Applications and word problems

This subtopic collects the practical, story-based problems that translate the sector and segment formulas into real-world settings. They are the most frequently tested style in the board exam , usually $3$- or $4$-mark questions with a small diagram or a verbal description.

The recipe is always the same: identify the radius, identify the angle, identify which formula (arc length, sector area, segment area, or a combination). Then compute.

Clock hand problems

A clock has two hands , minute and hour , both rotating around the same centre but at different speeds. In one full revolution ($360^\circ$):

• The minute hand goes around once every $60$ minutes. Angular speed $= 6^\circ$ per minute.
• The hour hand goes around once every $12$ hours $= 720$ minutes. Angular speed $= 0.5^\circ$ per minute, or $30^\circ$ per hour.

Common questions:

• Area swept by the minute hand in $T$ minutes: $(6T/360) \pi r^2 = T\pi r^2/60$.
• Distance traversed by tip of minute hand in $T$ minutes: $(6T/360)(2\pi r) = T\pi r/30$.
• Angle between the hands at time $h$:$m$: $|30h - 5.5 m|$ degrees (or $360^\circ$ minus that, whichever is smaller).

Tethered animal problems

An animal is tied to a fixed point by a rope of length $\ell$. The grazing area depends on the surrounding constraints:

• Open field: full circle, area $\pi \ell^2$.
• Tied at a corner of a square or rectangle: quarter circle, area $\pi \ell^2/4$.
• Tied at a corner of an equilateral triangle: $60^\circ$ sector, area $\pi \ell^2/6$.
• Tied at a corner of a regular polygon with interior angle $\alpha$: sector of angle $360^\circ - \alpha$ on the outside, or $\alpha$ on the inside.

If the rope is longer than a side, the goat may also graze around the next corner , break the area into multiple sectors with different radii.

Wheel problems

A wheel of radius $r$ rolling on the ground without slipping:

• Circumference $= 2\pi r$.
• Distance covered in one full revolution $= 2\pi r$.
• Number of revolutions to cover distance $d$: $d/(2\pi r)$.

For speed-related problems, convert units carefully: e.g., $66$ km/h $= 66 \times 1000/60$ m/min $= 1100$ m/min.

Wiper problems

A windscreen wiper is an arm of length $L$ attached at one end to a pivot, with a blade of length $b$ at the far end. The wiper rotates through an angle $\theta$. The swept area is an annular sector:

$A = \frac{\theta}{360^\circ} \pi (L^2 - (L - b)^2)$

if the blade extends from the tip of the arm inward to length $L - b$ from the pivot (typical for car wipers).

If there are two wipers and their arcs do not overlap, the total swept area is twice this.

Worked examples

Example 1. The minute hand of a clock is $14$ cm long. Find the area swept in $5$ minutes.

In $5$ minutes the hand turns $5 \times 6^\circ = 30^\circ$. Area $= (30/360)(22/7)(196) = (1/12)(616) = 51.33$ cm$^2$.

Example 2. A wheel of diameter $77$ cm rolls a distance of $121$ m without slipping. Find the number of revolutions.

Circumference $= \pi d = (22/7)(77) = 242$ cm. Distance $= 12100$ cm. Revolutions $= 12100/242 = 50$.

Example 3. A goat is tied at the corner of a square plot of side $10$ m by a rope of length $7$ m. Find the area the goat can graze.

The corner has $90^\circ$ interior angle. Grazing area $= (90/360) \pi (7)^2 = (1/4)(22/7)(49) = 38.5$ m$^2$.

Example 4. A goat is tied at the corner of an equilateral triangular plot of side $5$ m by a rope of length $4$ m. Find the area the goat can graze (assuming the rope is shorter than the side).

The corner has $60^\circ$ interior angle. Area $= (60/360) \pi (4)^2 = (1/6)(22/7)(16) = 352/42 \approx 8.38$ m$^2$.

Example 5. A wiper has a blade of length $25$ cm attached to a $10$-cm arm. It sweeps through $115^\circ$. Find the area swept.

Outer radius $= 35$ cm, inner radius $= 10$ cm.

$A = (115/360) \pi (35^2 - 10^2) = (115/360)(22/7)(1125) = (115/360)(24750/7) \approx 1129.5$ cm$^2$.

Example 6. A horse is tied at one corner of a rectangular field of $30$ m by $20$ m by a rope of length $14$ m. Find the area available to graze.

Quarter circle at the corner: $(1/4)(22/7)(196) = 154$ m$^2$.

If the rope exceeded $20$ m (the shorter side), the horse could graze around the next corner , extra sector with reduced radius. Here $14 < 20$, so only the quarter circle.

Example 7. At $4$:$00$ the hour hand is at the $4$ o'clock position. Find the angle between the hands. (Use $|30h - 5.5m|$ formula.)

$|30 \times 4 - 5.5 \times 0| = 120^\circ$.

Try it yourself

1. The minute hand of a clock is $15$ cm long. Find the area swept in $20$ minutes.
2. A wheel of radius $35$ cm rolls $44$ m. Find the number of revolutions.
3. A wheel makes $1000$ revolutions while covering $11$ km. Find the radius.
4. A goat tied at the corner of a square plot of side $20$ m with rope of length $14$ m. Find the grazing area.
5. A goat tied at the corner of an equilateral triangle plot of side $10$ m with rope of length $5$ m. Find the grazing area.
6. Find the angle between the hour and minute hands at $5$:$30$.
7. A wiper has an arm of $20$ cm and a blade of $25$ cm. It sweeps through $90^\circ$. Find the area swept.
8. A car's tyre has radius $40$ cm. At $36$ km/h, find the revolutions per minute.
9. The hour hand of a clock is $6$ cm. Find the area swept between $6$ am and $6$ pm.
10. A pendulum of length $80$ cm swings through $30^\circ$. Find the arc length traced by its tip.
11. A goat tied to a corner of a hexagonal plot (interior angle $120^\circ$) of side $10$ m by a rope of $5$ m. Find the grazing area.
12. A semi-circular platform of radius $7$ m has tiles of side $0.5$ m. Find the number of tiles needed.

Pitfalls / Insight

(1) For clock-hand problems, convert minutes to degrees using $6^\circ$/min for the minute hand and $0.5^\circ$/min for the hour hand.

(2) For wheel problems, $C = \pi d$, so revolutions $= d/(\pi \cdot \text{diameter})$. Get the units right: km vs m vs cm.

(3) For tethered animals, always identify the interior angle available to the animal. A rectangular corner gives $90^\circ$, an equilateral triangle gives $60^\circ$, a regular hexagon gives $120^\circ$ at each interior corner.

(4) If the rope length exceeds the side of the plot, the animal can swing around the next corner , break the grazing area into multiple sectors with different radii (rope - side) for the second sector.

(5) Always state your assumed value of $\pi$ at the start, and stick to it throughout the problem.