 10 × 6° = 60°.

8. Why do we subtract the triangle from the sector to find a segment? The sector is made of the segment plus the triangle formed by the radii and chord. Removing the triangle leaves only the segment.

Common mistakes to avoid

• Using 2πr instead of πr² (or the other way round) in the sector and arc formulas.
• Forgetting square units for area.
• Subtracting the triangle from a major sector to find a major segment.
• Using the full angle θ instead of θ/2 when finding the triangle's height and half-chord.
• Answering for the minor part when the question asks for the major part.

Write the formula first and substitute slowly, and study this chapter with Joy for more sector and segment practice.