Chapter 11 · Areas Related to Circles

11.1 Sectors: Arc Length and Area

Two radii cut a circle into two sectors. The angle between them tells you what fraction of the circle each sector is — and that fraction, applied to the circumference and to the area, gives the arc length and the area of the sector.

A pizza slice, a fan of a wiper, a slice of a pie chart, the region a clock hand sweeps in ten minutes: all of these are parts of a circle bounded by two radii. To find the length of the curved edge or the amount of surface, you need not learn separate rules. A full turn is $360^\circ$ and a full circle has circumference $2\pi r$ and area $\pi r^2$. If a part uses an angle $\theta$ out of $360^\circ$, it takes the same fraction $\dfrac{\theta}{360}$ of the circumference and of the area.

In this section every numerical problem states the value of $\pi$ used: $\pi=\tfrac{22}{7}$ when the radius is a multiple of $7$ (the sevens cancel), and $\pi=3.14$ otherwise. Answers are given to two decimal places, with units: cm and m for lengths, cm² and m² for areas.

Arc, central angle and sector

📘 Definition 11.1 — Arc and central angle

Let $O$ be the centre of a circle and let $A$, $B$ be two points on it. The part of the circle between $A$ and $B$ is called an arc. The angle $\angle AOB$ between the radii $OA$ and $OB$ is the central angle (or angle at the centre) of the arc. Two points $A$, $B$ that are not ends of a diameter cut the circle into a shorter arc, the minor arc (central angle less than $180^\circ$), and a longer arc, the major arc (central angle $360^\circ-\theta$, greater than $180^\circ$).

📘 Definition 11.2 — Sector of a circle

A sector of a circle is the region enclosed by two radii and the arc between their ends. The sector bounded by the minor arc is the minor sector; the one bounded by the major arc is the major sector. If the central angle of a sector is $\theta$, the other sector of the same circle has central angle $360^\circ-\theta$.

A sector of angle $\theta=90^\circ$ is a quarter of the circle, one of $180^\circ$ is a semicircle, and one of $60^\circ$ is a sixth. The key step is to write the sector as a fraction: $\dfrac{\theta}{360}$.

📐 Theorem 11.1 — Length of an arc

In a circle of radius $r$, the arc whose central angle is $\theta$ degrees ($0\lt\theta\lt360$) has length $$l=\frac{\theta}{360}\times2\pi r.$$

Reason. Turn the radius $OA$ about $O$. Equal turns cut off equal lengths of the circle, so the length of an arc is proportional to its central angle. A full turn, $360^\circ$, gives the whole circumference $2\pi r$. Therefore the length for $\theta^\circ$ is $\dfrac{\theta}{360}$ of $2\pi r$. $\blacksquare$

📐 Theorem 11.2 — Area of a sector

In a circle of radius $r$, the sector whose central angle is $\theta$ degrees ($0\lt\theta\lt360$) has area $$A=\frac{\theta}{360}\times\pi r^2.$$

Reason. Cut the circle into narrow sectors by turning the radius through equal small angles: the pieces for equal angles are identical, so the area of a sector is proportional to its central angle, and $360^\circ$ gives the whole area $\pi r^2$. Hence the sector for $\theta^\circ$ has area $\dfrac{\theta}{360}\pi r^2$. $\blacksquare$

📐 Theorem 11.3 — Area of a sector from its arc length

A sector of radius $r$ and arc length $l$ has area $$A=\tfrac12\,l\,r.$$

Proof. By Theorem 11.2, $A=\dfrac{\theta}{360}\pi r^2=\left(\dfrac{\theta}{360}\times2\pi r\right)\times\dfrac r2=l\times\dfrac r2$, using Theorem 11.1 for $l$. $\blacksquare$

This is the "half base times height" rule of a triangle, with the arc as the base and the radius as the height. It is useful when the arc length is given and the angle is not. The perimeter of a sector is not the arc alone: it is the arc plus the two radii, $P=l+2r$.

⚠️ Common mistakes

Using the wrong circle formula: the arc length uses the circumference $2\pi r$ (a length, in cm), the sector uses the area $\pi r^2$ (in cm²). A "$2\pi r$" in an area, or a "$\pi r^2$" in a length, is always wrong.
Forgetting the two radii in the perimeter of a sector.
Writing the fraction as $\dfrac{\theta}{180}$ instead of $\dfrac{\theta}{360}$. Check: $\theta=360$ must give the whole circle.
Mixing units: if the radius is in metres, the area is in m².

🧭 Problem-solving strategy — sector problems

Draw the circle with the sector shaded. Mark $r$, $\theta$ and what is asked ($l$, $A$ or $P$).
Write down the value of $\pi$ you will use. Use $\tfrac{22}{7}$ if $r$ is a multiple of $7$.
Write the fraction $\dfrac{\theta}{360}$ first and simplify it ($\dfrac{60}{360}=\dfrac16$).
Multiply: by $2\pi r$ for a length, by $\pi r^2$ for an area.
If $\theta$ or $r$ is the unknown, write the same equation and solve it. Check with $A=\tfrac12lr$.

✏️ Example 1 — Arc and area of a 60° sector

A sector of a circle of radius $21$ cm has central angle $60^\circ$. Find the length of its arc and its area. What is the area of the major sector? (Use $\pi=\tfrac{22}{7}$.)

💡 Show solution▶

Fraction $\dfrac{\theta}{360}=\dfrac{60}{360}=\dfrac16$.

Arc $l=\dfrac16\times2\times\dfrac{22}{7}\times21=\dfrac16\times132=22$ cm.

Area $A=\dfrac16\times\dfrac{22}{7}\times21\times21=\dfrac16\times1386=231$ cm².

Major sector Its angle is $360^\circ-60^\circ=300^\circ$, i.e. $\dfrac56$ of the circle: $\dfrac56\times1386=1155$ cm².

Check $\tfrac12lr=\tfrac12\times22\times21=231$ ✓, and $231+1155=1386=\pi r^2$ ✓.

Choose a radius and sweep the central angle $\theta$ with the slider (or press Play). The red sector on the circle has angle $\theta$. The two bars on the right unroll the circumference and the area of the whole circle: the same fraction $\dfrac{\theta}{360}$ of each is coloured. Use $\pi=\tfrac{22}{7}$.

Angle $\theta$ (degrees)90

fraction $\theta/360$—

arc length $l$ (cm)—

area $A$ (cm²)—

perimeter $l+2r$ (cm)—

✏️ Example 2 — The tip of a minute hand

The minute hand of a clock is $14$ cm long. In $35$ minutes, how far does its tip travel, and what area does the hand sweep? (Use $\pi=\tfrac{22}{7}$.)

💡 Show solution▶

Angle The minute hand turns $360^\circ$ in $60$ minutes, i.e. $6^\circ$ each minute. In $35$ minutes: $\theta=35\times6=210^\circ$. Fraction $\dfrac{210}{360}=\dfrac7{12}$. The tip moves along an arc of a circle of radius $14$ cm.

Distance $l=\dfrac7{12}\times2\times\dfrac{22}{7}\times14=\dfrac7{12}\times88=\dfrac{154}{3}\approx51.33$ cm.

Area $A=\dfrac7{12}\times\dfrac{22}{7}\times14\times14=\dfrac7{12}\times616=\dfrac{1078}{3}\approx359.33$ cm².

Check $\tfrac12lr=\tfrac12\times51.33\times14=359.33$ ✓.

✏️ Example 3 — Finding the angle from the arc

In a circle of radius $7$ cm an arc has length $11$ cm. Find the central angle and the area of the sector. (Use $\pi=\tfrac{22}{7}$.)

💡 Show solution▶

Equation $l=\dfrac{\theta}{360}\times2\pi r$, so $11=\dfrac{\theta}{360}\times2\times\dfrac{22}{7}\times7=\dfrac{\theta}{360}\times44$.

Angle $\theta=\dfrac{11\times360}{44}=90^\circ$.

Area $A=\dfrac{90}{360}\times\dfrac{22}{7}\times49=\dfrac14\times154=38.5$ cm².

Check Theorem 11.3: $\tfrac12lr=\tfrac12\times11\times7=38.5$ ✓.

✏️ Example 4 — Perimeter of a sector

Find the perimeter and the area of a sector of angle $120^\circ$ in a circle of radius $21$ cm. (Use $\pi=\tfrac{22}{7}$.)

💡 Show solution▶

Fraction $\dfrac{120}{360}=\dfrac13$.

Arc $l=\dfrac13\times2\times\dfrac{22}{7}\times21=\dfrac13\times132=44$ cm.

Perimeter $P=l+2r=44+2\times21=86$ cm. (The two straight edges are radii.)

Area $A=\dfrac13\times\dfrac{22}{7}\times21^2=\dfrac13\times1386=462$ cm².

Check $\tfrac12lr=\tfrac12\times44\times21=462$ ✓.

Clock hands, wipers and ropes

Real problems almost always hide a sector. A clock hand sweeps a sector whose radius is the length of the hand. A windscreen wiper, a lawn sprinkler and a lighthouse beam sweep sectors too. A goat tied to the corner of a field by a rope can graze a sector of radius equal to the rope. When the thing that sweeps does not start at the centre (a wiper blade starts a little way from the pivot), the region is the difference of two sectors with the same angle: the big one minus the small one.

📐 Theorem 11.4 — Sector of a ring

Two arcs with the same centre $O$ and the same central angle $\theta$ have radii $R$ and $r$ with $R\gt r$. The region between them and the two straight edges (each of length $R-r$) has area $$A=\frac{\theta}{360}\,\pi\left(R^2-r^2\right).$$ Its perimeter is $\dfrac{\theta}{360}\,2\pi(R+r)+2(R-r)$.

Proof. The region is the sector of radius $R$ with the sector of radius $r$ removed, so $A=\dfrac{\theta}{360}\pi R^2-\dfrac{\theta}{360}\pi r^2$. For the perimeter add the two arcs, $\dfrac{\theta}{360}2\pi R+\dfrac{\theta}{360}2\pi r$, and the two straight edges. $\blacksquare$

✏️ Example 5 — A windscreen wiper

A wiper arm turns about a pivot $P$. The rubber blade is $15$ cm long and starts $10$ cm from the pivot, so it covers distances $10$ cm to $25$ cm from $P$. It swings through $120^\circ$. Find the area of glass the blade wipes. (Use $\pi=3.14$.)

💡 Show solution▶

Figure The blade sweeps the region between arcs of radii $R=25$ cm and $r=10$ cm, both with angle $\theta=120^\circ$. Fraction $\dfrac{120}{360}=\dfrac13$.

Difference $R^2-r^2=625-100=525$.

Area $A=\dfrac13\times3.14\times525=\dfrac{1648.5}{3}=549.5$ cm².

Check The big sector alone is $\dfrac13\times3.14\times625=654.17$ cm², the small one $\dfrac13\times3.14\times100=104.67$ cm²; $654.17-104.67=549.50$ cm² ✓.

✏️ Example 6 — A tethered cow

A cow is tied by a rope at one corner of a rectangular field $20$ m by $16$ m. (a) Find the area it can graze when the rope is $7$ m long. (b) By how much does the area increase if the rope is lengthened to $14$ m? Take $\pi=\tfrac{22}{7}$. The field is large enough for the rope to stay inside it.

💡 Show solution▶

Figure The corner of a rectangle is a right angle, so the cow can swing through $90^\circ$ inside the field: a quarter circle. A rope of $14$ m is shorter than both sides ($16$ m and $20$ m), so it stays inside the field.

(a) $A_1=\dfrac14\times\dfrac{22}{7}\times7^2=\dfrac14\times154=38.5$ m².

(b) $A_2=\dfrac14\times\dfrac{22}{7}\times14^2=\dfrac14\times616=154$ m². Increase $=154-38.5=115.5$ m².

Check Doubling the rope multiplies the area by $2^2=4$: $4\times38.5=154$ ✓.

Exercises

Arc length. Use $\pi=\tfrac{22}{7}$ unless another value is given.

1 Find the length of the arc of a circle of radius $14$ cm whose central angle is $45^\circ$.
2 Find the length of the arc of a circle of radius $21$ cm that subtends $150^\circ$ at the centre.
3 An arc of length $22$ cm belongs to a circle of radius $14$ cm. Find its central angle.
4 An arc of length $77$ cm belongs to a circle of radius $63$ cm. Find its central angle.
5 An arc of length $44$ cm subtends $60^\circ$ at the centre. Find the radius of the circle.
6 Find the perimeter of a sector of radius $14$ cm and central angle $90^\circ$.
7 A pendulum $21$ cm long swings through $60^\circ$. How far does the bob travel from one end of the swing to the other?
8 The minute hand of a clock is $7$ cm long. How far does its tip move in $20$ minutes?

💡 Show solutions▶

1 $l=\dfrac{45}{360}\times2\times\dfrac{22}{7}\times14=\dfrac18\times88=11$ cm.

2 $l=\dfrac{150}{360}\times132=\dfrac{5}{12}\times132=55$ cm.

3 $2\pi r=88$ cm. $\theta=\dfrac{22}{88}\times360=90^\circ$.

4 $2\pi r=2\times\dfrac{22}{7}\times63=396$ cm. $\theta=\dfrac{77}{396}\times360=70^\circ$.

5 $44=\dfrac16\times2\times\dfrac{22}{7}\times r=\dfrac{22}{21}\,r$, so $r=\dfrac{44\times21}{22}=42$ cm.

6 $l=\dfrac14\times88=22$ cm. $P=22+2\times14=50$ cm.

7 The bob moves along an arc of radius $21$ cm and angle $60^\circ$: $l=\dfrac16\times132=22$ cm.

8 In $20$ minutes the hand turns $20\times6=120^\circ$. $l=\dfrac13\times2\times\dfrac{22}{7}\times7=\dfrac{44}{3}\approx14.67$ cm.

Area of a sector.

1 Find the area of a sector of radius $6$ cm and angle $60^\circ$. (Use $\pi=3.14$.)
2 Find the area of a quadrant (angle $90^\circ$) of a circle of radius $14$ cm. (Use $\pi=\tfrac{22}{7}$.)
3 Find the area of a sector of radius $21$ cm and angle $120^\circ$. (Use $\pi=\tfrac{22}{7}$.)
4 A sector of a circle of radius $21$ cm has area $115.5$ cm². Find its central angle. (Use $\pi=\tfrac{22}{7}$.)
5 A sector has arc length $22$ cm and radius $14$ cm. Find its area using $\tfrac12lr$.
6 A sector of angle $45^\circ$ has area $77$ cm². Find the radius. (Use $\pi=\tfrac{22}{7}$.)
7 A horse is tied with a rope $10.5$ m long to a corner of a rectangular field $20$ m by $16$ m. Find the area it can graze. (Use $\pi=\tfrac{22}{7}$.)
8 Two arcs with the same centre have radii $21$ cm and $14$ cm and central angle $60^\circ$. Find the area between them. (Use $\pi=\tfrac{22}{7}$.)

💡 Show solutions▶

1 $A=\dfrac16\times3.14\times36=3.14\times6=18.84$ cm².

2 $A=\dfrac14\times\dfrac{22}{7}\times196=\dfrac14\times616=154$ cm².

3 $A=\dfrac13\times\dfrac{22}{7}\times441=\dfrac13\times1386=462$ cm².

4 Whole circle $=1386$ cm². $\theta=\dfrac{115.5}{1386}\times360=30^\circ$.

5 $A=\tfrac12\times22\times14=154$ cm².

6 $\dfrac18\times\dfrac{22}{7}r^2=77$, so $r^2=\dfrac{77\times8\times7}{22}=196$ and $r=14$ cm.

7 A rectangle's corner is $90^\circ$ and $10.5\lt16$, so the rope stays in the field: $A=\dfrac14\times\dfrac{22}{7}\times10.5^2=\dfrac14\times\dfrac{22}{7}\times110.25=\dfrac{346.5}{4}=86.625\approx86.63$ m².

8 $A=\dfrac16\times\dfrac{22}{7}\times(441-196)=\dfrac16\times\dfrac{22}{7}\times245=\dfrac{770}{6}\approx128.33$ cm².

III

Applications.

1 A wiper blade covers distances from $14$ cm to $28$ cm from the pivot and swings through $120^\circ$. Find the area of glass it cleans. (Use $\pi=\tfrac{22}{7}$.)
2 A garden sprinkler rotating through $135^\circ$ waters up to a distance of $8$ m. Find the area watered. (Use $\pi=3.14$.)
3 The hour hand of a clock is $6$ cm long. Find the area it sweeps from $3$ p.m. to $7$ p.m. (Use $\pi=3.14$.)
4 The minute hand of a clock is $10.5$ cm long. It sweeps an area of $57.75$ cm². For how many minutes has it been moving? (Use $\pi=\tfrac{22}{7}$.)
5 A camera on a circular cricket ground of radius $70$ m covers the sector between two fielders, $72^\circ$ apart as seen from the centre. Find the area covered. (Use $\pi=\tfrac{22}{7}$.)
6 A pizza of radius $14$ cm is cut into $8$ equal slices. Find the area of one slice, the length of its crust and the perimeter of the slice. (Use $\pi=\tfrac{22}{7}$.)
7 A fence $55$ m long runs along an arc of a circular garden of radius $21$ m. Find the central angle and the area of the sector enclosed. (Use $\pi=\tfrac{22}{7}$.)

💡 Show solutions▶

1 $A=\dfrac13\times\dfrac{22}{7}\times(28^2-14^2)=\dfrac13\times\dfrac{22}{7}\times588=\dfrac13\times1848=616$ cm².

2 $A=\dfrac{135}{360}\times3.14\times64=\dfrac38\times200.96=75.36$ m².

3 From $3$ p.m. to $7$ p.m. is $4$ hours, and the hour hand turns $30^\circ$ per hour: $\theta=120^\circ$. $A=\dfrac13\times3.14\times36=37.68$ cm².

4 Whole circle $=\dfrac{22}{7}\times10.5^2=346.5$ cm². Fraction $=\dfrac{57.75}{346.5}=\dfrac16$, i.e. $60^\circ$. At $6^\circ$ per minute: $10$ minutes.

5 $A=\dfrac{72}{360}\times\dfrac{22}{7}\times4900=\dfrac15\times15400=3080$ m².

6 Each slice has angle $45^\circ$, fraction $\dfrac18$. Area $=\dfrac18\times616=77$ cm². Crust $=\dfrac18\times88=11$ cm. Perimeter $=11+2\times14=39$ cm.

7 $2\pi r=132$ m. $\theta=\dfrac{55}{132}\times360=150^\circ$. $A=\tfrac12\times55\times21=577.5$ m².

Challenge.

1 A sector of radius $14$ cm has perimeter $72$ cm. Find its central angle and area. (Use $\pi=\tfrac{22}{7}$.)
2 The minute hand of a clock is $12$ cm long. Between $6{:}10$ and $6{:}40$, find the distance moved by its tip and the area swept. (Use $\pi=3.14$.)
3 Two sectors of the same circle have angles $40^\circ$ and $120^\circ$. Find the ratio of their arc lengths and of their areas. Then two sectors of the same angle have radii $3$ cm and $6$ cm: find the ratio of their arc lengths and of their areas.
4 Using Theorems 11.1 and 11.2, show that for a sector of radius $r$ and angle $\theta$ the ratio $\dfrac{\text{area}}{\text{arc length}}$ equals $\dfrac r2$.

💡 Show solutions▶

1 $l=72-2\times14=44$ cm, and $2\pi r=88$ cm, so $\theta=\dfrac{44}{88}\times360=180^\circ$ (a semicircle). $A=\tfrac12\times44\times14=308$ cm².

2 $30$ minutes is $180^\circ$, a half circle. $l=\dfrac12\times2\times3.14\times12=37.68$ cm. $A=\dfrac12\times3.14\times144=226.08$ cm².

3 Same circle: both lengths and areas are proportional to the angle, so both ratios are $40:120=1:3$. Same angle, radii $3$ and $6$: lengths are proportional to $r$, ratio $1:2$; areas to $r^2$, ratio $1:4$.

4 $\dfrac{A}{l}=\dfrac{\frac{\theta}{360}\pi r^2}{\frac{\theta}{360}2\pi r}=\dfrac{r}{2}$, so $A=\tfrac12lr$ (the angle cancels). $\blacksquare$

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · 5 marksBoard · Case study

How this is examined: one- and two-mark items test whether you can choose the fraction $\dfrac{\theta}{360}$ and apply it to $2\pi r$ or $\pi r^2$. Three- and five-mark questions combine a sector with a clock hand, a wiper, a ring-shaped region or the perimeter. Always write the formula, the substitution and the unit, and state the value of $\pi$ you use.

E1. Board · 1 mark The length of the arc of a sector of angle $72^\circ$ in a circle of radius $5$ cm is (A) $\pi$ cm (B) $2\pi$ cm (C) $4\pi$ cm (D) $5\pi$ cm

💡 Show solution▶

Solve $l=\dfrac{72}{360}\times2\pi\times5=\dfrac15\times10\pi=2\pi$ cm.

Trap (C) $4\pi$ would be the arc of $144^\circ$, twice the angle. (D) $5\pi$ is the length of the semicircle ($180^\circ$).

Answer (B).

E2. Board · 1 mark Assertion (A): If the radius of a sector is doubled and its central angle is kept the same, its area becomes four times as large. Reason (R): The area of a sector is $\dfrac{\theta}{360}\pi r^2$, which is proportional to $r^2$.
(A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

💡 Show solution▶

Check R With $\theta$ fixed, $A=\left(\dfrac{\theta}{360}\pi\right)r^2$: a constant times $r^2$. True.

Check A Replacing $r$ by $2r$ gives $(2r)^2=4r^2$, so the area is multiplied by $4$. True, and R is exactly why.

Answer (A).

E3. Board · 2 marks The minute hand of a clock is $7$ cm long. Find the area swept by it in $10$ minutes. (Use $\pi=\tfrac{22}{7}$.)

💡 Show solution▶

Angle (1 mark) $10\times6^\circ=60^\circ$, fraction $\dfrac16$.

Area (1 mark) $A=\dfrac16\times\dfrac{22}{7}\times7^2=\dfrac16\times154=\dfrac{77}{3}\approx25.67$ cm².

E4. Board · 3 marks A sector of a circle of radius $21$ cm has an arc of length $33$ cm. Find its central angle and its area. (Use $\pi=\tfrac{22}{7}$.)

💡 Show solution▶

Angle (2 marks) $2\pi r=132$ cm, so $\theta=\dfrac{33}{132}\times360=90^\circ$.

Area (1 mark) $A=\tfrac12lr=\tfrac12\times33\times21=346.5$ cm². (Check: $\dfrac14\times1386=346.5$ ✓.)

E5. Board · 5 marks Two concentric arcs of radii $28$ cm and $21$ cm have the same central angle $60^\circ$. Find the area and the perimeter of the region between them. (Use $\pi=\tfrac{22}{7}$.)

💡 Show solution▶

Figure (1 mark) The region is bounded by two arcs and two straight edges of length $28-21=7$ cm.

Area (2 marks) $A=\dfrac16\times\dfrac{22}{7}\times(28^2-21^2)=\dfrac16\times\dfrac{22}{7}\times343=\dfrac{1078}{6}\approx179.67$ cm².

Arcs (1 mark) $\dfrac16\times2\times\dfrac{22}{7}\times28=\dfrac{88}{3}\approx29.33$ cm and $\dfrac16\times132=22$ cm.

Perimeter (1 mark) $29.33+22+2\times7\approx65.33$ cm.

E6. Board · Case study A circular park has radius $35$ m. A sprinkler fixed at the centre waters a sector of angle $72^\circ$ at a time. (Use $\pi=\tfrac{22}{7}$.)
(i) What fraction of the park is watered at a time? (1 mark)
(ii) Find the area of the watered region. (2 marks)
(iii) Find the length of the park boundary that lies in the watered region. (1 mark)

💡 Show solution▶

(i) $\dfrac{72}{360}=\dfrac15$. (1 mark)

(ii) $A=\dfrac15\times\dfrac{22}{7}\times35^2=\dfrac15\times3850=770$ m². (2 marks)

(iii) $l=\dfrac15\times2\times\dfrac{22}{7}\times35=\dfrac15\times220=44$ m. (1 mark)

🔬 Research–Teaching NexusMini project · optional, about 40 minutes

Research question. Can you check the rule "area of a sector $=\dfrac{\theta}{360}\times$ area of the circle" by cutting and weighing card, and how close do you get?

Why it matters. Pie charts, radar screens and speedometer dials are all sectors. A pie chart is drawn by turning each percentage into an angle, $1\%\leftrightarrow3.6^\circ$. Engineers and designers use the same fraction rule to estimate the material in a curved part when a precise drawing is not at hand.

What to do.

Draw a circle of radius $8$ cm on thick card with a compass and cut it out. Weigh it (a kitchen scale in grams works): call this $M$.
Draw a second identical circle. With a protractor mark a sector of $\theta=60^\circ$, then cut out only the sector and weigh it: call this $m_1$.
Compute the fraction $\dfrac{m_1}{M}$ and compare it with $\dfrac{60}{360}=\dfrac16\approx0.167$. Repeat with $\theta=90^\circ$ and $\theta=135^\circ$.
Use the sector's mass to estimate the area of the sector: $\text{area}\approx\dfrac{m}{M}\times\pi r^2$, taking $\pi=3.14$. Compare with $\dfrac{\theta}{360}\pi r^2$.
With a thread laid along the curved edge, measure the arc length and compare with $\dfrac{\theta}{360}\times2\pi r$.

Tools: Compass, protractor, thick card, scissors, a kitchen scale (grams), a thread and a ruler

Hand in: A table of $\theta$, $m$, $m/M$, $\theta/360$, measured and calculated arc length, and a short paragraph on the biggest source of error.

Going further. Take a survey of your class's favourite sports, draw the pie chart with a protractor, and check that all the sectors together add up to $360^\circ$.

Teacher/parent note: the weighing method works because the card has the same thickness everywhere, so mass is proportional to area. Differences of a few percent come from cutting errors and from the card edge; this is a good talking point about measurement error.

🧪 Self-check quiz

0/5 correct

Q1. The length of the arc of a sector of angle $30^\circ$ in a circle of radius $42$ cm is (take $\pi=\tfrac{22}{7}$)

Q2. The area of a quadrant of a circle of radius $14$ cm is (take $\pi=\tfrac{22}{7}$)

Q3. The radius of a sector is tripled and its angle is kept the same. Its area becomes

Q4. The perimeter of a sector of radius $7$ cm and angle $60^\circ$ is (take $\pi=\tfrac{22}{7}$)

Q5. A sector of a circle of radius $14$ cm has area $77$ cm². Its central angle is (take $\pi=\tfrac{22}{7}$)

← Chapter 11 — Areas Related to Circles 11.2 Segments of a Circle →