Arc length and sector area using radians: Worked examples

This page is the "leave no case behind" companion to the parent note. Before we solve anything, we map out every kind of situation the two formulas $s = r θ$ and $A = \frac{1}{2} r^{2} θ$ can throw at you. Then every worked example is tagged with which cell of the map it fills.

Two symbols we will use throughout, defined in plain words first:

The scenario matrix

Each row is a distinct case class. The last column names the example that covers it.

#	Case class	What's given	What's asked	Twist to watch	Example
1	Direct, radians	$r, θ$ (rad)	$s, A$	none — plug in	Ex 1
2	Angle in degrees	$r, θ$ (°)	$s, A$	must convert first	Ex 2
3	Back-solve for angle	$s, r$	$θ$	rearrange $θ = s / r$	Ex 3
4	Back-solve for radius	$A, θ$	$r$	square-root appears	Ex 4
5	Perimeter trap	$r, θ$	$P$	add the two radii	Ex 5
6	Zero / degenerate	$θ = 0$ , or $θ = 2 π$	check limits	edge behaviour	Ex 6
7	Real-world word problem	mixed	speed / length	reading units	Ex 7
8	Exam twist — two unknowns	$s$ and $A$	$r$ and $θ$	solve simultaneously	Ex 8

Recall Which formula "undoes" which?

$θ = s / r$ undoes $s = r θ$ ::: yes — same equation, divided by $r$ . To find $r$ from $A$ and $θ$ you take a ::: square root, because $A \propto r^{2}$ .

Ex 1 — Cell 1: direct, radians

Figure — Arc length and sector area using radians

Ex 2 — Cell 2: angle given in degrees

Worked example Statement

$r = 12$ m, angle $= 3 0^{\circ}$ . Find $s$ and $A$ .

Forecast: $3 0^{\circ}$ is one-twelfth of a full turn. So both $s$ and $A$ should be one-twelfth of the full circumference and area. Keep that as your sanity check.

Step 1 — convert. $3 0^{\circ} = 30 \times \frac{π}{180} = \frac{π}{6}$ rad $\approx 0.5236$ . Why this step? $s = r θ$ was derived with the full circle equal to $2 π$ . If you feed in $30$ , the cancellation breaks and the answer is nonsense. See Radian measure and degree conversion.

Step 2 — arc. $s = 12 \times \frac{π}{6} = 2 π \approx 6.283$ m. Why this step? Straight substitution now that $θ$ is in radians.

Step 3 — area. $A = \frac{1}{2} (12)^{2} \times \frac{π}{6} = \frac{1}{2} \times 144 \times \frac{π}{6} = 12 π \approx 37.70$ m².

Verify: Full circumference $= 2 π r = 24 π$ ; one-twelfth of that is $2 π$ ✔ (matches $s$ ). Full area $= π r^{2} = 144 π$ ; one-twelfth is $12 π$ ✔ (matches $A$ ). The forecast held.

Ex 3 — Cell 3: back-solve for the angle

Worked example Statement

An arc of length $s = 21$ cm lies on a circle of radius $r = 6$ cm. Find $θ$ , and then the sector area.

Forecast: $s = 21 > 6 = r$ , and each radius-length of arc is $1$ radian. $21$ is about $3.5$ radius-lengths, so expect $θ \approx 3.5$ rad (over half a turn, since half a turn is $π \approx 3.14$ ).

Step 1 — angle. $θ = \frac{s}{r} = \frac{21}{6} = 3.5$ rad. Why this step? This is just $s = r θ$ divided by $r$ . It is the definition of a radian: angle = arc ÷ radius.

Step 2 — area (smart route). $A = \frac{1}{2} r s = \frac{1}{2} \times 6 \times 21 = 63$ cm². Why this step? We already have $r$ and $s$ , so $A = \frac{1}{2} r s$ skips re-using $θ$ . (Check it matches: $\frac{1}{2} r^{2} θ = \frac{1}{2} \times 36 \times 3.5 = 63$ ✔.)

Verify: $3.5$ rad is between $π \approx 3.14$ and $2 π \approx 6.28$ , so it's a slice bigger than a half-circle but less than the whole disc — sensible for an arc that's $3.5$ radius-lengths long. ✔

Ex 4 — Cell 4: back-solve for the radius (a square root appears)

Ex 5 — Cell 5: the perimeter trap

Ex 6 — Cell 6: zero and degenerate inputs

Worked example Statement

Take $r = 5$ cm fixed. Evaluate $s$ and $A$ at the two extreme angles: $θ = 0$ and $θ = 2 π$ (a full turn). Do they match what a whole circle should give?

Forecast: At $θ = 0$ there is no slice at all, so both $s$ and $A$ should be $0$ . At $θ = 2 π$ the "sector" is the entire disc, so $s$ should equal the full circumference and $A$ the full disc area.

Step 1 — collapse case, $θ = 0$ . $s = 5 \times 0 = 0 cm, A = \frac{1}{2} (25) (0) = 0 cm^{2} .$ Why this step? Multiplying by $0$ zeroes both. Geometrically the slice has closed up to a line of zero width — no crust, no area. Correct.

Step 2 — full-circle case, $θ = 2 π$ . $s = 5 \times 2 π = 10 π \approx 31.42 cm, A = \frac{1}{2} (25) (2 π) = 25 π \approx 78.54 cm^{2} .$ Why this step? $10 π$ is exactly $2 π r$ (the circumference) and $25 π$ is exactly $π r^{2}$ (the disc area). The formulas agree with the whole circle at the boundary — a strong consistency check.

Verify: $2 π r = 2 π (5) = 10 π$ ✔ and $π r^{2} = π (25) = 25 π$ ✔. Both formulas behave correctly at the degenerate ends.

Ex 7 — Cell 7: real-world word problem

Worked example Statement

A wheel of radius $r = 0.30$ m rolls without slipping. A paint spot on its rim sweeps through an angle of $θ = 4$ rad. (a) How far along the ground does the wheel travel? (b) If it took $2$ seconds, what was the rim's average speed?

Forecast: "Rolls without slipping" means the ground distance equals the arc the rim sweeps. So this is just $s = r θ$ wearing a disguise.

Step 1 — ground distance = arc. $s = r θ = 0.30 \times 4 = 1.2$ m. Why this step? For rolling without slipping, the point of contact lays down exactly the arc length onto the ground — no sliding. So distance travelled $= s$ .

Step 2 — speed. Speed $= \frac{distance}{time} = \frac{1.2}{2} = 0.6$ m/s. Why this step? Average speed is distance over time. This links to Angular velocity, where $v = r ω$ mirrors $s = r θ$ exactly ( $ω = θ / t = 4/2 = 2$ rad/s, and $r ω = 0.30 \times 2 = 0.6$ m/s ✔).

Verify: $r ω = 0.30 \times 2 = 0.6$ m/s matches the direct calculation ✔. Units: m ÷ s = m/s ✔.

Arc length and sector area using radians (parent)
Radian measure and degree conversion — needed in Ex 2
Circumference and area of a circle — the boundary check in Ex 6
Small-angle approximations — the thin-slice limit
Angular velocity — the wheel in Ex 7
Segment area — next step: sector minus triangle

Worked examples — Arc length and sector area using radians

The scenario matrix

Ex 1 — Cell 1: direct, radians

Ex 2 — Cell 2: angle given in degrees

Ex 3 — Cell 3: back-solve for the angle

Ex 4 — Cell 4: back-solve for the radius (a square root appears)

Ex 5 — Cell 5: the perimeter trap

Ex 6 — Cell 6: zero and degenerate inputs

Ex 7 — Cell 7: real-world word problem

Ex 8 — Cell 8: exam twist, two unknowns from $s$ and $A$

Wrap-up: the whole matrix in one picture

Connections