Exercises — Arc length and sector area using radians

WHAT: we want arc length, so we reach for $s=r\theta$. WHY this formula: the angle is already in radians (it says "rad"), so no conversion is needed — plug straight in. $s=r\theta =4\times 2=8\text{ cm}.$ Sanity check: $\theta =2$ rad is a little under a third of a full turn ($2\pi \approx 6.28$), and the arc came out to twice the radius — reasonable.

WHAT: area of the "pizza slice," so use $A=\frac{1}{2}{r}^2\theta$. WHY the $\frac{1}{2}$: the disc area $\pi r^2$ divided by the full angle $2\pi$ leaves a factor of $\frac{1}{2}$ — it never disappears (unlike in arc length). $A=\frac{1}{2}(4{)}^2(2)=\frac{1}{2}\times 16\times 2=16{\text{ cm}}^2.$

WHAT: we know $s$ and $r$, want $\theta$. Rearrange $s=r\theta$. WHY: dividing both sides by $r$ isolates $\theta$, and this rearrangement $\theta =s/r$ is literally the definition of a radian — angle equals arc measured in radius-lengths. $\theta =\frac{s}{r}=\frac{21}{7}=3\text{ rad}.$

WHAT: arc length, but the angle is in degrees, so we must convert before touching $s=r\theta$. WHY convert: the formula was derived assuming a full turn is $2\pi$. Feed it $60$ and the cancellation breaks — you'd get nonsense. $\theta =60\times \frac{\pi }{180}=\frac{\pi }{3}\text{ rad}.$ $s=r\theta =12\times \frac{\pi }{3}=4\pi \approx 12.57\text{ m}.$ Sanity check: $\frac{\pi }{3}\approx 1.05$ rad — a "small-ish" number, exactly what a radian angle should look like (not 60).

WHAT: we know $A$ and $r$, want $\theta$. Rearrange $A=\frac{1}{2}{r}^2\theta$. WHY: multiply both sides by $2$ and divide by $r^2$ to free $\theta$. $\theta =\frac{2A}{r^2}=\frac{2\times 30}{25}=\frac{60}{25}=2.4\text{ rad}.$

WHAT (perimeter): the boundary of a sector is two straight radii plus the curved arc — look at the red outline in the figure. $P=2r+s=2(8)+6=22\text{ cm}.$ WHAT (area): we have $r$ and $s$ directly, so the smart move is $A=\frac{1}{2}rs$ — no need to find $\theta$ first. WHY this shortcut: since $s=r\theta$, we have $\frac{1}{2}{r}^2\theta =\frac{1}{2}r(r\theta )=\frac{1}{2}rs$. It saves a division. $A=\frac{1}{2}(8)(6)=24{\text{ cm}}^2.$

WHAT (a): $s=r\theta$ is linear in $r$. Replace $r$ by $3r$: $s_{\text{new}}=(3r)\theta =3(r\theta )=3s\Rightarrow \times 3.$ WHAT (b): $A=\frac{1}{2}{r}^2\theta$ depends on $r^2$. Replace $r$ by $3r$: $A_{\text{new}}=\frac{1}{2}(3r{)}^2\theta =\frac{1}{2}\cdot 9r^2\theta =9A\Rightarrow \times 9.$ WHY the difference: length lives in one direction, so it scales like $r$; area covers two directions, so it scales like $r^2=3^2=9$.

WHAT: set the two areas equal and solve for ${\theta }_2$. WHY: "equal area" is an equation $\frac{1}{2}{r}_1^2{\theta }_1=\frac{1}{2}{r}_2^2{\theta }_2$; the $\frac{1}{2}$ cancels on both sides. $r_1^2{\theta }_1=r_2^2{\theta }_2\Rightarrow 36\times 2=16\times {\theta }_2\Rightarrow {\theta }_2=\frac{72}{16}=4.5\text{ rad}.$ Sense check: the smaller circle needs a wider angle to sweep the same area — and indeed $4.5>2$.

Step 1 — isolate the arc. The perimeter is $P=s+2r$, so the arc is what's left after removing the two radii. $s=P-2r=28-2(10)=28-20=8\text{ cm}.$ Step 2 — angle. Use $\theta =s/r$ (the radian definition). $\theta =\frac{8}{10}=0.8\text{ rad}.$ Step 3 — area. Fastest via $A=\frac{1}{2}rs$ since we have both. $A=\frac{1}{2}(10)(8)=40{\text{ cm}}^2.$

WHAT: the wire is the whole boundary, so $P=r\theta +2r=24$. WHY factor: both terms share $r$, so factor it out to solve in one step. $r(\theta +2)=24\Rightarrow r=\frac{24}{1.4+2}=\frac{24}{3.4}\approx 7.0588\text{ cm}.$ Area: $A=\frac{1}{2}{r}^2\theta =\frac{1}{2}(7.0588{)}^2(1.4)\approx \frac{1}{2}(49.827)(1.4)\approx 34.88{\text{ cm}}^2.$

WHAT: total disc area minus the eaten sector. WHY subtract: "remains" = whole $-$ removed. Whole disc uses $\pi R^2$; the slice uses $\frac{1}{2}{R}^2\theta$. $A_{\text{disc}}=\pi (9{)}^2=81\pi \approx 254.47{\text{ cm}}^2.$ $A_{\text{slice}}=\frac{1}{2}(81)(0.6)=24.3{\text{ cm}}^2.$ $A_{\text{remain}}=81\pi -24.3\approx 230.17{\text{ cm}}^2.$

Step 1 — convert. $s=r\theta$ needs radians. $\theta =120\times \frac{\pi }{180}=\frac{2\pi }{3}\text{ rad}.$ Step 2 — arc length. $s=r\theta =30\times \frac{2\pi }{3}=20\pi \approx 62.83\text{ m}.$ Step 3 — time. Distance $÷$ speed. $t=\frac{s}{5}=\frac{62.83}{5}\approx 12.57\text{ s}.$ This links straight to Angular velocity: the runner's angular speed is $\omega =\theta /t=\frac{2\pi /3}{12.57}\approx 0.167$ rad/s, and $v=r\omega =30\times 0.167\approx 5$ m/s ✓.

Step 1 — radius from $A=\frac{1}{2}rs$. This formula uses only $A$ and $s$ (both given), so it hands us $r$ directly. $50=\frac{1}{2}r(20)=10r\Rightarrow r=\frac{50}{10}=5\text{ cm}.$ Step 2 — angle from $\theta =s/r$. $\theta =\frac{20}{5}=4\text{ rad}.$ Cross-check: $A=\frac{1}{2}{r}^2\theta =\frac{1}{2}(25)(4)=50$ ✓.

Step 1 — the main sweep. At the tie-corner the barn blocks a right-angle ($90^{\circ }$), leaving $360^{\circ }-90^{\circ }=270^{\circ }$. ${\theta }_1=270\times \frac{\pi }{180}=\frac{3\pi }{2}\text{ rad}.$ $A_1=\frac{1}{2}{r}^2{\theta }_1=\frac{1}{2}(10{)}^2\left(\frac{3\pi }{2}\right)=\frac{1}{2}(100)\left(\frac{3\pi }{2}\right)=75\pi {\text{ m}}^2.$ Step 2 — the two wrap-around quarter-circles. Rope left after reaching a neighbour corner is $10-6=4$ m; each sweeps a quarter turn ($\frac{\pi }{2}$ rad). $A_2=\frac{1}{2}(4{)}^2\left(\frac{\pi }{2}\right)=\frac{1}{2}(16)\left(\frac{\pi }{2}\right)=4\pi {\text{ m}}^2\text{ (each)}.$ Two of them: $2\times 4\pi =8\pi$. Step 3 — total. $A=75\pi +8\pi =83\pi \approx 260.75{\text{ m}}^2.$ WHY three sectors: each stretch of rope, after clearing a corner, becomes the radius of a new sector — the whole area is a sum of independent slices. Look at the shaded regions in the figure.