MathsCBSE Class 10 · Areas Related to Circles

Word Problems on Circles: Wheels, Clocks and Grazing Fields

Q: How do I know whether to use arc length or sector area?

Check what the story measures. A distance moved, a path or a fence is a length, so use the arc. A region swept, cleaned, watered or grazed is an area, so use the sector.

Q: Why does one revolution equal one circumference?

When a wheel rolls without slipping, every point on its rim touches the ground exactly once in a turn. So the ground covered equals the length of the rim, which is the circumference.

Q: What if the rope is longer than a side of the field?

Then the grazing region is no longer a simple sector and spills past the fence. Board questions in this chapter choose the rope shorter than the sides, so the region stays a clean sector.

Q: Are combinations of plane figures beyond sectors and segments in the syllabus?

The NCERT section on general areas of combinations of plane figures was removed. Shaded and word problems in 2026-27 are built from sectors and segments with squares, rectangles and triangles.

Q: Can I leave answers as fractions?

Yes, an exact fraction like 154/3 square cm is fine unless the question asks for a decimal. Always add the correct unit.

Updated 14 October 2026 · Fact-checked

These are real-life questions where a circle, sector or arc hides inside a story. A wheel covers one circumference per revolution, a clock hand sweeps a sector, a wiper cleans a sector or a sector ring, and a tied animal grazes a sector. The board asks you to name the shape, convert units and calculate.

Ask first: is the story asking for a length (use 2πr or an arc) or an area (use a sector)? Wheels give length, sweeping and grazing give area.

Illustration drawn for QuestionBuddy.

Understand Word Problems on Circles: Wheels, Clocks, Fields

A word problem on circles is an ordinary sector, arc or circumference question wrapped in a story. Your first job is translation. Ask yourself one question: is the story asking for a length or an area? A distance travelled, a fence, or the path of a tip is a length, so you use $2 π r$ or an arc. A region cleaned, watered, swept or grazed is an area, so you use $π r^{2}$ or a sector.

Wheels. When a wheel rolls once without slipping, every point of its rim touches the road once. So one revolution moves the wheel forward by exactly one circumference, $2 π r$ . For $n$ revolutions the distance is $n \times 2 π r$ . Turn this around and you get the number of revolutions: distance divided by circumference. Speed questions add one more step: convert km per hour into cm per minute before you divide.

Clocks. A hand turns through the whole $36 0^{\circ}$ in a fixed time. The minute hand takes 60 minutes, so it turns $6^{\circ}$ every minute. The hour hand takes 12 hours, so it turns $3 0^{\circ}$ every hour, or $\frac{1}{2}^{\circ}$ every minute. Once you have the angle, the area swept is a sector and the distance moved by the tip is an arc, both with radius equal to the length of the hand.

Wipers, sprinklers and grazing animals. Each one sweeps a sector. The radius is the blade length, spray range or rope length, and the angle is how far it turns. If the rubber of a wiper starts some distance from the pivot, the cleaned part is a sector of the big radius minus a sector of the small radius. A goat tied at the corner of a field can only reach inside the field, so the angle is the angle of that corner: $9 0^{\circ}$ for a rectangle or square, $6 0^{\circ}$ for an equilateral triangle. Tied at a point on a straight fence it gets $18 0^{\circ}$ .

All of this works only when the conditions are clean: the wheel rolls without slipping, the rope is shorter than the sides it lies along (so the sector does not spill out of the field), and the hands move steadily. Board questions always choose numbers that respect these conditions, and stating the shape in one line before calculating earns you a mark.

Word Problems on Circles: Wheels, Clocks, Fields: formulas to remember

Distance covered by a wheel: $Distance = n \times 2 π r$; $n$ is the number of revolutions; one revolution = one circumference.
Number of revolutions: $n = \frac{Distance}{2 π r}$; Distance and circumference must be in the same unit.
Speed from revolutions: $Speed = (revolutions per minute) \times 2 π r per minute$; Multiply by 60 for per hour; 1 km = 100000 cm.
Angle turned by the minute hand: $θ = 6^{\circ} \times (minutes)$; 15 min = $9 0^{\circ}$ , 20 min = $12 0^{\circ}$ , 10 min = $6 0^{\circ}$ .
Angle turned by the hour hand: $θ = 3 0^{\circ} \times (hours) = \frac{1}{2}^{\circ} \times (minutes)$; 3 hours = $9 0^{\circ}$ , 4 hours = $12 0^{\circ}$ .
Area swept by a hand, wiper or beam: $A = \frac{θ}{36 0 ^{\circ}} \times π r^{2}$; Radius = length of the hand, blade or beam range.
Distance moved by the tip: $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r$; Length, not area: uses $2 π r$ .
Area cleaned when the rubber starts away from the pivot: $A = \frac{θ}{36 0 ^{\circ}} \times π (R^{2} - r^{2})$; $R$ = outer end, $r$ = inner end of the rubber.
Grazing area at a corner: $A = \frac{θ}{36 0 ^{\circ}} \times π L^{2}$; $L$ = rope length; $θ = 9 0^{\circ}$ at a rectangle corner, $6 0^{\circ}$ at an equilateral triangle corner, $18 0^{\circ}$ on a straight fence.
Corners of a triangle together: $∠ A + ∠ B + ∠ C = 18 0^{\circ} \Rightarrow total = \frac{1}{2} π L^{2}$; Equal ropes at all three vertices graze a semicircle's worth of area.
Perimeter of a sector: $Perimeter = 2 r + \frac{θ}{36 0 ^{\circ}} \times 2 π r$; Use for fencing or border around a sector-shaped bed, plot or track.
Area of a segment: $Segment = \frac{θ}{36 0 ^{\circ}} \times π r^{2} - area of △ O A B$; Board questions use only $θ = 6 0^{\circ}$ (triangle $\frac{3}{4} r^{2}$ ), $9 0^{\circ}$ (triangle $\frac{1}{2} r^{2}$ ) or $12 0^{\circ}$ (triangle $\frac{3}{4} r^{2}$ ).

How to solve Word Problems on Circles: Wheels, Clocks, Fields questions

Use this routine on any wheel, clock, wiper or grazing question.

1Read the story and decide: length (distance, fence, tip) or area (swept, cleaned, grazed)?
2Write one line naming the shape, e.g. 'area grazed is a sector of radius 14 m and angle $9 0^{\circ}$ '.
3Find the radius: hand length, blade, rope, or half the wheel's diameter.
4Find the angle: from time for clock hands, from the corner of the field for grazing, from the question for wipers.
5Convert all units to one system (km to cm, hours to minutes) before dividing or multiplying.
6Calculate the full circle value first ( $2 π r$ or $π r^{2}$ ), then take the fraction $\frac{θ}{36 0 ^{\circ}}$ .
7For 'not grazed' or 'remaining' parts, subtract from the field or circle area.
8Write the final answer with units and, if possible, check it with a rough estimate.

Word Problems on Circles: Wheels, Clocks, Fields in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Wheel revolutions

Imagine your sports teacher uses a measuring wheel to mark out a cricket pitch and the school boundary. The wheel has a diameter of 35 cm and a counter that records every complete revolution.

(i) How many revolutions does the wheel make while measuring a 22 m pitch? (ii) Walking once round the boundary, the counter reads 250. How long is the boundary? (iii) A second measuring wheel records only 125 revolutions for the same boundary. Find its diameter. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Radius $= 17.5$ cm, so circumference $= 2 \times \frac{22}{7} \times 17.5 = 110$ cm.
22 m $= 2200$ cm, so revolutions $= \frac{2200}{110} = 20$ .
Boundary $= 250 \times 110 = 27500$ cm $= 275$ m.
Second wheel: circumference $= \frac{27500}{125} = 220$ cm.
Diameter $= \frac{220}{π} = 220 \times \frac{7}{22} = 70$ cm, twice the first wheel's, which is why it needs half as many revolutions.

Answer: (i) 20 revolutions (ii) 275 m (iii) 70 cm

Memory hook: One click of the wheel = one circumference of ground.

Grazing area

Suppose a farmer has a square vegetable plot of side 28 m. He ties one goat at one corner and another goat at the opposite corner, each with a 14 m rope, to eat the weeds near the corners.

Find the total area the goats can graze and the area of the plot they cannot reach. The farmer then moves the pegs to the midpoints of two opposite sides of the plot, keeping the same ropes. Find the new grazed area, and say whether the goats' regions now overlap. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Each goat grazes a quadrant: sector of radius 14 m and angle $9 0^{\circ}$ , area $\frac{1}{4} \times \frac{22}{7} \times 1 4^{2} = 154$ m $^{2}$ .
Opposite corners are $282 \approx 39.6$ m apart, more than $14 + 14 = 28$ m, so the quadrants do not overlap. Total grazed $= 308$ m $^{2}$ .
Plot area $= 2 8^{2} = 784$ m $^{2}$ , so not reached $= 784 - 308 = 476$ m $^{2}$ .
At the midpoint of a side, each goat grazes a semicircle of radius 14 m (the side blocks the other half): $\frac{1}{2} \times \frac{22}{7} \times 196 = 308$ m $^{2}$ each.
The midpoints of opposite sides are 28 m apart, exactly $14 + 14$ , so the semicircles only touch at the centre of the plot and do not overlap. New grazed area $= 616$ m $^{2}$ , twice the earlier area.

Answer: Grazed 308 m², not reached 476 m². With pegs at the midpoints of opposite sides, the goats graze 616 m², and the two semicircles just touch without overlapping.

Memory hook: A goat at a square corner eats a quarter pizza.

Tricks to remember Word Problems on Circles: Wheels, Clocks, Fields

Short memory aids. Say each one aloud once, then use it on a question the same day.

Six and a half
Minute hand: 6 degrees a minute. Hour hand: half a degree a minute.
Use it when: Turn any time interval into a central angle instantly.
Walk or paint?
If you could walk along it, use $2 π r$ . If you could paint it, use $π r^{2}$ .
Use it when: Decide between arc length and sector area in a story.
Corner decides the slice
The goat's slice angle equals the corner angle of the field.
Use it when: Grazing questions at rectangle, square or triangle corners.
Roll once, move once
One revolution of a wheel = one circumference of road.
Use it when: All revolution, distance and speed problems.

Word Problems on Circles: Wheels, Clocks, Fields: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Minute hand: area and tip distance · 2 marks · Very short answerBoard level

The minute hand of a wall clock is 21 cm long. Find the area swept by it between 7:10 a.m. and 7:30 a.m., and the distance moved by its tip in that time. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Time $= 20$ min, so angle $= 20 \times 6^{\circ} = 12 0^{\circ}$ , a third of a turn. (½)
Area swept $= \frac{1}{3} \times \frac{22}{7} \times 2 1^{2} = \frac{1}{3} \times 1386 = 462$ cm $^{2}$ . (½)
Distance moved by tip is an arc: $\frac{1}{3} \times 2 \times \frac{22}{7} \times 21$ . (½)
$= \frac{1}{3} \times 132 = 44$ cm. (½)

Answer: Area swept = 462 cm², distance moved by the tip = 44 cm

Example 2: Horse tied at the corner of a rectangular field · 3 marks · Short answerHarder board level

A horse is tied to a peg at one corner of a rectangular field measuring 30 m by 20 m with a rope 14 m long. Find (i) the area the horse can graze, (ii) the area it cannot graze, and (iii) the length of rope that would let the horse graze twice the area in (i), checking that the grazing region would still lie inside the field. (Take $π = \frac{22}{7}$ , $2 = 1.414$ )

Try it first, then show the solutionSolution

The corner is $9 0^{\circ}$ and $14 < 20$ , so the grazing region is a quadrant of radius 14 m.
(i) Area $= \frac{1}{4} \times \frac{22}{7} \times 196 = 154$ m $^{2}$ . (1)
(ii) Field $= 30 \times 20 = 600$ m $^{2}$ ; not grazed $= 600 - 154 = 446$ m $^{2}$ . (1)
(iii) Twice the area: $\frac{1}{4} \times \frac{22}{7} \times L^{2} = 308$ , so $L^{2} = 392$ and $L = 142 \approx 19.8$ m. (½)
Since $19.8 < 20$ and $19.8 < 30$ , the quadrant still fits inside the field, so the area really doubles. (½)

Answer: (i) 154 m² (ii) 446 m² (iii) about 19.8 m ( $142$ m), which still keeps the quadrant inside the field.

Example 3: Two wipers with rubber away from the pivot · 3 marks · Short answerBoard level

A bus has two windscreen wipers that do not overlap. On each wiper the rubber blade starts 7 cm from the pivot and ends 28 cm from the pivot. Each wiper turns through $12 0^{\circ}$ . Find the total area cleaned in one sweep of both wipers. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Region cleaned by one wiper = sector of radius 28 cm minus sector of radius 7 cm, angle $12 0^{\circ}$ . (1)
Area $= \frac{120}{360} \times \frac{22}{7} \times (2 8^{2} - 7^{2}) = \frac{1}{3} \times \frac{22}{7} \times 735$ . (½)
$= \frac{1}{3} \times 2310 = 770$ cm $^{2}$ . (½)
Both wipers (no overlap): $2 \times 770 = 1540$ cm $^{2}$ . (1)

Answer: 1540 cm²

Example 4: Bicycle wheel: distance, revolutions and speed · 5 marks · Long answerBoard level

The wheel of a bicycle has a diameter of 63 cm. (i) Find the distance covered in 500 revolutions. (ii) How many revolutions does the wheel make to cover 2.97 km? (iii) If the wheel makes 200 revolutions per minute, find the speed of the bicycle in km per hour. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Circumference $= \frac{22}{7} \times 63 = 198$ cm. (1)
(i) Distance $= 500 \times 198 = 99000$ cm $= 990$ m. (1)
(ii) $2.97$ km $= 297000$ cm. (½)
Revolutions $= \frac{297000}{198} = 1500$ . (1)
(iii) Distance per minute $= 200 \times 198 = 39600$ cm $= 396$ m. (½)
Per hour $= 396 \times 60 = 23760$ m $= 23.76$ km, so speed $= 23.76$ km/h. (1)

Answer: (i) 990 m (ii) 1500 revolutions (iii) 23.76 km/h

Example 5: Case study: the school hall clock · 4 marks · Case-basedHOTS / competency

A large clock in a school hall has a minute hand 42 cm long and an hour hand 28 cm long. (Take $π = \frac{22}{7}$ )
(i) Through what angle does the hour hand turn from 2:00 p.m. to 4:30 p.m.? (1 mark)
(ii) Find the distance moved by the tip of the minute hand in 35 minutes. (1 mark)
(iii) Find the area swept by the minute hand in 15 minutes and by the hour hand in 3 hours, and the ratio of the two areas. (2 marks)

Try it first, then show the solutionSolution

(i) 2.5 hours $\times 3 0^{\circ} = 7 5^{\circ}$ . (1)
(ii) Angle $= 35 \times 6^{\circ} = 21 0^{\circ}$ ; arc $= \frac{210}{360} \times 2 \times \frac{22}{7} \times 42 = \frac{7}{12} \times 264 = 154$ cm. (1)
(iii) Minute hand, 15 min $= 9 0^{\circ}$ : $\frac{1}{4} \times \frac{22}{7} \times 4 2^{2} = \frac{1}{4} \times 5544 = 1386$ cm $^{2}$ . (½)
Hour hand, 3 h $= 9 0^{\circ}$ : $\frac{1}{4} \times \frac{22}{7} \times 2 8^{2} = \frac{1}{4} \times 2464 = 616$ cm $^{2}$ . (½)
Ratio $= 1386 : 616 = 9 : 4$ , which equals $4 2^{2} : 2 8^{2}$ since the angles are equal. (1)

Answer: (i) 75° (ii) 154 cm (iii) 1386 cm² and 616 cm², ratio 9 : 4

Example 6: Fencing and area of a sector-shaped bed · 3 marks · Short answerBoard level

A flower bed is in the shape of a sector of a circle of radius 21 m with a central angle of $12 0^{\circ}$ . Find the length of fencing needed all round the bed and the area of the bed. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Arc length $= \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{3} \times 132 = 44$ m. (1)
Perimeter of the sector $= 2 r + arc = 42 + 44 = 86$ m of fencing. (1)
Area $= \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times \frac{22}{7} \times 2 1^{2} = \frac{1}{3} \times 1386 = 462$ m $^{2}$ . (1)

Answer: 86 m of fencing; area 462 m².

Example 7: A segment-shaped flower patch · 4 marks · Long answerHarder board level

A circular lawn has radius 21 m. A straight path runs along a chord AB that subtends $12 0^{\circ}$ at the centre O. The smaller piece of lawn cut off by the chord (the minor segment) is turned into a flower patch, edged along both the arc and the chord. Find (i) the area of the flower patch and (ii) the total length of edging. (Take $π = \frac{22}{7}$ , $3 = 1.73$ )

Try it first, then show the solutionSolution

Area of sector OAB $= \frac{1}{3} \times \frac{22}{7} \times 441 = 462$ m $^{2}$ . (1)
In $△ O A B$ draw $O M ⊥ A B$ . Then $∠ A O M = 6 0^{\circ}$ , so $O M = 21 cos 6 0^{\circ} = 10.5$ m and $A M = 21 sin 6 0^{\circ} = \frac{21 3}{2}$ m, giving $A B = 213$ m. Area of $△ O A B = \frac{1}{2} \times 213 \times 10.5 = \frac{441 3}{4} \approx 190.73$ m $^{2}$ . (1)
(i) Area of the flower patch $= 462 - 190.73 = 271.27$ m $^{2}$ . (1)
(ii) Arc AB $= \frac{1}{3} \times 132 = 44$ m and chord $A B = 213 \approx 36.33$ m, so edging $= 44 + 36.33 = 80.33$ m. (1)

Answer: (i) About 271.27 m² (ii) About 80.33 m of edging.

Where students lose marks in Word Problems on Circles: Wheels, Clocks, Fields

The slips examiners see most often, and what to write instead.

Mistake: Not converting units in wheel problems
Speed is given in km/h but the circumference (here, a wheel of circumference 220 cm) is in cm.
Wrong working
Revolutions per minute $= \frac{66}{220} = 0.3$
Right working
66 km/h $= \frac{6600000}{60} = 110000$ cm/min; $\frac{110000}{220} = 500$ rev/min
Do this: Convert everything to cm and minutes before dividing.
Usually costs: 1 to 2 marks
Mistake: Using area when length is asked (or the reverse)
Both formulas look alike with $\frac{θ}{36 0 ^{\circ}}$ .
Wrong working
Distance moved by tip of a 21 cm hand in 20 min $= \frac{1}{3} \times \frac{22}{7} \times 2 1^{2} = 462$ cm
Right working
Distance $= \frac{1}{3} \times 2 \times \frac{22}{7} \times 21 = 44$ cm
Do this: Distance, fence, tip means $2 π r$ ; swept, cleaned, grazed means $π r^{2}$ .
Usually costs: 1 to 2 marks
Mistake: Wrong angle for clock hands
Mixing the $6^{\circ}$ and $3 0^{\circ}$ rates.
Wrong working
Hour hand in 3 hours: $3 \times 6^{\circ} = 1 8^{\circ}$
Right working
Hour hand in 3 hours: $3 \times 3 0^{\circ} = 9 0^{\circ}$
Do this: Minute hand $6^{\circ}$ per minute; hour hand $3 0^{\circ}$ per hour.
Usually costs: Whole question's marks
Mistake: Using a full circle for an animal tied at a corner
Forgetting the fences block the rope.
Wrong working
Rope 14 m at rectangle corner: area $= \frac{22}{7} \times 196 = 616$ m $^{2}$
Right working
Quadrant: $\frac{1}{4} \times 616 = 154$ m $^{2}$
Do this: Angle = corner angle of the field.
Usually costs: 1 to 2 marks
Mistake: Taking diameter as radius
Wheel sizes are usually given as diameters.
Wrong working
Diameter 63 cm: circumference $= 2 \times \frac{22}{7} \times 63 = 396$ cm
Right working
Circumference $= \frac{22}{7} \times 63 = 198$ cm
Do this: Underline 'diameter' and halve it, or use $π d$ .
Usually costs: All later marks

Exam tips

Write a one-line statement of the shape and its radius and angle before calculating; examiners reward it.
Show unit conversions on their own line in wheel and speed questions.
In clock questions, write the angle from time clearly, e.g. '20 min = $12 0^{\circ}$ '.
For 'not grazed' or 'remaining area', show the field area and the subtraction as separate steps.
Case-based clock and wheel questions often link parts; reuse your circumference or full area from the first part.

Practice questions on Word Problems on Circles: Wheels, Clocks, Fields

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A wheel of diameter 42 cm rolls along a road. How many revolutions does it make to cover 264 m? (Take $π = \frac{22}{7}$ )
(a) 100
(b) 200
(c) 400
(d) 2000
Hint
Find the circumference in cm and convert 264 m to cm.
Answer
(b) 200. Circumference $= \frac{22}{7} \times 42 = 132$ cm; $\frac{26400}{132} = 200$ .
Q2 · 3 marks · Short answerHarder board level
Three goats are tied, one at each vertex of an equilateral triangular field of side 20 m, each with a rope 7 m long. Find the area of the field the goats cannot graze. (Take $π = \frac{22}{7}$ and $3 = 1.73$ )
Hint
Each goat grazes a $6 0^{\circ}$ sector; the three angles add to $18 0^{\circ}$ .
Answer
Grazed $= \frac{180}{360} \times \frac{22}{7} \times 49 = 77$ m $^{2}$ . Field $= \frac{3}{4} \times 400 = 100 \times 1.73 = 173$ m $^{2}$ . Not grazed $= 173 - 77 = 96$ m $^{2}$ .
Q3 · 3 marks · Short answerBoard level
The tip of the minute hand of a clock moves 66 cm in 15 minutes. Find the length of the minute hand and the area it sweeps in those 15 minutes. (Take $π = \frac{22}{7}$ )
Hint
15 minutes is $9 0^{\circ}$ , so 66 cm is a quarter of the circumference.
Answer
$\frac{1}{4} \times 2 π r = 66 \Rightarrow 2 π r = 264 \Rightarrow r = 42$ cm. Area $= \frac{1}{4} \times \frac{22}{7} \times 1764 = 1386$ cm $^{2}$ .
Q4 · 5 marks · Long answerHarder board level
Each wheel of a car has a diameter of 84 cm and makes 300 revolutions per minute. (i) Find the speed of the car in km/h. (ii) How long does the car take to travel 11.88 km at this speed? (iii) How many revolutions does each wheel make in that time? (Take $π = \frac{22}{7}$ )
Hint
Circumference first, then cm per minute, then km per hour.
Answer
Circumference $= \frac{22}{7} \times 84 = 264$ cm. (i) $300 \times 264 = 79200$ cm/min $= 792$ m/min $= 47520$ m/h $= 47.52$ km/h. (ii) $\frac{11.88}{47.52} = \frac{1}{4}$ h $= 15$ min. (iii) $15 \times 300 = 4500$ revolutions (check: $\frac{1188000}{264} = 4500$ ).
Q5 · 4 marks · Case-basedHOTS / competency
A cow is tied with a 14 m rope to a peg at the midpoint of the 40 m side of a rectangular field measuring 40 m by 30 m. Riya says the grazing area is $π \times 1 4^{2} = 616$ m $^{2}$ . (i) Find her error and the correct area. (ii) What rope length would double the correct grazing area, and is that rope still short enough to stay inside the field? (Take $π = \frac{22}{7}$ , $2 = 1.414$ )
Hint
The peg is on a straight fence, so the cow reaches only a half-turn.
Answer
(i) The fence blocks half the circle; region is a semicircle: $\frac{1}{2} \times 616 = 308$ m $^{2}$ . (ii) Need $\frac{1}{2} \times \frac{22}{7} L^{2} = 616 \Rightarrow L^{2} = 392 \Rightarrow L = 142 \approx 19.8$ m. Since $19.8 < 20$ (half of 40) and $< 30$ , the semicircle stays inside the field.
Q6 · 3 marks · Short answerHarder board level
A square garden ABCD has side 28 m. A flower bed in the shape of a quadrant of a circle with centre A and radius 14 m is dug at corner A, and the rest of the garden is grassed. Find (i) the area of the grassed part and (ii) the perimeter of the grassed part. (Take $π = \frac{22}{7}$ )
Hint
The grassed part is bounded by the parts of AB and AD outside the bed, by BC and CD, and by the arc of the bed.
Answer
(i) Square $= 784$ m $^{2}$ , quadrant $= \frac{1}{4} \times \frac{22}{7} \times 196 = 154$ m $^{2}$ , grassed area $= 784 - 154 = 630$ m $^{2}$ . (1½)
(ii) Boundary: $(28 - 14) + 28 + 28 + (28 - 14) = 84$ m of straight edges plus the arc $\frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = 22$ m, so the perimeter is $84 + 22 = 106$ m. (1½)
Q7 · 4 marks · Long answerHOTS / competency
The front wheel of a cart has a diameter 14 cm less than that of the back wheel. In covering 132 m, the front wheel makes 50 more revolutions than the back wheel. Find the diameter of each wheel, and check your answer. (Take $π = \frac{22}{7}$ )
Hint
Let the front diameter be d cm. Number of revolutions = distance ÷ circumference; the difference is 50.
Answer
Let the front diameter be d cm, so the back diameter is $d + 14$ cm. 132 m $= 13200$ cm. Revolutions: $\frac{13200}{\frac{22}{7} d} - \frac{13200}{\frac{22}{7} ( d + 14 )} = 50$ , that is $\frac{4200}{d} - \frac{4200}{d + 14} = 50$ . So $4200 \times 14 = 50 d (d + 14)$ , giving $d^{2} + 14 d - 1176 = 0$ , $(d + 42) (d - 28) = 0$ , and $d = 28$ cm (a diameter is positive). Front wheel 28 cm, back wheel 42 cm. Check: circumferences 88 cm and 132 cm give $\frac{13200}{88} = 150$ and $\frac{13200}{132} = 100$ revolutions, a difference of 50.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Distance covered in one revolution of a wheel?Show answer
One circumference, $2 π r$ or $π d$ .
2.Angle turned by the minute hand per minute?Show answer
$6^{\circ}$
3.Angle turned by the hour hand per hour?Show answer
$3 0^{\circ}$
4.Shape grazed by an animal tied at a rectangle corner with a short rope?Show answer
A quadrant ( $9 0^{\circ}$ sector) of radius equal to the rope.
5.1 km in cm?Show answer
100000 cm
6.Area cleaned when the rubber runs from $r$ to $R$ from the pivot?Show answer
$\frac{θ}{36 0 ^{\circ}} \times π (R^{2} - r^{2})$

Word Problems on Circles: Wheels, Clocks, Fields: frequently asked questions

How do I know whether to use arc length or sector area?

Check what the story measures. A distance moved, a path or a fence is a length, so use the arc. A region swept, cleaned, watered or grazed is an area, so use the sector.

Why does one revolution equal one circumference?

When a wheel rolls without slipping, every point on its rim touches the ground exactly once in a turn. So the ground covered equals the length of the rim, which is the circumference.

What if the rope is longer than a side of the field?

Then the grazing region is no longer a simple sector and spills past the fence. Board questions in this chapter choose the rope shorter than the sides, so the region stays a clean sector.

Are combinations of plane figures beyond sectors and segments in the syllabus?

The NCERT section on general areas of combinations of plane figures was removed. Shaded and word problems in 2026-27 are built from sectors and segments with squares, rectangles and triangles.

Can I leave answers as fractions?

Yes, an exact fraction like 154/3 square cm is fine unless the question asks for a decimal. Always add the correct unit.