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1. Introduction

This chapter extends the study of circles from their basic properties to the computation of lengths and areas of circular parts. Students learn the formulas for the circumference and area of a circle, the length of an arc, the area of a sector, and the area of a segment. These quantities appear constantly in design, engineering, and everyday calculations involving wheels, discs, and circular paths.

The chapter begins with the familiar formulas for circumference and area of a circle, then generalises them to sectors and arcs through the concept of proportionality with the central angle. A sector is the region enclosed by two radii and the arc between them, while a segment is the region enclosed by an arc and its chord. The difference between these two regions is a common source of both marks and confusion.

In addition to these core quantities, the chapter covers practical applications such as the areas of circular rings, areas of combinations of plane figures, and problems involving roads and paths around circular parks. Careful reading of the problem and correct identification of the required region are essential for solving these questions correctly.

2. Circumference and Area of a Circle

For a circle of radius r: Circumference = 2 x pi x r Area = pi x r^2

The value of pi is approximately 22/7 or 3.14. In problems, the given value of pi should be used; if not specified, 22/7 is commonly used.

Example: For a circle of radius 7 cm, Circumference = 2 x (22/7) x 7 = 44 cm. Area = (22/7) x 49 = 154 cm^2.

Area of a Semicircle

Area = (1/2) x pi x r^2 Perimeter = pi x r + 2r (arc plus the diameter)

Circular Ring

If a ring has outer radius R and inner radius r, then: Area of ring = pi(R^2 - r^2)

These basic quantities are the building blocks of every problem in the chapter.

3. Length of an Arc

If a sector of a circle of radius r subtends an angle theta at the centre, the length of its arc is proportional to theta: Length of arc l = (theta/360) x 2 x pi x r

Example: Arc length for a sector of radius 14 cm with central angle 60 degrees: l = (60/360) x 2 x (22/7) x 14 = (1/6) x 88 = 44/3 cm.

The arc length formula is derived from the fact that a full circle subtends 360 degrees, so a sector of angle theta covers the fraction theta/360 of the whole circumference.

4. Area of a Sector

The area of a sector of a circle of radius r subtending an angle theta at the centre is: Area of sector = (theta/360) x pi x r^2

Alternative Forms

Area of sector = (1/2) x r^2 x theta (with theta in radians), but at Class 10 level the (theta/360) form is standard. Also: Area of sector = (1/2) x l x r, where l is the arc length.

Example: Area of a sector of radius 6 cm and angle 30 degrees: Area = (30/360) x pi x 36 = (1/12) x 36 pi = 3 pi cm^2.

The sector area formula is used for problems on slices of pie charts, fans, and other circular divisions.

5. Area of a Segment

A segment of a circle is the region between an arc and its chord. The area of a minor segment (angle theta at the centre) is: Area of segment = Area of sector - Area of the isosceles triangle formed by the two radii and the chord Area of segment = (theta/360) x pi x r^2 - (1/2) x r^2 x sin theta

Example: For r = 7 cm and theta = 60 degrees: Sector area = (60/360) x (22/7) x 49 = 77/6 cm^2. Triangle area = (1/2) x 49 x sin 60 = (1/2) x 49 x (sqrt(3)/2) = (49 sqrt(3))/4 cm^2. Segment area = sector area - triangle area.

The area of the major segment is the remaining part of the circle: Area of major segment = pi x r^2 - area of minor segment.

6. Areas of Combinations of Plane Figures

Many board questions involve figures made by combining sectors, triangles, squares, and semicircles. The strategy is:

  1. Identify the region whose area is required.
  2. Decompose it into standard figures (sectors, triangles, rectangles, semicircles).
  3. Compute each standard area separately.
  4. Add or subtract the parts as appropriate.

Common configurations include a square with semicircles drawn on its sides, a sector with an inscribed triangle, a circular ring, and a rectangle with semicircular ends. The key to these problems is a clear figure and correct identification of the difference between the regions.

Quick Revision Tables

Table 1: Core Formulas

Quantity Formula Remarks
Circumference 2 pi r Full circle
Area of circle pi r^2 Full circle
Area of semicircle (1/2) pi r^2 Half circle
Area of ring pi(R^2 - r^2) Outer radius R, inner r
Arc length (theta/360) x 2 pi r Sector angle theta
Sector area (theta/360) x pi r^2 Sector angle theta
Segment area Sector area - triangle area Minor segment

Table 2: Formulas with Sector Angle

Angle theta Fraction of circle Arc length Sector area
360 1 2 pi r pi r^2
180 1/2 pi r (1/2) pi r^2
90 1/4 (pi r)/2 (1/4) pi r^2
60 1/6 (pi r)/3 (1/6) pi r^2

Mind Map

graph TD A["Areas Related to Circles"] --> B["Basic Formulas"] A --> C["Sector"] A --> D["Segment"] A --> E["Combined Figures"] B --> B1["Circumference = 2 pi r"] B --> B2["Area = pi r^2"] C --> C1["Arc length = (theta/360) 2 pi r"] C --> C2["Sector area = (theta/360) pi r^2"] D --> D1["Segment = Sector - Triangle"] D --> D2["Major segment = circle - minor"] E --> E1["Add or subtract standard areas"]

Important Diagrams (SVG)

Diagram 1: Sector and Segment of a Circle

Sector and Segment of a Circle O Sector Segment Sector area = (theta/360) pi r^2; Arc length = (theta/360) 2 pi r Segment area = sector area - triangle area Golden Rule: A segment is a sector minus its triangle; the major segment is the rest of the circle.

Diagram 2: Circular Ring and its Area

Circular Ring (Annulus) O R (outer) r (inner) Area of ring = pi(R^2 - r^2) Example: R = 14, r = 7 gives area = pi(196 - 49) = 147 pi cm^2 Golden Rule: For a ring, subtract the inner circle area from the outer circle area, never the radii first.

Common Mistakes

  1. Using the formula (theta/360) x pi r^2 for arc length, or (theta/360) x 2 pi r for sector area; the two formulas must not be interchanged.
  2. Forgetting to take the central angle in degrees when the formula uses 360; angles given in other forms must be converted first.
  3. Confusing the minor and major segments and subtracting the sector from the circle incorrectly for the major segment.
  4. In the segment formula, forgetting to subtract the triangle area from the sector area.
  5. Using pi as 3.14 in one part and 22/7 in another without being consistent, leading to different answers.
  6. In combined figures, adding areas where the required region is a difference, such as the shaded region outside a square but inside a circle.
  7. Reporting the arc length and sector area in the wrong units, or mixing cm and cm^2 for area and length.

Exam Tips

  1. Write the formulas for arc length and sector area before substituting, clearly marking theta/360.
  2. Draw the figure and shade the required region; identify whether it is a sector, segment, or combination before computing.
  3. For segment problems, compute the sector area and triangle area separately, then subtract.
  4. Use consistent values of pi throughout a problem; state which value you are using.
  5. In ring problems, compute pi(R^2 - r^2) directly rather than finding the difference of the two circle areas separately.
  6. Remember area of major segment = area of circle - area of minor segment.
  7. Practise combined figures from previous years, as these are the most challenging but also the most rewarding questions in the chapter.

Conclusion

Areas related to circles bring together the formulas of circumference, area, arcs, sectors, and segments into a coherent toolkit for measuring circular regions. The proportionality of arc length and sector area with the central angle makes these calculations straightforward, while segment problems add the geometric flavour of subtracting triangles. Combined-figure problems then test the ability to decompose complex regions into standard shapes. This chapter is both practical and scoring, appearing regularly in board examinations in the form of short and long answer questions, and it lays the groundwork for mensuration in higher classes.