Areas Related to Circles — Class 10 Mathematics
1. Circle — Quick Recap
| Measure | Formula |
|---|---|
| Circumference (perimeter) | 2πr |
| Area | πr² |
| Diameter | d = 2r |
Where π = 22/7 (approximate) or 3.14159... . 'Use π = 22/7 for exam problems unless stated otherwise. This simplifies calculations when radius is a multiple of 7.'
2. Sector of a Circle
A SECTOR is the region between TWO RADII and the included ARC. It looks like a 'pizza slice.' Minor sector: The SMALLER sector (central angle < 180°). Major sector: The LARGER sector (central angle > 180°).
Area of a Sector
Area = (θ/360°) × πr², where θ = central angle in degrees.
Derivation: The entire circle (360°) has area πr². A sector with angle θ takes (θ/360°) of the whole circle. So its area = (θ/360°) × πr².
Length of an Arc
Arc length = (θ/360°) × 2πr. The arc is a FRACTION of the circumference, proportional to the angle.
Worked Examples
Example 1 — Sector Area: Radius = 14 cm, central angle = 60°. Find the area of the sector. (Use π = 22/7). Area = (60/360) × (22/7) × 14 × 14 = (1/6) × 22 × 2 × 14 = (1/6) × 616 = 102.67 cm².
Example 2 — Arc Length: Same circle. Arc length = (60/360) × 2 × (22/7) × 14 = (1/6) × 88 = 14.67 cm.
Example 3 — Sector from Arc: An arc of length 22 cm subtends 60° at the centre. Find the radius. 22 = (60/360) × 2πr = (1/6) × 2 × (22/7) × r = (22/21)r. r = 22 × 21/22 = 21 cm.
3. Segment of a Circle
A SEGMENT is the region between a CHORD and the corresponding ARC. Minor segment: The SMALLER region (cut by a chord, arc is minor). Major segment: The LARGER region.
Area of a Segment
Area of minor segment = Area of sector − Area of triangle (formed by the two radii and the chord).
Area = (θ/360°)×πr² − (1/2)r²×sin θ. For the common case of θ=90°: Area = (1/4)πr² − (1/2)r² = r²(π/4 − 1/2).
Worked Example
Radius = 10 cm, central angle = 90°. Find area of minor segment. Area of sector = (90/360)×π×100 = (1/4)×3.14×100 = 78.5 cm². Area of triangle = ½×10×10×sin90° = 50 cm² (right isosceles triangle). Area of segment = 78.5 − 50 = 28.5 cm².
Alternate formula for 60°: Area of sector = (60/360)πr² = πr²/6. Triangle is EQUILATERAL (since both radii and chord are equal when θ=60°). Area of equilateral triangle = (√3/4)r². Segment area = πr²/6 − (√3/4)r².
4. Areas of Combinations of Plane Figures
Many exam problems involve SHADED REGIONS that are combinations of circles, sectors, triangles, rectangles, etc. Strategy: 1. Identify the individual shapes. 2. Find the area of each. 3. ADD areas of non-overlapping parts. SUBTRACT overlapping parts. 4. The final expression gives the required area.
Worked Examples — Combinations
Example 1 — Circle Inscribed in Square: A circle of radius 7 cm is inscribed in a square. Find the area of the remaining part (square − circle). Side of square = 2r = 14 cm. Area of square = 196 cm². Area of circle = π×49 = 154 cm² (π=22/7). Remaining = 196−154 = 42 cm².
Example 2 — Four Circles in a Square: Four equal circles of radius 7 cm are placed touching each other in a square. Find the shaded area between them. Each circle's area = 154 cm². Total area of 4 circles = 616 cm². Side of enclosing square = 4r = 28 cm. Area of square = 784 cm². But circles overlap with square edges — actually, in the square of side 2r×2 = 28 cm, four circles of radius 7 cm each take up 4 × ¼πr² at the corners... Actually: The circles are placed such that each touches two sides and two other circles. Area left = Area of square − Area of 4 quarter-circles = (14+14)² − π(7)² = 28² − 154 = 784−154 = 630 cm². Wait, that's not right. Let me reconsider.
Circles of radius 7 arranged in a 2×2 grid within a square of side 28. Each circle touches adjacent circles. The space between the four circles at the centre = a curved diamond shape. Area = Square − 4×(quarter circles) = 28² − 4×(¼π×7²) = 784 − 154 = 630 cm²... but this gives area outside the circles, not between them. Area BETWEEN circles = (side² − 4 full circles)/4... hmm.
Actually: The region between four touching circles = square of side 14 (joining the centres) − one full circle = 14² − 154 = 196−154 = 42 cm². 'When four equal circles are arranged touching each other, the area of the central gap = (2r)² − πr².'
5. Key Formulas Summary
| Quantity | Formula |
|---|---|
| Circumference | 2πr |
| Area of circle | πr² |
| Arc length (angle θ°) | (θ/360) × 2πr |
| Area of sector (angle θ°) | (θ/360) × πr² |
| Area of minor segment | Area(sector) − Area(triangle) |
| Perimeter of sector | 2r + arc length |
6. Common Mistakes
- Using diameter instead of radius: All formulas use r. If d is given, convert to r = d/2 FIRST.
- Arc length vs Sector area: Arc is a LENGTH (cm). Sector is an AREA (cm²). Don't mix units.
- Forgetting 2r in sector perimeter: Perimeter of a sector = arc + 2 radii. Not just the arc.
- Segment area sign: The triangle area is ALWAYS subtracted from the sector area for minor segment. Major segment = area of circle − minor segment.
7. AP Exam Focus
| Topic | Marks |
|---|---|
| Area and circumference of circle | 2-3 |
| Sector area and arc length | 4-5 |
| Segment area | 3-4 |
| Combined figure problems | 4-5 |
Quick Self-Test
- Circumference of circle of radius 14 cm? (Answer: 2πr = 2×(22/7)×14 = 88 cm.)
- Area of sector with radius 21 cm, angle 60°? (Answer: (1/6)×(22/7)×441 = 231 cm².)
- Arc length subtending 90° at centre, radius 7 cm? (Answer: (1/4)×2×(22/7)×7 = 11 cm.)
- Perimeter of semicircle of radius 7 cm? (Answer: πr + 2r = 22 + 14 = 36 cm.)
- Area of a ring (annulus) with outer radius 10 cm and inner radius 6 cm? (Answer: π(10²−6²) = π(100−36) = 64π ≈ 201.14 cm².)
More Combination Problems
Example — Square with Four Corner Circles: A square of side 14 cm has a circle inscribed within (touching all sides). Find the area of the four corner spaces outside the circle. Side = 14 cm. Radius of inscribed circle = 7 cm. Area of square = 196 cm². Area of circle = π×49 = 154 cm². Area of 4 corners = 196−154 = 42 cm².
Example — Two Concentric Circles: Find the area of the circular path (track) between two concentric circles of radii 21 m and 14 m. Track width = 21−14 = 7 m. Area = π(21²−14²) = π(441−196) = 245π m² ≈ 770 m² (π=22/7).
Example — Flower Design: Six equal circles of radius 7 cm are arranged around a central circle of the same radius, all touching. Find the area of the hexagon formed by joining centres and the space between. The centres form a regular hexagon of side 14 cm (2r). Area of hexagon = (3√3/2)×14² = 509.22 cm². This is beyond the Class 10 scope but shows how circles relate to polygons.
Example — Clock Face: A clock has minute hand of length 14 cm. Find the area swept by the minute hand in 5 minutes. In 60 min, hand sweeps 360°. In 5 min: (5/60)×360° = 30°. Area swept = (30/360)×π×14² = (1/12)×(22/7)×196 = (1/12)×616 = 51.33 cm².
Area of Segment — Common Angles
| Angle θ | Sector Area | Triangle Area | Segment Area |
|---|---|---|---|
| 60° | πr²/6 | (√3/4)r² | r²(π/6 − √3/4) |
| 90° | πr²/4 | r²/2 | r²(π/4 − 1/2) |
| 120° | πr²/3 | (√3/4)r² | r²(π/3 − √3/4) |
AP Exam Tips for Areas Related to Circles
- Always state whether you're using π = 3.14 or π = 22/7 BEFORE starting calculations.
- If radius is a multiple of 7, use π = 22/7 for exact answers. Otherwise, use 3.14 or leave in terms of π.
- For shaded region problems: number the regions, write area of each, add/subtract as needed.
- The MOST common 5-mark question: area of a segment + area of a triangle → combined figure.
Quick Test — Challenge Problems
- A horse is tied to a peg at one corner of a 20m × 15m rectangular field with a 7m rope. Find the area the horse can graze. (Answer: ¼ × π × 7² = ¼×154 = 38.5 m² — only a quarter circle at the corner.)
- A bicycle wheel of diameter 70 cm makes 5000 revolutions. Distance covered? (Answer: Circumference = πd = 220 cm. Distance = 220 × 5000 = 1,100,000 cm = 11 km.)
