🎯⭐ INTERACTIVE LESSON

Arc Length and Sector Area

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Arc Length and Sector Area - Complete Interactive Lesson

Part 1: Radians: The Natural Angle

🌀 Arc Length and Sector Area

Part 1 of 7 — Radians: The Natural Angle

Topics in This Part

Section
What Is a Radian?
Why Radians Make Arc Length Easy
Converting Between Degrees and Radians

🔑 Key Concept: A radian measures an angle by the arc it sweeps on a circle. One radian is the angle whose arc length equals the radius. This single idea makes the whole topic — arc length, sector area, and circular speed — fall into place.

What Is a Radian?

Picture a circle of radius $r$ . If you walk along the edge a distance of exactly one radius, the angle you sweep at the center is one radian.

$1 \text{ radian} = \text{the angle whose arc length} = r$

Because the full circumference is $2\pi r$ , walking all the way around sweeps $\dfrac{2\pi r}{r} = 2\pi$ radians. So:

Fraction of circle	Degrees	Radians
Full turn	$360^\circ$	$2\pi$
Half turn	$180^\circ$

🔑 The bridge between systems: $\;180^\circ = \pi \text{ radians}$ . Every conversion comes from this one equation.

Why Radians Make Arc Length Easy

In radians, the angle and the arc are directly proportional to the radius. Walking $\theta$ radians around a circle of radius $r$ covers an arc of

$s = r\theta$

That clean formula is only true when $\theta$ is in radians. In degrees you'd carry an awkward factor of $\frac{}{}$ everywhere — which is exactly why radians are the natural unit for circular measurement.

Concept Check 🎯

Converting Degrees ↔ Radians

Start from $180^\circ = \pi$ radians and multiply by the right conversion factor:

$\text{degrees} \to \text{radians: } \times \frac{\pi}{180^\circ} \qquad \text{radians} \to \text{degrees: } \times \frac{180^\circ}{\pi}$

Match the Angles 🔽

Choose the correct equivalent for each angle.

Convert It 🧮

Convert each angle. For radian answers, give the coefficient of $\pi$ as a fraction (e.g. write $\frac{2}{3}$ for $\frac{2\pi}{3}$ ).

Part 2: Arc Length: $s = r\theta$

🌀 Arc Length and Sector Area

Part 2 of 7 — Arc Length: $s = r\theta$

🔑 The Idea: An arc is a piece of a circle's edge. Its length is the radius times the central angle, when the angle is measured in radians: $s = r\theta$

The Arc Length Formula

Symbol	Meaning	Unit

Part 3: Sector Area: $A = \tfrac{1}{2}r^2\theta$

🌀 Arc Length and Sector Area

Part 3 of 7 — Sector Area: $A = \tfrac{1}{2}r^2\theta$

🔑 The Idea: A sector is a "pizza slice" of a circle — the region between two radii and the arc between them. Its area is

Part 4: The Degree Formulas (and the "Fraction" Method)

🌀 Arc Length and Sector Area

Part 4 of 7 — The Degree Formulas (and the "Fraction" Method)

🔑 The Idea: You don't always have to convert. There are direct degree formulas built on the same "fraction of the circle" logic — a $\theta^\circ$ sector is $\dfrac{\theta}{360}$ of the whole circle.

Part 5: Angular & Linear Speed

🌀 Arc Length and Sector Area

Part 5 of 7 — Angular & Linear Speed

🔑 The Idea: When something spins — a wheel, a clock hand, a record — it has two speeds. Angular speed $\omega$ is how fast the angle changes; linear speed $v$ is how fast a point on the edge travels. They are linked by $v = r\omega$ .

Two Kinds of Speed

Quantity	Symbol	Definition

Part 6: Applications & Solving Backwards

🌀 Arc Length and Sector Area

Part 6 of 7 — Applications & Solving Backwards

🔑 The Idea: Real problems often give you the arc or area and ask for the radius or angle. The same two formulas work — just rearrange. And many shapes are combinations of sectors and triangles.

Solving for the Unknown

Find the angle: An arc of length $15$ sits on a circle of radius $6$ . What central angle (in radians) does it subtend? $\theta = \frac{s}{r} = \frac{15}{6} = 2.5 \text{ radians}$

Part 7: Mixed Practice & Mastery Check

🌀 Arc Length and Sector Area

Part 7 of 7 — Mixed Practice & Mastery Check

You can now (1) convert between degrees and radians, (2) find arc length with $s = r\theta$ , (3) find sector area with $A = \frac{1}{2}r^2\theta$ , (4) use the degree formulas, and (5) handle angular and linear speed. Let's put it all together.

💡 Think of $\theta$ as "how many radius-lengths of arc." If $\theta = 3$ radians on a circle of radius $5$ , you've traveled $3$ radius-lengths $= 3 \cdot 5 = 15$ units of arc.

Example — degrees to radians: Convert $60^\circ$ . $60^\circ \cdot \frac{\pi}{180^\circ} = \frac{60\pi}{180} = \frac{\pi}{3}$

Example — radians to degrees: Convert $\dfrac{3\pi}{4}$ . $\frac{3\pi}{4} \cdot \frac{180^\circ}{\pi} = \frac{3 \cdot 180^\circ}{4} = 135^\circ$

⚠️ Pick the factor that cancels the unit you start with. Degrees in the denominator cancels degrees; $\pi$ in the denominator cancels radians.

1) $45^\circ = \,?\,\pi$ radians (enter the fraction) 2) $\dfrac{5\pi}{6}$ radians $= \,?^\circ$ 3) $120^\circ = \,?\,\pi$ radians (enter the fraction)

$\boxed{\,s = r\theta\,}$

Worked Example: Find the arc length cut off by a $\dfrac{\pi}{3}$ angle on a circle of radius $12$ cm.

$s = r\theta = 12 \cdot \frac{\pi}{3} = \frac{12\pi}{3} = 4\pi \approx 12.57 \text{ cm}$

⚠️ Non-negotiable rule: $\theta$ must be in radians. If you're given degrees, convert first — otherwise $s = r\theta$ gives nonsense.

When the Angle Is in Degrees

Worked Example: A circle has radius $9$ m. Find the arc length for a $40^\circ$ central angle.

Step 1 — Convert to radians: $40^\circ \cdot \frac{\pi}{180^\circ} = \frac{40\pi}{180} = \frac{2\pi}{9}$

Step 2 — Apply $s = r\theta$ : $s = 9 \cdot \frac{2\pi}{9} = 2\pi \approx 6.28 \text{ m}$

💡 Notice how the radius $9$ cancelled the $9$ in the denominator. Keeping $\pi$ exact until the end keeps the arithmetic clean.

Concept Check 🎯

Compute the Arc 🧮

Use $s = r\theta$ . Give exact answers as a coefficient of $\pi$ (e.g. enter $6$ for $6\pi$ ).

1) $r = 8$ , $\theta = \dfrac{\pi}{4}$ : $\;s = \,?\,\pi$ 2) $r = 15$ , $\theta = \dfrac{2\pi}{5}$ : $\;s = \,?\,\pi$ 3) $r = 6$ , $\theta = 60^\circ$ (convert first!): $\;s = \,?\,\pi$

Solve for the Missing Piece 🔽

The formula $s = r\theta$ can be rearranged. Pick the right rearrangement and value.

A = \frac{1}{2}r^2\theta \quad (\theta \text{ in radians})

Where the Sector Formula Comes From

A full circle ( $\theta = 2\pi$ ) has area $\pi r^2$ . A sector is the fraction $\dfrac{\theta}{2\pi}$ of the whole circle:

$A = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{\theta \, r^2}{2} = \frac{1}{2}r^2\theta$

Worked Example: Find the area of a sector with radius $10$ and central angle $\dfrac{\pi}{5}$ .

$A = \frac{1}{2}r^2\theta = \frac{1}{2}(10)^2 \cdot \frac{\pi}{5} = \frac{1}{2}(100)\cdot\frac{\pi}{5} = \frac{100\pi}{10} = 10\pi \approx 31.4$

⚠️ The radius is squared here — a common slip is forgetting the square or forgetting the $\dfrac{1}{2}$ . Both factors matter.

Concept Check 🎯

Sector Area From Degrees

Worked Example: A sprinkler sprays water over a $90^\circ$ wedge with reach $20$ ft. What area does it water?

Step 1 — Convert: $90^\circ = \dfrac{\pi}{2}$ radians.

Step 2 — Apply the formula: $A = \frac{1}{2}r^2\theta = \frac{1}{2}(20)^2 \cdot \frac{\pi}{2} = \frac{1}{2}(400)\cdot\frac{\pi}{2} = 100\pi \approx 314 \text{ ft}^2$

💡 Sanity check: A full circle of radius $20$ has area $400\pi$ . A $90^\circ$ sector is one quarter of that: $\dfrac{400\pi}{4} = 100\pi$ ✓

Compute the Sector Area 🧮

Use $A = \dfrac{1}{2}r^2\theta$ . Give exact answers as a coefficient of $\pi$ (e.g. enter $9$ for $9\pi$ ).

1) $r = 4$ , $\theta = \dfrac{\pi}{2}$ : $\;A = \,?\,\pi$ , : , :

Compare the Two Formulas 🔽

Match each quantity to its correct formula or fact.

Degree Versions

If $\theta$ is in degrees, the sector is $\dfrac{\theta}{360^\circ}$ of the circle, so:

$s = \frac{\theta}{360^\circ} \cdot 2\pi r \qquad\qquad A = \frac{\theta}{360^\circ} \cdot \pi r^2$

Quantity	Radian formula	Degree formula
Arc length	$s = r\theta$	$s = \dfrac{\theta}{360}\cdot 2\pi r$

💡 Two valid roads, same destination. Convert-then-use-radians, or use the degree formula directly. Pick whichever feels cleaner for the numbers in front of you.

Worked Example — Both Roads

A circle has radius $18$ . Find the arc length for a $30^\circ$ central angle.

Road A — convert first: $30^\circ = \dfrac{\pi}{6}$ , so $s = r\theta = 18 \cdot \frac{\pi}{6} = 3\pi \approx 9.42$

Road B — degree formula directly: $s = \frac{30}{360}\cdot 2\pi(18) = \frac{1}{12}\cdot 36\pi = 3\pi \approx 9.42$

✅ Same answer, $3\pi$ . The fraction $\dfrac{30}{360} = \dfrac{1}{12}$ tells you the arc is one-twelfth of the full circumference .

Concept Check 🎯

Use the Degree Formula 🧮

Give exact answers as a coefficient of $\pi$ (e.g. enter $5$ for $5\pi$ ).

1) Arc length, $\theta = 60^\circ$ , $r = 12$ : $\;s = \,?\,\pi$ 2) Sector area, $\theta = 90^\circ$ , $r = 8$ : $\;A = \,?\,\pi$ 3) Sector area, $\theta = 45^\circ$ , $r = 4$ : $\;A = \,?\,\pi$

Pick the Fraction 🔽

Each degree measure represents what fraction of a full circle?

Because arc length is $s = r\theta$ , dividing by time $t$ gives the link:

$\frac{s}{t} = r\cdot\frac{\theta}{t} \;\Longrightarrow\; \boxed{v = r\omega}$

🔑 A point farther from the center (larger $r$ ) has a greater linear speed even though every point shares the same angular speed $\omega$ . That's why the outer edge of a merry-go-round whips faster than the middle.

Worked Example — Angular Speed

A wheel makes $3$ full revolutions per second. Find its angular speed in radians per second.

Each revolution is $2\pi$ radians, so: $\omega = 3 \text{ rev/s} \cdot 2\pi \text{ rad/rev} = 6\pi \text{ rad/s} \approx 18.85 \text{ rad/s}$

Worked Example — Linear Speed

A wheel of radius $0.5$ m spins at $\omega = 6\pi$ rad/s. How fast does a point on the rim travel? $v = r\omega = 0.5 \cdot 6\pi = 3\pi \text{ m/s} \approx 9.42 \text{ m/s}$

⚠️ For $v = r\omega$ , the angular speed $\omega$ must be in radians per unit time. Convert revolutions or degrees to radians first.

Concept Check 🎯

Speed Calculations 🧮

Give exact answers as a coefficient of $\pi$ where indicated (e.g. enter $6$ for $6\pi$ ).

1) A wheel spins at $5$ rev/s. Angular speed $\omega = \,?\,\pi$ rad/s 2) $r = 2$ m, $\omega = 4\pi$ rad/s. Linear speed $v = \,?\,\pi$ m/s 3) $r = 3$ m, $\omega = 10$ rad/s. Linear speed $v = \,?$ m/s (plain number)

Speed Relationships 🔽

Find the radius from area: A sector of area $24\pi$ has central angle $\dfrac{\pi}{3}$ . Find $r$ . $A = \frac{1}{2}r^2\theta \;\Rightarrow\; 24\pi = \frac{1}{2}r^2\cdot\frac{\pi}{3} = \frac{\pi r^2}{6}$ $r^2 = \frac{24\pi \cdot 6}{\pi} = 144 \;\Rightarrow\; r = 12$

💡 When solving for $r$ from area, you'll take a square root at the end (radius is positive, so keep the $+$ root).

Concept Check 🎯

A Combined Shape: The Segment

A segment is the region between a chord and its arc — it's the sector minus the triangle formed by the two radii.

$A_{\text{segment}} = \underbrace{\frac{1}{2}r^2\theta}_{\text{sector}} - \underbrace{\frac{1}{2}r^2\sin\theta}_{\text{triangle}} = \frac{1}{2}r^2(\theta - \sin\theta)$

Worked Example: Radius $10$ , central angle $\dfrac{\pi}{2}$ . $A = \frac{1}{2}(10)^2\left(\frac{\pi}{2} - \sin\frac{\pi}{2}\right) = 50\left(\frac{\pi}{2} - 1\right) = 25\pi - 50 \approx 28.5$

💡 The triangle's area uses $\frac{1}{2}r^2\sin\theta$ — the "two sides and included angle" formula, where both sides are the radius.

Solve Backwards 🧮

1) Arc $s = 21$ on radius $r = 7$ . Central angle $\theta = \,?$ rad (plain number) 2) Sector area $50\pi$ with angle $\theta = \dfrac{\pi}{4}$ . Find $r$ . $\;r = \,?$ (plain number) 3) Sector area $9\pi$ with radius $r = 6$ . Find $\theta$ as a coefficient of $\pi$ . $\;\theta = \,?\,\pi$

Choose the Right Move 🔽

For each goal, pick the correct rearranged formula.

Quick Reference

Goal	Formula	Angle unit
Degrees → radians	$\times \dfrac{\pi}{180}$	—
Radians → degrees	$\times \dfrac{180}{\pi}$	—
Arc length	$s = r\theta$	radians
Sector area	$A = \frac{1}{2}r^2\theta$	radians
Arc length (degrees)	$s = \dfrac{\theta}{360}\cdot 2\pi r$	degrees
Sector area (degrees)	$A = \dfrac{\theta}{360}\cdot \pi r^2$	degrees
Linear speed	$v = r\omega$	$\omega$ in rad/time

⚠️ The #1 rule of this topic: the radian formulas ( $s = r\theta$ , $A = \frac{1}{2}r^2\theta$ , ) demand . Always check the unit of your angle before plugging in.

Mixed Practice 🎯

Mixed Drill 🧮

Give answers as a coefficient of $\pi$ unless noted (e.g. enter $3$ for $3\pi$ ).

1) Convert $135^\circ$ to radians: $\,?\,\pi$ (enter the fraction) 2) Arc length, $r = 21$ , $\theta = \dfrac{2\pi}{3}$ : $\,?\,\pi$ 3) Linear speed, $r = 4$ m, $\omega = 2.5$ rad/s: $\,?$ m/s (plain number)

Concept Wrap-Up 🔽

Exit Quiz ✅

Answer all three to finish the lesson.

\theta = \dfrac{2\pi}{9}

A = \frac{1}{2} (10)^{2} (\frac{π}{2} - sin \frac{π}{2}) = 50 (\frac{π}{2} - 1) = 25 π - 50 \approx 28.5

Angular speed	$\omega$	angle swept per time	radians/second
Linear speed	$v$	arc distance per time	meters/second