Areas Related to Circles

Areas Related to Circles

Areas Related to Circles carries 4 to 6 marks in the board paper (4 in the 2026-27 sample paper, 5 in both 2026 papers and 6 in 2025).

Expect one or two 1-mark MCQs (sector area, arc length, a wheel or a clock hand), often a 2- or 3-mark sum on a sector or a segment, and in recent papers a whole case study on sectors (a brooch, a circular field). There has been no 5-mark question from this chapter, so the case-based questions here are the long practice.

Where marks are usually lost:

• mixing up arc length $\frac{\theta}{360^\circ} \times 2\pi r$ with sector area $\frac{\theta}{360^\circ} \times \pi r^2$;
• leaving out the two radii (or the diameter) when finding a perimeter;
• taking the triangle in a segment as $\frac{1}{2}r^2$ when the angle is $60^\circ$ or $120^\circ$, not $90^\circ$;
• not converting km or m to cm before dividing, in wheel problems.

Revise in 5 Minutes

Formulas (angle $\theta$ in degrees, radius $r$)

Triangle $OAB$ in a segment (the only angles in the syllabus)

For $120^\circ$: drop $OM \perp AB$; then $OM = \frac{r}{2}$ and $AM = \frac{\sqrt{3}}{2} r$.

Handy facts

• Major sector angle $= 360^\circ - \theta$.
• Semicircle perimeter $= \pi r + 2r$; quadrant perimeter $= \frac{\pi r}{2} + 2r$.
• Minute hand turns $6^\circ$ a minute; hour hand turns $30^\circ$ an hour.
• Distance in one turn of a wheel = its circumference; revolutions = distance ÷ circumference (same units).
• Fencing cost given? Circumference = total cost ÷ rate per metre; then find $r$ and the area $\pi r^2$ (for a ploughing or levelling cost).
• Radius doubled: circumference $\times 2$, area $\times 4$.
• With $\pi = \frac{22}{7}$, radii 7, 14, 21, 28, 35 give clean answers: circumference 44, 88, 132, 176, 220; area 154, 616, 1386, 2464, 3850.

Traps

• Using $2\pi r$ (arc) where $\pi r^2$ (area) is needed, or the other way round.
• Leaving out the two radii from a sector's perimeter, or the chord from a segment's perimeter.
• Using $\frac{1}{2} r^2$ for the triangle when the angle is $60^\circ$ or $120^\circ$.
• Giving the radius when the question asks for the diameter.

How to use this page: try each question on paper first, then read the answer. The marks against each step show what an examiner looks for. The 1-mark MCQs and Assertion-Reason questions are in the quiz at the end, together with questions that test how well you understand the chapter; every quiz answer comes with its explanation.

Short Answer Questions (2 and 3 Marks)

Question 1 (2 marks)

The area of a sector of a circle is $154 \text{ cm}^2$ and its central angle is $40^\circ$. Find the radius of the circle. (Use $\pi = \frac{22}{7}$)

Answer.

1. $\frac{40^\circ}{360^\circ} \times \frac{22}{7} \times r^2 = 154$ — 1 mark
2. $r^2 = \frac{154 \times 9 \times 7}{22} = 441$, so $r = 21$ cm — 1 mark

Question 2 (2 marks)

Aarav cycles from his home in Pune to his school, 11 km away, in 25 minutes. The radius of each wheel of his bicycle is 35 cm. How many revolutions does each wheel make per minute? (Use $\pi = \frac{22}{7}$)

Answer.

1. Distance in one revolution $= 2 \times \frac{22}{7} \times 35 = 220$ cm; total distance $= 11 \text{ km} = 1100000$ cm — 1 mark
2. Revolutions $= \frac{1100000}{220} = 5000$ in 25 minutes, so $\frac{5000}{25} = 200$ revolutions per minute — 1 mark

Question 3 (2 marks)

Kavya was asked to find the area of a sector of a circle of radius 6 cm with central angle $150^\circ$, in terms of $\pi$. She wrote:

Area $= \frac{150^\circ}{360^\circ} \times 2\pi \times 6 = 5\pi \text{ cm}^2$.

Is her answer correct? If not, find the error and the correct area.

Answer.

1. Not correct: she used $2\pi r$, which gives the arc length ($5\pi$ cm), not the area — 1 mark
2. Area $= \frac{150^\circ}{360^\circ} \times \pi \times 6^2 = \frac{5}{12} \times 36\pi = 15\pi \text{ cm}^2$ — 1 mark

Question 4 (2 marks)

A sector of a circle of radius 9 cm has a central angle of $80^\circ$. Find the area of the sector and the length of the major arc, in terms of $\pi$.

Answer.

1. Area $= \frac{80^\circ}{360^\circ} \times \pi \times 81 = 18\pi \text{ cm}^2$ — 1 mark
2. Major arc: angle $360^\circ - 80^\circ = 280^\circ$, length $= \frac{280^\circ}{360^\circ} \times 2\pi \times 9 = 14\pi$ cm — 1 mark

Question 5 (2 marks)

Ravi says: "In a circle of radius 14 cm, a sector with central angle $60^\circ$ has an area of $\frac{308}{3} \text{ cm}^2$. If I keep the angle the same and double the radius, the area of the sector will also double." Check both of his statements. (Use $\pi = \frac{22}{7}$)

Answer.

Model answer:

The first statement is correct. Area of the sector $= \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{6} \times 616 = \frac{308}{3}$ cm², which is about 102.7 cm².

The second statement is wrong. The area of a sector is $\frac{\theta}{360^\circ} \times \pi r^2$, so it depends on the square of the radius. If the radius is doubled and the angle stays the same, the area becomes $2^2 = 4$ times. With radius 28 cm the area is $4 \times \frac{308}{3} = \frac{1232}{3}$ cm², not $2 \times \frac{308}{3}$.

(It is the arc length that doubles when the radius is doubled.)

Marking scheme:

1. Area $= \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = \frac{616}{6} = \frac{308}{3} \text{ cm}^2$; the first statement is correct — 1 mark
2. Area depends on $r^2$: with radius 28 cm the area is $4 \times \frac{308}{3} = \frac{1232}{3} \text{ cm}^2$, four times, not double; the second statement is wrong — 1 mark

Question 6 (3 marks)

A chord $AB$ of a circle with centre $O$ and radius 42 cm subtends an angle of $120^\circ$ at $O$. Find the area and the perimeter of the minor segment. (Use $\pi = \frac{22}{7}$ and $\sqrt{3} = 1.73$)

Answer.

Model answer:

Area of sector $OAB$ $= \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 42 \times 42 = \frac{1}{3} \times 5544 = 1848$ cm². Length of arc $AB$ $= \frac{1}{3} \times 2 \times \frac{22}{7} \times 42 = 88$ cm.

Draw $OM \perp AB$. It bisects $\angle AOB$ and $AB$, so $\angle AOM = 60^\circ$. In right triangle $OMA$: $OM = 42 \cos 60^\circ = 21$ cm and $AM = 42 \sin 60^\circ = 21\sqrt{3}$ cm. So $AB = 42\sqrt{3} = 42 \times 1.73 = 72.66$ cm.

Area of $\triangle OAB$ $= \frac{1}{2} \times AB \times OM = \frac{1}{2} \times 42\sqrt{3} \times 21 = 441\sqrt{3} = 441 \times 1.73 = 762.93$ cm².

Area of the minor segment $= 1848 - 762.93 = 1085.07$ cm².

Perimeter of the minor segment = arc $AB$ + chord $AB$ $= 88 + 72.66 = 160.66$ cm.

Marking scheme:

1. Sector $= \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 42 \times 42 = 1848 \text{ cm}^2$; arc $= \frac{1}{3} \times 2 \times \frac{22}{7} \times 42 = 88$ cm — 1 mark
2. $OM \perp AB$: $OM = 21$ cm, $AM = 21\sqrt{3}$ cm, so $AB = 42\sqrt{3} = 72.66$ cm and $\triangle OAB$ $= \frac{1}{2} \times 42\sqrt{3} \times 21 = 441\sqrt{3} = 762.93 \text{ cm}^2$ — 1 mark
3. Minor segment $= 1848 - 762.93 = 1085.07 \text{ cm}^2$; perimeter $= 88 + 72.66 = 160.66$ cm — 1 mark

Question 7 (3 marks)

A circular window in a school in Shillong has radius 18 cm. A chord $AB$ subtends an angle of $60^\circ$ at the centre $O$, and the part between the chord and the minor arc (shaded in the figure) is made of coloured glass. Find the perimeter and the area of the coloured glass piece, in terms of $\pi$ and $\sqrt{3}$.

Answer.

1. $OA = OB$ and $\angle AOB = 60^\circ$, so $\triangle OAB$ is equilateral and $AB = 18$ cm; arc $AB = \frac{60^\circ}{360^\circ} \times 2\pi \times 18 = 6\pi$ cm — 1 mark
2. Perimeter of the glass piece $= (6\pi + 18)$ cm — 1 mark
3. Area $= \frac{60^\circ}{360^\circ} \times \pi \times 18^2 - \frac{\sqrt{3}}{4} \times 18^2 = (54\pi - 81\sqrt{3}) \text{ cm}^2$ — 1 mark

Question 8 (3 marks)

A round pizza of diameter 28 cm is cut into 8 equal slices by cuts through its centre. Find (i) the area of the top of each slice, (ii) the length of the crust (the curved edge) of each slice, (iii) the perimeter of each slice. (Use $\pi = \frac{22}{7}$)

Answer.

1. Each slice is a sector of radius 14 cm and angle $\frac{360^\circ}{8} = 45^\circ$; area $= \frac{1}{8} \times \frac{22}{7} \times 14 \times 14 = 77 \text{ cm}^2$ — 1 mark
2. Crust $= \frac{1}{8} \times 2 \times \frac{22}{7} \times 14 = 11$ cm — 1 mark
3. Perimeter $= 11 + 14 + 14 = 39$ cm — 1 mark

Question 9 (3 marks)

The area of a sector of a circle is $462 \text{ cm}^2$ and the length of its arc is 44 cm. Find the radius of the circle, the central angle of the sector, and the perimeter of the corresponding major sector. (Use $\pi = \frac{22}{7}$)

Answer.

1. Area $= \frac{1}{2} l r$ with arc $l = 44$: $462 = \frac{1}{2} \times 44 \times r$, so $r = 21$ cm — 1 mark
2. Circumference $= 2 \times \frac{22}{7} \times 21 = 132$ cm; $\theta = \frac{44}{132} \times 360^\circ = 120^\circ$ — 1 mark
3. Major arc $= 132 - 44 = 88$ cm; perimeter of major sector $= 88 + 21 + 21 = 130$ cm — 1 mark

Question 10 (3 marks)

A farmer near Karnal fenced his circular field at ₹ 24 per metre and paid ₹ 5280 in all. Find the radius of the field, its area, and the cost of ploughing it at ₹ 0.50 per m². (Use $\pi = \frac{22}{7}$)

Answer.

1. Circumference $= \frac{5280}{24} = 220$ m; $2 \times \frac{22}{7} \times r = 220$, so $r = 35$ m — 1 mark
2. Area $= \frac{22}{7} \times 35 \times 35 = 3850 \text{ m}^2$ — 1 mark
3. Cost of ploughing $= 3850 \times 0.50 =$ ₹ 1925 — 1 mark

Long Answer and Case-Based Questions

Question 11 (4 marks)

Diya's family bought a round wall clock for their new flat in Bhopal. The face of the clock is a circle of radius 21 cm, and its minute hand is 10.5 cm long. The twelve hour marks are equally spaced along the edge of the face. (Use $\pi = \frac{22}{7}$)

(i) Through what angle does the minute hand turn from 4:10 p.m. to 4:50 p.m.? (1 mark)

Answer.

1. 40 minutes: $\frac{40}{60} \times 360^\circ = 240^\circ$ — 1 mark

(ii) Find the distance moved by the tip of the minute hand in that time. (1 mark)

Answer.

1. $\frac{240^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 10.5 = \frac{2}{3} \times 66 = 44$ cm — 1 mark

(iii) Find the area swept by the minute hand in that time. (2 marks)

Answer.

1. Angle $240^\circ$, radius 10.5 cm (the minute hand, not the face) — 1 mark
2. Area $= \frac{240^\circ}{360^\circ} \times \frac{22}{7} \times 10.5 \times 10.5 = \frac{2}{3} \times 346.5 = 231 \text{ cm}^2$ — 1 mark

OR

(iii) Lines drawn from the centre of the face to the twelve hour marks divide the face into 12 equal sectors. Find the area of one such sector and the length of its arc. (2 marks)

Answer.

1. Each sector has angle $30^\circ$ and radius 21 cm; area $= \frac{1}{12} \times \frac{22}{7} \times 21 \times 21 = \frac{1386}{12} = 115.5 \text{ cm}^2$ — 1 mark
2. Arc $= \frac{1}{12} \times 2 \times \frac{22}{7} \times 21 = \frac{132}{12} = 11$ cm — 1 mark

Question 12 (4 marks)

For a craft fair in Kutch, Hetal makes hand fans (pankhas). Each fan is a sector $OAB$ of a circle of radius 28 cm with $\angle AOB = 90^\circ$, fixed to a handle at $O$, as shown in the figure. (Use $\pi = \frac{22}{7}$)

(i) Find the length of the arc $AB$ of one fan. (1 mark)

Answer.

1. $\frac{90^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 28 = 44$ cm — 1 mark

(ii) Find the area of cloth used for one fan. (1 mark)

Answer.

1. $\frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 28 \times 28 = 616 \text{ cm}^2$ — 1 mark

(iii) A lace is stitched once along the whole boundary of the fan (the arc $AB$ and the radii $OA$ and $OB$). Find the length of lace needed for one fan and its cost at ₹ 15 per 10 cm. (2 marks)

Answer.

1. Lace $= 44 + 28 + 28 = 100$ cm — 1 mark
2. Cost $= \frac{100}{10} \times 15 =$ ₹ 150 — 1 mark

OR

(iii) Hetal also makes a small fan: a sector of radius 14 cm with the same angle of $90^\circ$. How much less cloth does the small fan need than the big one? (2 marks)

Answer.

1. Small fan $= \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 \text{ cm}^2$ — 1 mark
2. Less cloth $= 616 - 154 = 462 \text{ cm}^2$ (half the radius needs a quarter of the cloth) — 1 mark

Question 13 (4 marks)

A giant wheel at the Dussehra mela in Mysuru has a radius of 14 m. It has 8 cabins fixed at equal distances on its rim, and each cabin is joined to the centre $O$ by a spoke, as shown in the figure. $P$ and $Q$ are two neighbouring cabins. (Use $\pi = \frac{22}{7}$)

(i) Find $\angle POQ$, the angle between the spokes of two neighbouring cabins. (1 mark)

Answer.

1. $\angle POQ = \frac{360^\circ}{8} = 45^\circ$ — 1 mark

(ii) How far does a cabin travel along the rim in going from the position of $P$ to the position of $Q$? (1 mark)

Answer.

1. Arc $PQ = \frac{45^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 14 = \frac{88}{8} = 11$ m — 1 mark

(iii) During one ride the wheel makes 5 complete rounds. Find the distance travelled by a cabin during the ride. If a cabin moves at a steady 2.2 m per second, how long does the ride last? (2 marks)

Answer.

1. Distance $= 5 \times 2 \times \frac{22}{7} \times 14 = 5 \times 88 = 440$ m — 1 mark
2. Time $= \frac{440}{2.2} = 200$ seconds, that is, 3 minutes 20 seconds — 1 mark

OR

(iii) Find the area of the sector of the wheel between the spokes of two cabins that have two other cabins between them. (2 marks)

Answer.

1. There are 3 gaps between them, so the angle is $3 \times 45^\circ = 135^\circ$ — 1 mark
2. Area $= \frac{135^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = \frac{3}{8} \times 616 = 231 \text{ m}^2$ — 1 mark

Question 14 (4 marks)

A carpenter in Saharanpur makes a round dining table with a top of radius 70 cm and centre $O$. A part of the top folds down: it is the region cut off by a straight hinge $AB$, where the chord $AB$ subtends a right angle at $O$ (shaded in the figure). (Use $\pi = \frac{22}{7}$ and $\sqrt{2} = 1.41$)

(i) Find the area of the whole table top. (1 mark)

Answer.

1. $\frac{22}{7} \times 70 \times 70 = 15400 \text{ cm}^2$ — 1 mark

(ii) Find the length of the hinge $AB$. (1 mark)

Answer.

1. $AB = \sqrt{70^2 + 70^2} = 70\sqrt{2} = 70 \times 1.41 = 98.7$ cm — 1 mark

(iii) Find the area of the part that folds down. (2 marks)

Answer.

1. Quadrant $OAB = \frac{1}{4} \times 15400 = 3850 \text{ cm}^2$; $\triangle OAB = \frac{1}{2} \times 70 \times 70 = 2450 \text{ cm}^2$ — 1 mark
2. Folding part (minor segment) $= 3850 - 2450 = 1400 \text{ cm}^2$ — 1 mark

OR

(iii) A metal strip is fixed all round the edge of the folding part (along the arc and along the hinge). Find the length of the strip. (2 marks)

Answer.

1. Arc $AB = \frac{1}{4} \times 2 \times \frac{22}{7} \times 70 = 110$ cm — 1 mark
2. Strip $= 110 + 98.7 = 208.7$ cm — 1 mark