Big picture
Areas related to circles deals with circular gardens, sprinkler patterns, tracks and curved tiles. A single diagram can ask for a curved boundary, an enclosed region or a cost. These are different quantities. A successful solution begins by deciding what the question measures, rather than choosing the first familiar formula.
Draw the circle, mark its centre, radius and central angle, and shade the required region. Separate length from area, and keep π exact until the question asks for a particular approximation. Combining sectors with triangles and squares is the main reasoning skill in this chapter.
Learning goals
Distinguish arcs, chords, sectors and segments.
Calculate circular lengths, areas and complete boundaries.
Find segments by subtracting appropriate triangles.
Solve shaded-region and concentric-path problems.
Apply circle measurements to costs, wheels and clocks.
Check answers using units and whole-circle bounds.
Core ideas
1. Whole-circle measurements
A radius joins the centre to the circle. A diameter passes through the centre and equals 2r. The circumference is the complete boundary, C = 2πr = πd. The enclosed area is A = πr². Length uses cm or m, whereas area uses cm² or m².
Worked example. A mat has diameter 18 cm. Radius = 18 ÷ 2 = 9 cm. Circumference = 2 × π × 9 = 18π cm. Area = π × 9² = 81π cm². Incorrectly using 18 as the radius gives 324π cm², four times the correct area. Doubling radius doubles circumference but quadruples area because the radius is squared.
2. Arc length
An arc is part of the circumference; a chord is the straight segment joining two circle points. A central angle θ represents θ/360 of a complete turn. Therefore arc length = (θ/360) × 2πr. The minor and major arcs together form the full circumference.
Worked example. Radius = 12 cm and minor angle = 75°. Minor arc = 75/360 × 24π = 5π cm. Major angle = 360 − 75 = 285°. Major arc = 24π − 5π = 19π cm. Their sum, 24π cm, checks the calculation. Do not substitute the straight chord for a curved arc in perimeter questions.
3. Sector area and complete perimeter
A sector is bounded by two radii and their intercepted arc. Sector area = (θ/360) × πr². Its complete perimeter includes both radii: P = arc length + 2r. Area and arc use the same angular fraction but different whole-circle formulas.
Worked example. A sprinkler reaches 9 m and sweeps 80°. Area = 80/360 × 81π = 18π m². Arc = 80/360 × 18π = 4π m. Complete sector perimeter = 18 + 4π m. If the question asks only for the curved edge, the answer is 4π m, not the complete perimeter.
If arc length is l, combining the formulas gives sector area = ½rl. Here ½ × 9 × 4π = 18π m², confirming the earlier method. This version is useful when an arc is given without a central angle.
4. Special sectors and inverse problems
Shape | Area | Complete perimeter |
|---|---|---|
Semicircle | ½πr² | πr + 2r |
Quadrant | ¼πr² | ½πr + 2r |
General sector | (θ/360)πr² | (θ/360)2πr + 2r |
Worked example. A semicircular bed has radius 7 m. With π = 22/7, area = ½ × 22/7 × 49 = 77 m². The curved edge is 22 m and diameter 14 m, giving complete perimeter 36 m. Half of the circumference omits the straight diameter.
Inverse example. A sector of radius 15 cm has area 30π cm². Write (θ/360) × 225π = 30π. Cancel π and rearrange: θ = 30 × 360 ÷ 225 = 48°. Substitution checks the result. State the equation before rearranging so the role of the unknown angle remains clear.
5. Segments and triangle subtraction
A segment is bounded by a chord and its arc. Imagine centre O, radii OA and OB and chord AB. The curved cap between the chord and minor arc is the minor segment. For the usual minor angle up to 180°, minor segment area = sector area − triangle AOB area.
Right-angle example. Radius = 10 cm and central angle = 90°. Sector = ¼ × 100π = 25π cm². The triangle has perpendicular sides 10 cm, so area = ½ × 10 × 10 = 50 cm². Minor segment = 25π − 50 cm². Major segment = 100π − (25π − 50) = 75π + 50 cm².
60° example. Radius = 8 cm. The central triangle is equilateral, side 8 cm. Its altitude = √(64 − 16) = 4√3 cm. Triangle area = ½ × 8 × 4√3 = 16√3 cm². Sector = 64π/6 = 32π/3 cm². Segment = 32π/3 − 16√3 cm².
120° example. Radius = 6 cm. Bisect the angle and draw the perpendicular from centre to chord. Centre-to-chord distance = 6 cos 60° = 3 cm. Half-chord = 6 sin 60° = 3√3 cm. Triangle area = ½ × 6√3 × 3 = 9√3 cm². Sector = ⅓ × 36π = 12π cm². Segment = 12π − 9√3 cm².
6. Annuli, paths and costs
An annulus lies between concentric circles. Its area is π(R² − r²). For a path outside a garden, outer radius equals garden radius plus path width. The path itself is not a circle whose radius is its width.
Worked example. Garden radius = 10 m and external path width = 2 m. Outer radius = 12 m. Path area = π(144 − 100) = 44π m². At ₹30 per m², cost = ₹1,320π. With π = 3.14, area = 44 × 3.14 = 138.16 m², and cost = 138.16 × 30 = ₹4,144.80. Pricing by square metre requires area, not circumference.
7. Composite areas and exposed boundaries
Express a shaded region as a sum of non-overlapping pieces or a larger region minus smaller parts. A diagram's unshaded complement requires the opposite subtraction, so label what is actually asked before calculating.
Worked example. A circle is inscribed in a square of side 16 cm. Diameter = 16 cm, radius = 8 cm. Square area = 256 cm². Circle area = 64π cm². Total corner area = 256 − 64π cm². Symmetry divides it into four equal pieces, each 64 − 16π cm².
For one corner's perimeter, count two half-sides, totalling 16 cm, and a quadrant arc, 4π cm. Complete perimeter = 16 + 4π cm. Internal shared edges do not automatically belong to a composite figure's outside boundary. Trace the boundary separately from the area calculation.
8. Wheels, clocks and reasonableness
A wheel rolling without slipping travels one circumference per revolution. Worked example. Radius = 35 cm and π = 22/7. Circumference = 2 × 22/7 × 35 = 220 cm = 2.2 m. For 30 revolutions, distance = 30 × 2.2 = 66 m. Convert cm into m consistently.
A minute hand turns 360° in 60 minutes. In 20 minutes, angle = 360 × 20/60 = 120°. For length 6 cm, tip distance = 120/360 × 12π = 4π cm. Swept area = 120/360 × 36π = 12π cm². The tip traces an arc, while the hand sweeps a sector.
A sector smaller than a full circle must have smaller area than πr². A minor segment must have smaller area than its containing sector. Negative area usually means the subtraction order or triangular area is incorrect. Unit checks and bounds often reveal mistakes before a final answer is written.
A reliable approach to questions
Label the centre, radius, angle and required boundary or region.
Decide whether length, area, perimeter or cost is required.
Decompose the figure into circles, sectors and triangles.
Convert units and use the specified approximation consistently.
Show formula, substitution, simplification and final units.
Check magnitude and trace only exposed boundaries.
Connections that matter
Trigonometry gives heights and chord lengths in segment calculations.
Surface areas and volumes uses circular bases and curved surfaces.
Ratios explain the angle fraction shared by arcs and sectors.
Common mistakes
Using diameter as radius: divide it by two before squaring.
Giving only an arc for sector perimeter: add exposed radii.
Confusing sector and segment: identify radii versus chord boundaries.
Treating path width as radius: subtract the concentric-circle areas.
Revision checklist
I can distinguish arcs, sectors and segments.
I can find sector area and complete perimeter.
I can subtract the correct triangle for a segment.
I can solve paths, costs and shaded figures.
I can model wheels and clock hands using circular measures.
I can check units, bounds and exposed boundaries.