Real-Life Applications of Sectors

Clock Hands — Angles per Minute

A clock hand is a rotating radius, so the region it sweeps is a sector. The only new skill is finding the angle it turns.

• The minute hand goes right round ($360^\circ$) in 60 minutes, so it turns $\dfrac{360^\circ}{60} = 6^\circ$ per minute.
• The hour hand goes round ($360^\circ$) in 12 hours $=720$ minutes, so it turns $0.5^\circ$ per minute (or $30^\circ$ per hour).

Area swept by a hand of length $r$ in $t$ minutes $= \dfrac{\theta}{360}\pi r^2$, with $\theta = 6t$ for the minute hand.

[Board Important] Minute hand in 5 min $\to 30^\circ$; in 10 min $\to 60^\circ$; in 15 min $\to 90^\circ$ (a quadrant).

Wipers and Fans

A windscreen wiper blade of length $r$ sweeping through an angle $\theta$ cleans a sector of area $\dfrac{\theta}{360}\pi r^2$. If there are two wipers that do not overlap, double it.

The same sector idea covers a fan or any blade rotating through an angle.

Key Point: "Area cleaned/covered by a sweeping blade" $=$ area of a sector with radius $=$ blade length and angle $=$ sweep angle. Watch for "two blades" (multiply by 2) or "overlap" (subtract the shared part).

Grazing Animals Tied by a Rope

An animal tied by a rope of length $r$ to a peg at a corner of a field grazes a region shaped like a sector. At the corner of a square field the inside angle is $90^\circ$, so the grazing region is a quarter circle: $$\text{Grazing area} = \dfrac14\pi r^2 \quad(\text{if the rope does not exceed the side}).$$

If the rope is lengthened, the increase in grazing area is the difference of the two quarter circles: $\dfrac14\pi(r_2^2 - r_1^2)$.

[Board Important] At a corner of an equilateral-triangle field the angle is $60^\circ$ (a sextant $\tfrac16$ circle); at a corner of a regular hexagon it is $120^\circ$. Always use the interior angle at the corner.

Solved Examples

Example 1: Minute hand sweep

The minute hand of a clock is 14 cm long. Find the area it sweeps in 5 minutes. $\left(\pi=\dfrac{22}{7}\right)$

Solution:

1. In 5 minutes the hand turns $\theta = 6\times 5 = 30^\circ$.
2. Area $= \dfrac{30}{360}\times\dfrac{22}{7}\times 14^2 = \dfrac{1}{12}\times\dfrac{22}{7}\times 196 = \dfrac{22\times 196}{84} \approx 51.33$ cm$^2$.

Final Answer: $\approx 51.33$ cm$^2$.

Takeaway: Minute hand angle $= 6^\circ\times$ (minutes).

Example 2: Windscreen wipers

A car has two wipers that do not overlap. Each blade is 25 cm long and sweeps through $115^\circ$. Find the total area cleaned per sweep. $\left(\pi=\dfrac{22}{7}\right)$

Solution:

1. One wiper cleans $\dfrac{115}{360}\times\dfrac{22}{7}\times 25^2 = \dfrac{115}{360}\times\dfrac{22}{7}\times 625 \approx 627.48$ cm$^2$.
2. Two wipers: $2\times 627.48 = 1254.96$ cm$^2$.

Final Answer: $\approx 1254.96$ cm$^2$.

Takeaway: Two non-overlapping blades $\Rightarrow$ multiply one sector by 2.

Example 3: Grazing horse

A horse is tied to a peg at a corner of a square field of side 15 m by a 5 m rope. Find (i) the grazing area, and (ii) the increase if the rope is 10 m instead. $\left(\pi=3.14\right)$

Solution:

1. Corner of a square $\Rightarrow$ quarter circle. With $r=5$: area $= \dfrac14\times 3.14\times 25 = 19.625$ m$^2$.
2. With $r=10$: area $= \dfrac14\times 3.14\times 100 = 78.5$ m$^2$.
3. Increase $= 78.5 - 19.625 = 58.875$ m$^2$.

Final Answer: (i) 19.625 m$^2$; (ii) increase 58.875 m$^2$.

Takeaway: Corner of a square field $\Rightarrow \tfrac14\pi r^2$ (both rope lengths are $\le$ the side).

Example 4: Lighthouse beam

A lighthouse spreads red light over a sector of angle $80^\circ$ to a distance of 16.5 km. Find the area warned. $\left(\pi=3.14\right)$

Solution:

1. Area $= \dfrac{80}{360}\times 3.14\times 16.5^2 = \dfrac{2}{9}\times 3.14\times 272.25$.
2. $= \dfrac{2}{9}\times 854.865 \approx 189.97$ km$^2$.

Final Answer: $\approx 189.97$ km$^2$.

Takeaway: The beam is a sector with radius $=$ range and angle $=$ spread.

Example 5: Umbrella ribs

An umbrella has 8 equally spaced ribs, modelled as a flat circle of radius 45 cm. Find the area between two consecutive ribs. $\left(\pi=\dfrac{22}{7}\right)$

Solution:

1. 8 equal sectors $\Rightarrow$ each angle $= \dfrac{360^\circ}{8} = 45^\circ$.
2. Area $= \dfrac{45}{360}\times\dfrac{22}{7}\times 45^2 = \dfrac18\times\dfrac{22}{7}\times 2025 = \dfrac{22\times 2025}{56} \approx 795.54$ cm$^2$.

Final Answer: $\approx 795.54$ cm$^2$.

Takeaway: $n$ equal ribs/sectors $\Rightarrow$ each angle $= \dfrac{360^\circ}{n}$.