Worked examples — Circumference and area of a circle

This is the hands-on companion to Circumference and area of a circle. The parent note gave you the two engines:

$C=2\pi rA=\pi r^2$

Here we do not learn new formulas. Instead we hunt down every kind of situation a problem can put in front of you — and solve one of each, slowly, out loud. If you meet a circle question on an exam and panic, it will be because it lives in one of the cells below. So let us fill every cell.

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The scenario matrix

Read this table as: "What am I given, and what makes this case different from the others?"

#	Case class	What you are given	The twist	Example
A	Forward, from radius	radius $r$	none — plug in	Ex 1
B	Forward, from diameter	diameter $d$	must halve first	Ex 2
C	Backward, from circumference	$C$	solve for $r$	Ex 3
D	Backward, from area	$A$	square-root step	Ex 4
E	Ring / annulus	two radii	subtract areas	Ex 5
F	Degenerate	$r=0$ or $r\to \infty$	limiting behaviour	Ex 6
G	Real-world word problem	rotations, distance	translate words to $C$	Ex 7
H	Exam twist (ratio)	scale factor	how $C$ and $A$ scale	Ex 8

Every column of "twist" is a place students slip. We will hit all eight.

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Case A — Forward from radius

Forecast: Circumference is a distance, so its answer will be in cm. Area is a space, so cm². Which do you expect to be the bigger number? Guess before reading.

1. Circumference. $C=2\pi r=2\pi (12)=24\pi \approx 75.40\text{ cm}$. Why this step? We are given $r$ directly, so the machine $2\pi r$ runs with no preparation.
2. Area. $A=\pi r^2=\pi (12{)}^2=144\pi \approx 452.39{\text{ cm}}^2$. Why this step? Area needs $r$ squared; the exponent is what makes the answer two-dimensional.

Verify: Units check — cm for $C$, cm² for $A$. Sanity: $C\approx 75$, and area $\approx 452$; the area number is larger, which fits because squaring 12 (→144) grows faster than doubling it (→24). Good.

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Case B — Forward from diameter

Forecast: There is a trap here. Will you plug 25 into $\pi r^2$? Pause — the formula wants $r$, not $d$.

1. Halve the diameter. $r=\frac{d}{2}=\frac{25}{2}=12.5\text{ mm}$. Why this step? $A=\pi r^2$ is written in terms of the radius. The single most common mistake is skipping this line.
2. Apply the area formula. $A=\pi r^2=\pi (12.5{)}^2=156.25\pi \approx 490.87{\text{ mm}}^2$. Why this step? Now the input is genuinely a radius, so the machine is safe.

Verify: If we had wrongly used $r=25$ we'd get $625\pi$ — that is $4\times$ too big. Why exactly 4? Because doubling the radius quadruples the area ($2^2=4$). Catching that factor of 4 is the sanity check that tells you the halving step was necessary.

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Case C — Backward from circumference

Forecast: We know $C=2\pi r$. This time $C$ is known and $r$ is the mystery. We must undo the multiplication.

1. Write the forward relation. $C=2\pi r$. Why this step? You always start from the equation that connects the two quantities.
2. Divide both sides by $2\pi$. $r=\frac{C}{2\pi }=\frac{400}{2\pi }=\frac{200}{\pi }\approx 63.66\text{ m}$. Why this step? $r$ was multiplied by $2\pi$, so dividing by $2\pi$ isolates it — the inverse operation of multiplication.

Verify: Plug back: $2\pi (63.66)\approx 400.0\text{ m}$. ✓ Units: metres in, metres out (dividing a length by the pure number $2\pi$ keeps it a length). Good.

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Case D — Backward from area (the square-root case)

Forecast: Going backward from area is harder than from circumference, because $r$ is squared. What operation undoes squaring?

1. Start from the area formula. $A=\pi r^2$. Why this step? It links the given ($A$) to the unknown ($r$).
2. Isolate $r^2$. $r^2=\frac{A}{\pi }=\frac{314}{\pi }\approx 99.95$. Why this step? Divide out $\pi$ to leave $r^2$ alone.
3. Take the positive square root. $r=\sqrt{99.95}\approx 10.00\text{ cm}$. Why this step? Square root undoes squaring — the tool that answers "which length, squared, gives this area?" We keep only the positive root because a radius cannot be negative.
4. Now the circumference. $C=2\pi r=2\pi (10)\approx 62.83\text{ cm}$.

Verify: Feed $r=10$ into $\pi r^2=100\pi \approx 314.16$ — matches the given 314 (which was itself a rounding of $100\pi$). ✓ Notice we needed two inverse steps (divide, then square-root); area problems always cost one step more than circumference problems.

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Case E — Ring / annulus (subtract two circles)

Forecast: The paving is not a circle — it is a circle with a bite taken out. So one area alone will not do. What two areas do you need?

1. Find both radii. Inner $r_1=3\text{ m}$. Outer $r_2=3+1.5=4.5\text{ m}$. Why this step? The paving's outer edge sits $1.5\text{ m}$ beyond the fountain edge, so we add the width to reach $r_2$.
2. Area of the big disc. $A_2=\pi r_2^2=\pi (4.5{)}^2=20.25\pi {\text{ m}}^2$.
3. Area of the fountain (the hole). $A_1=\pi r_1^2=\pi (3{)}^2=9\pi {\text{ m}}^2$.
4. Subtract. $A_{\text{ring}}=A_2-A_1=20.25\pi -9\pi =11.25\pi \approx 35.34{\text{ m}}^2$. Why this step? The ring is exactly "everything inside the outer circle, minus everything inside the inner circle." Subtraction removes the hole.

Verify: The ring must be smaller than the big disc ($35.34<63.6$) and it is. Also $11.25\pi$ is positive — if it came out negative you would have swapped $r_1$ and $r_2$. ✓ See the shaded region below.

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Case F — Degenerate and limiting inputs

Forecast: A "circle" of radius 0 is just a single point. Do the formulas agree with that picture?

1. Radius zero. $C=2\pi (0)=0$ and $A=\pi (0{)}^2=0$. Why this step? A point has no length around it and no space inside — both should collapse to zero, and they do. The formulas behave sensibly at the degenerate input.
2. Compare growth. $C=2\pi r$ is linear in $r$; $A=\pi r^2$ is quadratic in $r$. Why this step? Doubling $r$ doubles $C$ (factor $2$) but multiplies $A$ by $2^2=4$. As $r\to \infty$, the ratio $\frac{A}{C}=\frac{\pi r^2}{2\pi r}=\frac{r}{2}$ grows without bound.
3. Read the meaning. For big circles, area dwarfs circumference; this is why large containers hold far more than their rim suggests, and why a giant pizza is a better deal per cm of crust.

Verify: At $r=1$: $C=2\pi \approx 6.28$, $A=\pi \approx 3.14$, so here $C>A$. At $r=10$: $C\approx 62.8$, $A\approx 314$, now $A>C$. The crossover happens when $2\pi r=\pi r^2\Rightarrow r=2$. ✓ The graph below shows the linear line crossing the quadratic curve at $r=2$.

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Case G — Real-world word problem (rotations)

Forecast: A rolling wheel covers one circumference of ground per turn. So the hidden quantity we must compute first is $C$.

1. One turn = one circumference. $C=\pi d=\pi (60)=60\pi \text{ cm}\approx 188.50\text{ cm}$. Why this step? Rolling without slipping lays the rim flat on the road; the distance per rotation is the perimeter. (See Wheel and Circular Motion.)
2. Convert the trip to the same units. $1\text{ km}=1000\text{ m}=100000\text{ cm}$. Why this step? You can only divide distances that share units.
3. Divide total distance by distance-per-turn. $\text{turns}=\frac{100000}{60\pi }\approx \frac{100000}{188.50}\approx 530.5$. Why this step? Each turn eats one $C$ of road, so the number of turns is how many $C$s fit into the trip.

Verify: $530.5$ turns $\times 188.5\text{ cm}\approx 100000\text{ cm}=1\text{ km}$. ✓ Units cancel cleanly: $\frac{\text{cm}}{\text{cm/turn}}=\text{turns}$. About 530 full rotations.

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Case H — Exam twist: scaling

Forecast: Scaling problems trip people because circumference and area scale by different factors. Guess both before reading.

1. Set up the ratio for circumference. Let $r_P=r$, so $r_Q=2r$. Then $\frac{C_Q}{C_P}=\frac{2\pi (2r)}{2\pi r}=2$. Why this step? Circumference is linear in $r$, so the length scales by the same factor as the radius — here $\times 2$. (This is the Similar Figures rule for lengths.)
2. Set up the ratio for area. $\frac{A_Q}{A_P}=\frac{\pi (2r{)}^2}{\pi r^2}=\frac{4r^2}{r^2}=4$. Why this step? Area is quadratic in $r$, so it scales by the square of the length factor: $2^2=4$.

Verify: Concrete numbers — take $r_P=5$: $C_P=10\pi$, $A_P=25\pi$; and $r_Q=10$: $C_Q=20\pi$ (indeed $2\times$), $A_Q=100\pi$ (indeed $4\times$). ✓ Rule to remember: length scales by $k$, area by $k^2$.

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Recall

Recall Which case is which?

Given the circumference and asked for the radius — which inverse step? ::: Divide by $2\pi$: $r=C/(2\pi )$. Given the area and asked for the radius — which extra step vs. the circumference case? ::: A square root, because $r$ is squared in $A=\pi r^2$. A ring's area — one formula or two? ::: Two: $\pi r_2^2-\pi r_1^2$ (big disc minus the hole). If radius triples, area multiplies by? ::: $3^2=9$. Distance a wheel rolls in one turn? ::: Its circumference, $C=\pi d=2\pi r$.

See also: Radius and Diameter · Pi (π) and Irrational Numbers · Perimeter and Area · Arc Length · Sector Area · Radians · Integration