Worked examples — Circles — centre, radius, diameter, chord, arc, sector, segment

This page is a firing range for circle problems. The parent note (topic index) gave you the tools; here we point them at every kind of target — the friendly ones, the tricky degenerate ones, the word problem, and the exam twist that tries to make you slip.

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The scenario matrix

Everything on this page uses these symbols, all built in the parent note:

• $r$ ::: the radius — distance from centre $O$ to any point on the circle.
• $d=2r$ ::: the diameter.
• $\theta$ ::: the central angle — the angle at $O$ between two radii, measured in radians unless we say "degrees". See Radian Measure.
• Arc length $s=r\theta$; sector area $=\frac{1}{2}{r}^2\theta$; segment area $=\frac{1}{2}{r}^2(\theta -\sin \theta )$.

Cell	Case class	What makes it dangerous	Example
A	Basic parts from $r$	none — warm-up	Ex 1
B	Angle given in degrees	must convert or the formula lies	Ex 2
C	Reflex / major arc ($\theta >180°$)	which arc? subtract from $2\pi$	Ex 3
D	Minor segment (chord cuts arc, $\theta <\pi$)	subtract the triangle, watch $\sin$	Ex 4
D'	Major segment ($\theta >\pi$)	which segment? use the reflex angle	Ex 4b
E	Chord length unknown, use Pythagoras	perpendicular bisects the chord	Ex 5
F	Degenerate: $\theta \to 0$ and $\theta \to 2\pi$	limits, sanity of formulas	Ex 6
G	Real-world word problem	translate words → $r,\theta$	Ex 7
H	Exam twist: work backwards from area to angle	inverse, hidden $\sin$	Ex 8

We now hit each cell.

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Cell A — Basic parts from the radius

Forecast: Guess: will the area be bigger or smaller than $150{\text{m}}^2$? Jot a number before reading on.

1. Diameter. $d=2r=2(7)=14$ m. Why this step? The diameter is a chord through the centre, so it is two radii end-to-end.
2. Circumference. $C=2\pi r=2\pi (7)=14\pi \approx 43.98$ m. Why this step? $\pi$ is defined as circumference $÷$ diameter, so circumference $=\pi d=2\pi r$.
3. Area. $A=\pi r^2=\pi (7{)}^2=49\pi \approx 153.94{\text{m}}^2$. Why this step? Filling the disk with thin rings and adding them (Integration) gives $\pi r^2$; see the parent note.

Verify: Units are m, m, m² — correct for length, length, area. And $153.94>150$, matching a quick "$\pi \times 49\approx 3.14\times 49$" estimate.

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Cell B — Angle in degrees (the conversion trap)

Forecast: If you forgot to convert and plugged $\theta =60$ into $s=r\theta$, would your arc be too big or too small? Guess now.

1. Convert to radians. $\theta =60°\times \frac{\pi }{180}=\frac{\pi }{3}$. Why this step? $s=r\theta$ and $\frac{1}{2}{r}^2\theta$ are only true in radians (that's how radians were built). Plugging $60$ raw would give an arc of $360$ cm — absurd, since the whole circumference is only $\approx 37.7$ cm.
2. Arc length. $s=r\theta =6\times \frac{\pi }{3}=2\pi \approx 6.28$ cm. Why this step? Arc is the fraction $\frac{\theta }{2\pi }$ of the full circumference.
3. Sector area. $A=\frac{1}{2}{r}^2\theta =\frac{1}{2}(36)\frac{\pi }{3}=6\pi \approx 18.85{\text{cm}}^2$. Why this step? Same fraction of the full disk. See Sectors and Pizza Problem.

Verify: $60°$ is $\frac{1}{6}$ of the circle. Full area $=\pi (6{)}^2=36\pi$; a sixth is $6\pi$ ✓. Full circumference $=12\pi$; a sixth is $2\pi$ ✓.

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Cell C — Reflex / major arc (which arc do they mean?)

Forecast: The minor angle is $360°-250°=110°$. Which arc is longer — is $250°$ the big one or the small one? Look at the figure: the plum arc is the reflex (major) one.

1. Identify the correct angle. The major arc corresponds to $\theta =250°$ (the reflex angle). Why this step? $s=r\theta$ uses the angle that actually subtends the arc you want. Using $110°$ would give the minor arc instead.
2. Convert. $\theta =250°\times \frac{\pi }{180}=\frac{25\pi }{18}$. Why this step? Formula needs radians.
3. Arc length. $s=r\theta =10\times \frac{25\pi }{18}=\frac{250\pi }{18}=\frac{125\pi }{9}\approx 43.63$ cm. Why this step? Direct application once the angle is right.

Verify: The two arcs must add to the whole circumference. Minor arc $=10\times \frac{110\pi }{180}=\frac{55\pi }{9}\approx 19.20$ cm. Sum $=\frac{125\pi }{9}+\frac{55\pi }{9}=\frac{180\pi }{9}=20\pi =2\pi (10)$ = full circumference ✓.

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Cell D — Minor segment (chord slices off a cap)

Forecast: The segment is the sector minus a triangle. Will it be more or less than half the sector? Guess.

1. Convert. $\theta =\angle AOB=90°\times \frac{\pi }{180}=\frac{\pi }{2}$. Why this step? Both sector and $\sin$ pieces expect radians (the $\sin$ argument especially). From here on we write $\frac{\pi }{2}$, never $90°$, inside the formulas.
2. Sector area (region bounded by radii $OA$, $OB$ and the minor arc). $A_{\text{sector}}=\frac{1}{2}{r}^2\theta =\frac{1}{2}(64)\frac{\pi }{2}=16\pi \approx 50.27{\text{cm}}^2$. Why this step? This is the full "pizza slice" including the pointy triangle.
3. Triangle area (triangle $OAB$, with the two radii $OA=OB=r$ meeting at angle $\theta$ at $O$). $A_{△OAB}=\frac{1}{2}(OA)(OB)\sin \theta =\frac{1}{2}{r}^2\sin \theta =\frac{1}{2}(64)\sin \left(\frac{\pi }{2}\right)=\frac{1}{2}(64)(1)=32{\text{cm}}^2$. Why this step? $\frac{1}{2}ab\sin \theta$ is the area of a triangle from two sides ($a=OA$, $b=OB$) and their included angle $\theta$; the height off one radius is $r\sin \theta$. We feed $\sin$ the radian value $\frac{\pi }{2}$ (which equals $1$), staying consistent with step 1.
4. Segment. $A_{\text{seg}}=A_{\text{sector}}-A_{△OAB}=16\pi -32\approx 18.27{\text{cm}}^2$. Why this step? The chord $AB$ separates the slice into triangle $OAB$ (inner) + segment (outer cap); remove the triangle.

Verify: $18.27<\frac{1}{2}(50.27)=25.13$, so the segment is less than half the sector — sensible for a mild $90°$ cut. Units cm² ✓.

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Cell D' — Major segment (the other side of the same chord)

Forecast: The whole disk splits into the tiny minor cap ($\approx 18.27{\text{cm}}^2$) plus the huge major part. So the major segment should be nearly the whole disk. Predict roughly.

1. Use the reflex angle. The major segment is subtended by the reflex central angle ${\theta }^{\prime }=360°-90°=270°=\frac{3\pi }{2}$. Why this step? The same formula $\frac{1}{2}{r}^2(\theta -\sin \theta )$ gives whichever segment the angle subtends; the major cap belongs to the reflex angle ${\theta }^{\prime }$.
2. Apply the segment formula with ${\theta }^{\prime }$. $A_{\text{major seg}}=\frac{1}{2}{r}^2({\theta }^{\prime }-\sin {\theta }^{\prime })=\frac{1}{2}(64)\left(\frac{3\pi }{2}-\sin \frac{3\pi }{2}\right)$. Why this step? $\sin \frac{3\pi }{2}=-1$, so the triangle term adds here — a good sanity flag that the formula still works past $\pi$.
3. Evaluate. $A_{\text{major seg}}=32\left(\frac{3\pi }{2}-(-1)\right)=32\left(\frac{3\pi }{2}+1\right)=48\pi +32\approx 182.80{\text{cm}}^2$. Why this step? Plug the numbers straight in.

Verify: Minor segment $+$ major segment must equal the whole disk. $(16\pi -32)+(48\pi +32)=64\pi =\pi (8{)}^2$ = full disk area ✓. And $182.80$ is indeed nearly all of $64\pi \approx 201.06{\text{cm}}^2$ ✓.

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Cell E — Unknown chord length (Pythagoras enters)

Forecast: A chord closer to the centre is longer; one far away is shorter. At distance $0$ it would be a diameter ($26$ cm). So expect an answer somewhere below $26$. Guess a number.

1. Name the pieces and drop a perpendicular. Let the chord have endpoints $A$ and $B$ on the circle. Drop a perpendicular from centre $O$ to the chord, meeting it at the foot $M$; the given distance is $OM=5$ cm. Why this step? The perpendicular from the centre bisects the chord (proved by congruent triangles in the parent note), so $M$ is the midpoint of $AB$ and $AM=MB=\frac{1}{2}AB$.
2. Name the right triangle $OMA$. Its vertices are the centre $O$, the foot $M$, and the endpoint $A$. Its sides are: $OM=5$ (the given distance), $OA=r=13$ (a radius to the endpoint $A$), and $AM$ = half the chord. Why this step? Because $OM\perp AB$, the angle at $M$ is $90°$, so triangle $OMA$ is right-angled with hypotenuse $OA$ — exactly the setting for Pythagorean Theorem.
3. Apply Pythagoras in $OMA$. $AM=\sqrt{O{A}^2-O{M}^2}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$ cm. Why this step? Hypotenuse $OA$, legs $OM$ and $AM$: legs² sum to hypotenuse².
4. Double it. Chord $AB=2\times AM=24$ cm. Why this step? $M$ is the midpoint (step 1), so the full chord is twice one half.

Verify: $24<26$ (less than the diameter) ✓, and it's a clean $5$–$12$–$13$ right triangle. Sanity: if the distance were $0$, we'd get $2\sqrt{169}=26$ = diameter ✓.

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Cell F — Degenerate limits ($\theta \to 0$ and $\theta \to 2\pi$)

Forecast: As the chord shrinks to a point ($\theta \to 0$), the cap should vanish. As the chord swings all the way around ($\theta \to 2\pi$), the "segment" should become the whole disk. Predict the two areas.

The figure above plots $A_{\text{seg}}$ against $\theta$: read it left-to-right. The teal curve starts glued to zero on the far left (the orange dot at $\theta =0$), stays almost flat for small $\theta$ — that is the ${\theta }^3$ crawl of step 2 — then swings up to meet the plum dashed line $\pi r^2$ at the far right (the orange dot at $\theta =2\pi$). Every claim below is just a point you can put your finger on in that picture.

1. Tiny angle $\theta \to 0$. $\sin \theta \to 0$ and $\theta \to 0$, so $A_{\text{seg}}\to \frac{1}{2}{r}^2(0-0)=0$. Why this step? A vanishing central angle means the two radii collapse together — there is no cap. Zero is exactly right, and it is the left-hand orange dot sitting on the axis.
2. Check it's not "just barely" zero. For small $\theta$, $\sin \theta \approx \theta -\frac{{\theta }^3}{6}$, so $\theta -\sin \theta \approx \frac{{\theta }^3}{6}$. Thus $A_{\text{seg}}\approx \frac{r^2{\theta }^3}{12}$ — it dies like ${\theta }^3$, very fast. Why this step? Shows the cap shrinks faster than the sector (which dies like $\theta$), matching the near-flat start of the teal curve just right of the origin.
3. Full turn $\theta \to 2\pi$. $\sin (2\pi )=0$, so $A_{\text{seg}}\to \frac{1}{2}{r}^2(2\pi -0)=\pi r^2=100\pi \approx 314.16$. Why this step? The chord has collapsed to a point again but the arc now wraps the whole circle; the "segment" is the entire disk, area $\pi r^2$. In the figure this is the right-hand orange dot landing exactly on the plum dashed line. ✓

Verify: $\theta \to 0$ gives $0$; $\theta \to 2\pi$ gives $\pi (10{)}^2=100\pi \approx 314.16$ = full disk area ✓. Both extremes match intuition and both are marked on the figure.

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Cell G — Real-world word problem

Forecast: The cleaned region is a sector. Bigger sweep and longer blade → more area. Rough guess before computing?

1. Model it. The wiper is a radius $r=45$ cm; its sweep is a sector of angle $120°$. Why this step? The blade's far tip traces an arc; the region swept is a "pizza slice" — a sector.
2. Convert. $\theta =120°\times \frac{\pi }{180}=\frac{2\pi }{3}$. Why this step? Sector-area formula needs radians.
3. Sector area. $A=\frac{1}{2}{r}^2\theta =\frac{1}{2}(45{)}^2\frac{2\pi }{3}=\frac{1}{2}(2025)\frac{2\pi }{3}=675\pi \approx 2120.58{\text{cm}}^2$. Why this step? Direct sector formula.

Verify: $120°$ is $\frac{1}{3}$ of a full turn, so the answer should be $\frac{1}{3}$ of $\pi r^2=\pi (2025)=2025\pi$. A third is $675\pi$ ✓. Units cm² ✓. (If the blade started $10$ cm from the pivot you'd subtract a smaller sector — but here it sweeps from the pivot.)

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Cell H — Exam twist: work backwards from area to angle

Forecast: We normally go angle → area. Here we're handed the area and must run the machine in reverse. Will the angle be big or small compared to $90°$? Guess.

1. Set up the inverse. From $A=\frac{1}{2}{r}^2\theta$, solve for $\theta$: $\theta =\frac{2A}{r^2}$. Why this step? Every forward formula can be rearranged; here $\theta$ is the unknown, so isolate it.
2. Substitute. $\theta =\frac{2(3\pi )}{6^2}=\frac{6\pi }{36}=\frac{\pi }{6}$ radians. Why this step? Plug the given numbers into the rearranged formula.
3. Convert to degrees. $\theta =\frac{\pi }{6}\times \frac{180}{\pi }=30°$. Why this step? The question asks for degrees; multiply radians by $\frac{180}{\pi }$.
4. Arc length. $s=r\theta =6\times \frac{\pi }{6}=\pi \approx 3.14$ cm. Why this step? Now that $\theta$ is known, arc length is a one-liner.

Verify: Forward-check: sector area $=\frac{1}{2}(6{)}^2\cdot \frac{\pi }{6}=\frac{1}{2}(36)\frac{\pi }{6}=3\pi$ ✓ — back to the given. $30°<90°$ matches "small". Units cm ✓.

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Recall drills

Recall Which arc does

$s=r\theta$ give — minor or major? Whichever arc the angle $\theta$ you plug in subtends. Use $\theta$ for the minor arc, $2\pi -\theta$ for the major arc.

Recall Why must

$\theta$ be in radians for $s=r\theta$ and $\frac{1}{2}{r}^2\theta$? Because a radian is defined so that "arc equals radius times angle"; with that definition the constant of proportionality is exactly $1$. In degrees the constant changes to $\frac{\pi }{180}$, so the clean formulas break unless you convert first.

Recall A chord is

$5$ cm from the centre of a radius-$13$ circle. Length? $2\sqrt{13^2-5^2}=2\sqrt{144}=24$ cm (perpendicular from centre bisects the chord → Pythagoras).

Recall Segment area formula and what each piece is?

$\frac{1}{2}{r}^2(\theta -\sin \theta )$: sector $\frac{1}{2}{r}^2\theta$ minus triangle $\frac{1}{2}{r}^2\sin \theta$.

Recall How do you get the major segment area for a chord whose minor central angle is

$\theta$? Use the reflex angle $2\pi -\theta$ in the same formula: $\frac{1}{2}{r}^2((2\pi -\theta )-\sin (2\pi -\theta ))$. Minor $+$ major segment $=\pi r^2$.

Related tools you may need to chain in: Circle Equations · Circle Theorems · Circumference and Area of Circles · Tangent and Secant Lines.