Exercises — Circumference and area of a circle

WHAT we want: distance around $\Rightarrow$ that is $C$, so use $C=2\pi r$. WHY $C=2\pi r$ and not $\pi r^2$: "around" is a length (one-dimensional), so the formula must give length units, not squared units. $C=2\pi r=2\pi (5)=10\pi \text{ cm}$ Decimal: $10\times 3.14159\approx 31.4\text{ cm}$. Answer: $C=10\pi \approx 31.4\text{ cm}$.

WHAT we want: space inside $\Rightarrow$ that is $A$, so use $A=\pi r^2$. WHY squared: area needs two lengths multiplied (like length $\times$ width). Both come from $r$, giving $r\times r=r^2$, so the units end up ${\text{cm}}^2$. $A=\pi r^2=\pi (5{)}^2=25\pi {\text{ cm}}^2\approx 78.5{\text{ cm}}^2$ Answer: $A=25\pi \approx 78.5{\text{ cm}}^2$.

WHAT we want: around again $\Rightarrow C$. We were handed $d$ directly, so use the twin form $C=\pi d$ — no need to halve first. $C=\pi d=\pi (2.6)=2.6\pi \approx 8.17\text{ cm}$ Answer: $C=2.6\pi \approx 8.17\text{ cm}$.

Fence = around = $C$: $C=2\pi r=2\pi (7)=14\pi \approx 43.98\text{ m}$ Ground = inside = $A$: $A=\pi r^2=\pi (7{)}^2=49\pi \approx 153.94{\text{ m}}^2$ Answer: fence $\approx 43.98\text{ m}$; area $\approx 153.94{\text{ m}}^2$. (Fence in metres, area in metres squared — two different jobs.)

WHY circumference: one full turn of a wheel lays its whole rim onto the road, so one turn $=$ one circumference. This is exactly the idea in Wheel and Circular Motion. $C=\pi d=\pi (60)=60\pi \text{ cm}$ 100 turns: each turn adds one $C$, so multiply: $\text{distance}=100\times 60\pi =6000\pi \text{ cm}$ Convert cm} \to m} (divide by $100$): $6000\pi \text{ cm}=60\pi \text{ m}\approx 188.5\text{ m}$ Answer: $\approx 188.5\text{ m}$.

WHY rearrange: we know the around number and want the across-to-centre number. So flip $C=2\pi r$ to solve for $r$. $C=2\pi r\Rightarrow r=\frac{C}{2\pi }=\frac{1.57}{2\times 3.14}=\frac{1.57}{6.28}=0.25\text{ m}$ Answer: $r\approx 0.25\text{ m}$ (that is $25\text{ cm}$).

Set up the two radii (see figure — inner pink circle, outer blue circle):

• inner (pond): $r_1=10\text{ m}$
• outer (pond + path): $r_2=10+2=12\text{ m}$ — we add because the path grows outward from the rim.

WHY subtract: the path is the big disk with the pond-disk punched out. So $A_{\text{path}}=A_{\text{outer}}-A_{\text{inner}}$. $A_2=\pi (12{)}^2=144\pi A_1=\pi (10{)}^2=100\pi$ $A_{\text{path}}=144\pi -100\pi =44\pi \approx 138.2{\text{ m}}^2$ Answer: $\approx 138.2{\text{ m}}^2$. This "big minus small" ring is called an annulus.

WHY not just half of $C$: the perimeter walks all the way round the boundary. Half of a circle's rim is the curved bit, but the flat cut edge (the diameter) is also part of the boundary and must be added.

• Curved part $=\frac{1}{2}C=\frac{1}{2}(2\pi r)=\pi r=4\pi \text{ cm}$
• Straight part $=d=2r=8\text{ cm}$ $P=4\pi +8\approx 12.566+8=20.57\text{ cm}$ Answer: $P=4\pi +8\approx 20.57\text{ cm}$. (Compare with plain Perimeter and Area of straight shapes: same idea, "add up every edge you walk.")

WHY a fraction of $\pi r^2$: a quarter is one of four equal slices of the whole disk, so it holds $\frac{1}{4}$ of the area. This is the seed of Sector Area. $A=\frac{1}{4}\pi r^2=\frac{1}{4}\pi (6{)}^2=\frac{36\pi }{4}=9\pi \approx 28.27{\text{ m}}^2$ Answer: $9\pi \approx 28.27{\text{ m}}^2$.

Link the shapes: a circle touching all four sides has diameter equal to the square's side, so $d=10\Rightarrow r=5\text{ cm}$. WHY subtract: the leftover corners $=$ (square area) − (circle area). $A_{\text{square}}=10\times 10=100{\text{ cm}}^2$ $A_{\text{circle}}=\pi (5{)}^2=25\pi \approx 78.54{\text{ cm}}^2$ $A_{\text{corners}}=100-25\pi \approx 21.46{\text{ cm}}^2$ Answer: $100-25\pi \approx 21.46{\text{ cm}}^2$.

(a) Circle. Its rim is the whole string: $C=2\pi r=20\Rightarrow r=\frac{20}{2\pi }=\frac{10}{\pi }$. $A_{\text{circle}}=\pi r^2=\pi {\left(\frac{10}{\pi }\right)}^2=\pi \cdot \frac{100}{{\pi }^2}=\frac{100}{\pi }\approx 31.83{\text{ cm}}^2$ (b) Square. Its four sides use the string: $4s=20\Rightarrow s=5$. $A_{\text{square}}=5^2=25{\text{ cm}}^2$ Compare: $31.83-25=6.83{\text{ cm}}^2$ more for the circle. WHY the circle wins: this is the parent note's headline fact — for a fixed perimeter, the circle encloses the most area. Here it beats the square by about $6.83{\text{ cm}}^2$. Answer: the circle, by $\approx 6.83{\text{ cm}}^2$.

WHY take a square root: area carries $r^2$, so to get back to $r$ we must undo the square — divide by $\pi$, then square-root. $A=\pi r^2\Rightarrow r^2=\frac{A}{\pi }=\frac{452.39}{3.14159}=144\Rightarrow r=\sqrt{144}=12\text{ cm}$ $d=2r=24\text{ cm}$ Answer: $r=12\text{ cm}$, $d=24\text{ cm}$.

Reason from the formulas (before numbers):

• $C=2\pi r$ is linear in $r$: replace $r$ by $2r$ and $C$ becomes $2\pi (2r)=2\cdot (2\pi r)$ — doubles.
• $A=\pi r^2$ is squared in $r$: replace $r$ by $2r$ and $A$ becomes $\pi (2r{)}^2=4\pi r^2$ — quadruples ($\times 4$).

Check with numbers: $r=3:C=6\pi ,A=9\pi r=6:C=12\pi ,A=36\pi$ $C$ went $6\pi \to 12\pi$ ($\times 2$ ✓). $A$ went $9\pi \to 36\pi$ ($\times 4$ ✓). Answer: circumference $\times 2$, area $\times 4$. (This scaling law is the heart of Similar Figures: scale a length by $k$, area scales by $k^2$.)

First wire (context): $2\pi r=12\pi \Rightarrow r=6\text{ cm}$ — not needed further, just confirming the wire length is $12\pi$. Second wire is an Arc Length: an arc is a fraction of a full rim. A full circle of radius $8$ has rim $2\pi (8)=16\pi$. Our arc is $12\pi$ long, so it is the fraction $\frac{12\pi }{16\pi }=\frac{3}{4}$ of the whole circle. Turn fraction into angle: a full circle is $360^{\circ }$, so $\theta =\frac{3}{4}\times 360^{\circ }=270^{\circ }$ Answer: $270^{\circ }$. (In Radians that fraction $\frac{3}{4}$ times $2\pi$ gives $\frac{3\pi }{2}$ rad — same three-quarter turn.)

WHY a ratio, not a difference: "what percentage" asks how the small area compares to the big area, so divide. $\frac{A_{\text{bull}}}{A_{\text{board}}}=\frac{\pi (2{)}^2}{\pi (10{)}^2}=\frac{4\pi }{100\pi }=\frac{4}{100}=0.04$ The $\pi$ cancels — the answer depends only on the ratio of radii, squared: ${\left(\frac{2}{10}\right)}^2=\frac{1}{25}=0.04$. $0.04\times 100=4\%$ Answer: the bullseye is $4\%$ of the board.

WHY rings: a disk is made of nested rings. A ring at radius $x$ has length $2\pi x$ (its own circumference) and tiny thickness $dx$, so its sliver of area is $2\pi xdx$. Summing all slivers is exactly an integral — see Integration. $A={\int }_0^{R}2\pi xdx=2\pi {\left[\frac{x^2}{2}\right]}_0^R=2\pi \cdot \frac{R^2}{2}=\pi R^2$ Check the idea, not just the algebra: the biggest rings (near $x=R$) are long and contribute most; the tiny central rings contribute almost nothing. That "grows with $x$" behaviour is why the total lands on $R^2$, not $R$. Answer: confirmed, $A=\pi R^2$.