Circles Class 10 ICSE — A Das Gupta Chapter 23

Written by Greya LakshmiPublished on 28 September 2026

Circles Class 10 ICSE is covered in A Das Gupta’s Mathematics book, Chapter 23, on the circumference and area of a circle. The chapter sets out the formulas for the circle, the semi-circle and the circular ring, then moves to sectors and segments. Three supporting results close the chapter: wheel rotations, clock-hand angles, and the incircle and circumcircle of an equilateral triangle. The working rule throughout is to take $\pi = \frac{22}{7}$ unless a question says otherwise.

Download Circumference and Area of a Circle Class 10 ICSE Mathematics Chapter 23 PDF

Download the Circumference and Area of a Circle Class 10 ICSE Mathematics chapter PDF from the A. Das Gupta textbook to revise offline or print for practice. It is the full chapter as printed in the book — 17 pages, 2 sections, 47 questions.

What is inside Chapter 23

17 pages · 2 sections · 47 exercise questions

Section	Title
1	Some Important Formulae
2	Results on Sectors and Segments

EXERCISE 23: 47 questions

1 Some Important Formulae

For a circle of radius $r$ , the circumference is the distance once around the boundary, and the area is the region inside it. The chapter gives the circumference as $2\pi r$ units, or $\pi d$ units where $d$ is the diameter. It gives the area as $\pi r^2$ square units. The circumference formula works because $\pi$ is the ratio of a circle’s circumference to its diameter.

A semi-circle is half a circle, so its area is $\frac{1}{2}\pi r^2$ . Its perimeter has two parts. The curved edge is $\pi r$ , and the straight diameter adds $2r$ . The full perimeter is therefore $\pi r + 2r$ . A common mistake is to write only $\pi r$ and drop the diameter term. Split the working as $\frac{1}{2}(2\pi r) + 2r$ to keep the diameter term visible.

A circular ring is the region between two concentric circles. If $R$ is the outer radius and $r$ the inner radius, the ring’s area is the outer circle’s area minus the inner circle’s area: $\pi R^2 - \pi r^2 = \pi(R^2 - r^2)$ . Factoring as $(R + r)(R - r)$ speeds up the arithmetic when both radii are simple numbers.

In this ICSE chapter, the default value is $\pi = \frac{22}{7}$ . A question that wants a decimal states it, such as ‘take $\pi = 3.14$ ‘. Rounding to the nearest whole unit happens only when the question asks for an approximate answer.

Worked pattern: a circle whose circumference exceeds its diameter by 44 cm. Let the radius be $r$ . Then $2\pi r - 2r = 44$ . With $\pi = \frac{22}{7}$ , this gives $\frac{30}{7}r = 44$ , so $r = \frac{154}{15} \approx 10.3$ cm. The circumference is $2\pi r \approx 64.5$ cm and the area is $\pi r^2 \approx 331.3$ cm².

Subtract the diameter, roughly 20.5 cm, from the circumference to confirm the 44 cm difference.

This chapter sits in the Mensuration unit of the book. The full Class 10 Maths book list shows the chapters that surround it.

2 Results on Sectors and Segments

Suppose an arc ACB of a circle of radius $r$ subtends an angle $\theta^\circ$ at the centre O. The arc length is that same fraction of the full circumference: $\frac{2\pi r\theta}{360}$ . The sector is the region bounded by the arc and the two radii OA and OB. Its area is the same fraction of the full circle: $\frac{\pi r^2\theta}{360}$ . This also equals $\frac{1}{2} \times r \times \text{arc length}$ .

That form mirrors the triangle rule $\frac{1}{2} \times \text{base} \times \text{height}$ , with the arc as the base and the radius as the height.

The sector’s perimeter adds the two radii to the arc: $\frac{2\pi r\theta}{360} + 2r$ . The segment is the region between the arc and the chord AB. Its area subtracts the triangle OAB from the sector: $\frac{\pi r^2\theta}{360} - \frac{1}{2}r^2\sin\theta$ . The triangle’s two equal sides are radii, so its area is $\frac{1}{2}r^2\sin\theta$ . The segment’s perimeter is the arc plus the chord. The major segment’s area is the full circle’s area minus the minor segment’s area.

The sine term is the trigonometric ratio used in the previous chapter, Heights and Distances.

Worked pattern: a sector of radius 21 cm with central angle 120°. The arc length is $2 \times \frac{22}{7} \times 21 \times \frac{120}{360} = 44$ cm. The sector area is $\frac{22}{7} \times 21^2 \times \frac{120}{360} = 462$ cm², which matches $\frac{1}{2} \times 21 \times 44$ . The perimeter is $44 + 42 = 86$ cm.

The minor segment’s area is about $462 - 191 = 271$ cm², since $\frac{1}{2} \times 21^2 \times \sin 120^\circ \approx 191$ .

The table lines up the formulas for the four figures in this chapter. In the sector and segment rows, $\theta$ is the central angle in degrees.

Figure	Area	Perimeter or boundary
Circle, radius $r$	$\pi r^2$	$2\pi r = \pi d$
Semi-circle, radius $r$	$\frac{1}{2}\pi r^2$	$\pi r + 2r$
Sector, radius $r$ , angle $\theta$	$\frac{\pi r^2\theta}{360}$	$\frac{2\pi r\theta}{360} + 2r$
Minor segment, radius $r$ , angle $\theta$	$\frac{\pi r^2\theta}{360} - \frac{1}{2}r^2\sin\theta$	$\text{arc } ACB + \text{chord } AB$

Notice the pattern: every sector and segment formula multiplies a full-circle quantity by $\frac{\theta}{360}$ . The segment formula then subtracts the triangle that the chord cuts off.

3. Rotations Made by a Wheel

A wheel covers exactly one circumference in one revolution. The distance moved in a single revolution is therefore $2\pi r$ . The number of rotations in a given time equals the distance moved in that time divided by the circumference: rotations $= \frac{\text{distance}}{\text{circumference}}$ .

Worked pattern: a wheel of diameter 70 cm has circumference $\frac{22}{7} \times 70 = 220$ cm. Over 1.1 km, or 110000 cm, it makes $\frac{110000}{220} = 500$ rotations. Convert kilometres to centimetres before dividing.

4. Angle Described by Clock Hands

Both hands turn uniformly around the clock face. The minute hand sweeps $360^\circ$ in 60 minutes, so it sweeps $30^\circ$ in 5 minutes. The hour hand sweeps $360^\circ$ in 12 hours, so it sweeps $30^\circ$ in 1 hour. These two rates are the building blocks for clock-angle questions. For example, in 20 minutes the minute hand sweeps $\frac{360}{60} \times 20 = 120^\circ$ .

5. Equilateral Triangle Results

For an equilateral triangle of side $a$ , the chapter records the height as $h = \frac{\sqrt{3}}{2}a$ and the area as $\frac{\sqrt{3}}{4}a^2$ . The incircle radius is one-third of the height, $r = \frac{a}{2\sqrt{3}}$ , and the circumcircle radius is two-thirds, $R = \frac{a}{\sqrt{3}}$ . The incircle and circumcircle share the centroid as their centre, which fixes the one-third and two-thirds split.

Worked pattern: with $a = 12$ cm, the height is $6\sqrt{3} \approx 10.4$ cm and the area is $36\sqrt{3} \approx 62.4$ cm². The incircle radius is $2\sqrt{3} \approx 3.5$ cm and the circumcircle radius is $4\sqrt{3} \approx 6.9$ cm.

What to practise from Circles Class 10 ICSE

Exercise 23 closes the chapter with 47 questions. The list below groups them by the skill each block tests, so you can aim practice at a weak area. Reverse problems and cost-per-square-metre field questions are the two patterns that recur in CISCE board papers; the grouping shows where each appears.

Cutting circular discs from a rectangular sheet (Q1).
Direct circumference and area from a given radius or diameter (Q2–Q4).
Reverse problems, where the circumference or area is given and the radius is found (Q5–Q7, Q10).
Semi-circular plates: area, perimeter and radius from the perimeter (Q8–Q9).
A wire bent into a square, triangle or circle keeps one perimeter and changes the enclosed area (Q11–Q12).
Cost of levelling and fencing a circular field at a given rate (Q13–Q15).
Shaded regions formed by two perpendicular diameters (Q16–Q17).
Two circles touching internally or externally, and concentric rings (Q18–Q25).
Circles circumscribing or inscribed in an equilateral triangle (Q26–Q27).
Combined figures using squares, rectangles, a trapezium, quadrants and semi-circles (Q28–Q42).
Wheel rotations, distance and speed (Q43–Q46).

Several questions in the exercise carry board-year tags. The official CISCE website publishes the current Class 10 Mathematics syllabus for reference.

Circles Class 10 ICSE frequently asked questions

What is the difference between circumference and area of a circle in circles class 10 ICSE?

Circumference is the length once around the circle, $2\pi r$ , and is measured in units of length. Area is the region the circle encloses, $\pi r^2$ , and is measured in square units. When the diameter is given, halve it before using either formula.

How do I find the perimeter of a semi-circle without forgetting the diameter?

Use $\pi r + 2r$ . The curved part contributes $\pi r$ and the straight diameter contributes $2r$ . A reliable check is to split the working as $\frac{1}{2}(2\pi r) + 2r$ , which keeps the diameter term visible.

What is the formula for the area of a minor segment of a circle?

Subtract the triangle from the sector: $\frac{\pi r^2\theta}{360} - \frac{1}{2}r^2\sin\theta$ . The first term is the sector area. The second term is the isosceles triangle formed by the two radii and the chord.

How do I calculate the number of rotations a wheel makes in circles class 10 ICSE?

Divide the distance the wheel travels by its circumference. One revolution covers exactly one circumference, so rotations $= \frac{\text{distance}}{\text{circumference}}$ . Convert all units to the same measure before dividing.

When should I use pi = 22/7 and when pi = 3.14?

Use $\pi = \frac{22}{7}$ unless the question states another value. Questions that want 3.14 or 3.142 say so in the question itself. Round the final answer only when the question asks for an approximate value.

The syllabus this chapter follows is set by the CISCE board.

Reference: ICSE Class 10 Mathematics textbook material, chapter 23: Circumference and Area of a Circle — CISCE-aligned.

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