Areas Related to Circles
CBSE Class 10 · Mathematics · Notes, formulas and practice questions
Areas Related to Circles is the Class 10 chapter that uses the circumference and area of a circle to find arc lengths, sector areas, segment areas and the areas of composite figures formed with squares, rectangles and triangles.
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The chapter rests on two basic results: C = 2πr and A = πr². From these, arc length and sector area are obtained by multiplying the whole circumference or area by the fraction of the central angle out of 360°. The same proportion turns a whole circle into a semicircle for 180°, a quadrant for 90°, or any sector whose boundary consists of two radii and an arc. Once this fraction idea is clear, most part-of-a-circle questions become straightforward applications.
A sector and a segment are closely related but have different boundaries. A sector is bounded by two radii and an arc, so its perimeter includes two straight sides. A segment is bounded by a chord and its arc. To find the area of a minor segment, subtract the area of the triangle formed by the two radii and the chord from the area of the corresponding sector. The major sector or major segment is the complement of the minor part inside the circle.
Composite-figure problems ask for the area of a shaded region or the remaining area after a circle, sector or ring is placed inside a square, rectangle or triangle. Begin by identifying which known length is the radius or diameter of each circle. For example, a circle inside a square that just touches all four sides has diameter equal to the side of the square. The required area is usually the difference of two areas, or the sum of simple areas that touch one another.
Numerical errors most often come from changing units and from using the wrong part of the circle. A wheel problem must compare the total distance with the circumference in the same unit, and a segment problem must use the central angle in degrees inside the sine ratio. If the question says take π = 22/7, substitute that value; otherwise leave the answer in terms of π. Drawing the figure and labelling every radius before calculating is the best way to avoid mistakes.
Key terms
- Circle and radius
- A circle is the set of points at a fixed distance from a fixed point called the centre. The fixed distance is the radius, written r. The diameter is twice the radius and is the longest chord of the circle.
- Circumference
- The circumference is the length of the complete circular boundary. It is the perimeter of the circle, equal to 2πr, where r is the radius.
- Arc of a circle
- An arc is any connected part of the circumference. If a chord divides the circle into two arcs, the shorter arc is the minor arc and the longer arc is the major arc.
- Central angle
- The central angle is the angle between two radii drawn to the endpoints of an arc or sector. The whole turn around the centre is 360°, so a central angle θ names the fraction θ/360 of the circle that is being used.
- Sector
- A sector is the region enclosed by two radii and the arc intercepted between them. If the sector angle is 180° it is a semicircle, and if it is 90° it is a quadrant.
- Segment
- A segment is the region enclosed by a chord and the arc cut off by that chord. Its boundary has one chord and one arc, not two radii.
- Minor and major parts
- For a chord that is not a diameter, one side of the chord encloses a smaller sector and segment; these are minor. The larger parts opposite them are major, and each major area equals the area of the whole circle minus the corresponding minor area.
- Concentric circles
- Concentric circles share the same centre but have different radii. The region between two concentric circles is a circular ring, and its area is the area of the outer circle minus the area of the inner circle.
Formula sheet
| What | Formula | Notes |
|---|---|---|
| Circumference of a circle | C = 2πr | r is the radius; C is the length of the circular boundary. |
| Area of a circle | A = πr² | r is the radius; A is the region enclosed by the circle. |
| Length of an arc | l = (θ/360) × 2πr | θ is the central angle in degrees, r is the radius. When θ = 360, the arc is the full circumference. |
| Area of a sector | A = (θ/360) × πr² | θ is the sector angle, also called the central angle, measured in degrees; r is the radius. |
| Perimeter of a sector | P = 2r + (θ/360) × 2πr | P adds the two radii and the curved arc length. θ is the sector angle in degrees and r is the radius. |
| Area of a minor segment | A = (θ/360) × πr² − ½r² sin θ | θ is the smaller central angle, in degrees, between the two radii to the endpoints of a chord; r is the radius. The second term is the area of the triangle formed by the chord and the two radii. |
| Area of a major sector | A = πr² − (θ/360) × πr² | r is the radius and θ is the minor sector angle in degrees. The same subtraction idea works for a major segment: circle area minus the minor segment area. |
| Area of a circular ring | A = π(R² − r²) | R is the outer radius and r is the inner radius of two concentric circles. |
Practice questions with answers
1. The circumference of a circle is 44 cm. Taking π = 22/7, find the area of the circle.
Let the radius be r. From C = 2πr, r = 44/(2 × 22/7) = 7 cm. Then A = πr² = (22/7) × 7² = 154 cm².
2. A wheel has diameter 70 cm. How many complete revolutions does it make in travelling 1.1 km? Use π = 22/7.
Radius = 70/2 = 35 cm, so one revolution covers 2πr = 2 × 22/7 × 35 = 220 cm = 2.2 m. The distance travelled is 1.1 km = 1100 m. Number of revolutions = 1100/2.2 = 500.
3. A sector of radius 21 cm has central angle 60°. Find the length of its arc. Take π = 22/7.
Arc length = (θ/360) × 2πr = (60/360) × 2 × 22/7 × 21 = (1/6) × 132 = 22 cm.
4. Find the area of a quadrant of a circle of radius 7 cm. Use π = 22/7.
A quadrant is a sector with central angle 90°. Area = (90/360) × (22/7) × 7² = (1/4) × 154 = 38.5 cm².
5. A chord of a circle of radius 7 cm subtends a right angle at the centre. Find the area of the minor segment. Take π = 22/7.
The minor segment area = sector area − triangle area. Sector area = (90/360) × (22/7) × 7² = 38.5 cm², and the triangle formed by the two radii and the chord is right isosceles with area ½ × 7 × 7 = 24.5 cm². Therefore the segment area = 38.5 − 24.5 = 14 cm².
6. A circular wire of radius 42 cm is bent into a square. Find the side of the square. Use π = 22/7.
The length of the wire is the circumference of the circle, 2πr = 2 × 22/7 × 42 = 264 cm. This becomes the perimeter of the square, so the side of the square = 264/4 = 66 cm.
7. A circular plate of radius 7 cm is cut out from a square sheet whose side is 14 cm. What area of the sheet remains? Use π = 22/7.
Square area = 14² = 196 cm². Circle area = (22/7) × 7² = 154 cm². Remaining area = 196 − 154 = 42 cm².
8. Find the area of the region enclosed between two concentric circles with radii 7 cm and 14 cm. Use π = 22/7.
This region is a circular ring. Its area = π(R² − r²) = (22/7)(14² − 7²) = (22/7) × (196 − 49) = (22/7) × 147 = 462 cm².
9. The minute hand of a clock is 7 cm long. Find the area swept by it in 20 minutes. Use π = 22/7.
The minute hand turns 6° every minute, so in 20 minutes it turns 120°. The swept area is a sector of radius 7 cm and angle 120°, so area = (120/360) × (22/7) × 7² = (1/3) × 154 = 154/3 cm².
10. What is meant by a minor segment and a major segment of a circle?
A chord that is not a diameter divides the circular region into two parts. The smaller part is the minor segment, and the larger part is the major segment. The area of the major segment equals the area of the whole circle minus the area of the minor segment.
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