Circumference and Circle Area

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Learning goals

Justify the common circumference-to-diameter ratio using similarity
Compute exact or rounded circle measures, including inverse radius and diameter problems
Bound circumference and area using inscribed and circumscribed regular polygons
Explain and apply exhaustion through arbitrarily narrow valid bounds
Derive the circle-area formula and distinguish finite rearrangements from limiting conclusions

Straight sides give bounds

The circumference, $C$ , is the length of a circle’s boundary. Its enclosed region is a disk; the disk’s area, $A$ , counts square units. A circle has positive radius $r$ and diameter $d=2r$ .

An inscribed polygon has its vertices on the circle. Its straight sides cut inside the curved boundary, so its perimeter $P$ is smaller than $C$ . A circumscribed polygon surrounds the circle with tangent sides. Its perimeter $Q$ is larger than $C$ .

The upper comparison uses convexity: cutting corners from an enclosing convex polygon cannot increase its perimeter, because the joining segment is shorter than the broken path it replaces. Finer inscribed paths remain below the enclosing perimeter as they approach the circular boundary.

\begin{aligned} P&<C,\\ C&<Q. \end{aligned}

For regular polygons, joining the center to each vertex divides the polygon into equal triangles. The apothem, $h$ , is the perpendicular distance from the center to a side. An outer tangent polygon with parallel sides is a dilation of the inner polygon by $r/h$ .

A regular polygon splits into triangles with the same perpendicular height h. Adding their bases gives the polygon perimeter P.

Take radius $1$ for the following snapshots. These are finite approximations, not circles with straight sides.

6-sided circumference bounds

The radius is 1 unit. Each polygon has 6 sides. Inner perimeter = 6 units. Outer perimeter ≈ 6.9282 units. Certified bounds: 6.0000 < C < 6.9283 units. These finite bounds do not determine an exact circumference.

Six-sided finite approximation: blue inscribed polygon and red dashed tangent polygon. Both are regular; the circle lies between them.

12-sided circumference bounds

The radius is 1 unit. Each polygon has 12 sides. Inner perimeter ≈ 6.2117 units. Outer perimeter ≈ 6.4308 units. Certified bounds: 6.2116 < C < 6.4308 units. These finite bounds do not determine an exact circumference.

Twelve-sided finite approximation. The inner and outer perimeters are closer than in the six-sided snapshot.

24-sided circumference bounds

The radius is 1 unit. Each polygon has 24 sides. Inner perimeter ≈ 6.2653 units. Outer perimeter ≈ 6.3193 units. Certified bounds: 6.2652 < C < 6.3194 units. These finite bounds do not determine an exact circumference.

Twenty-four-sided finite approximation. The remaining gap is small, but it has not vanished.

The table gives decimal enclosures, rounded outward: the inner perimeter was rounded down and the outer perimeter up. Thus each row remains a valid lower and upper bound for $C$ .

Sides	Lower bound for $C$	Upper bound for $C$
$6$	$6.0000$	$6.9283$
$12$	$6.2116$	$6.4308$
$24$	$6.2652$	$6.3194$

The radius is $1$ unit, so the entries are lengths in that unit. A smaller gap gives a more precise enclosure. None of these three rows gives $C$ exactly.

Select six sides in the figure below. Before increasing the number of sides, predict how each perimeter and the gap will change. Compare the lower and upper bounds, and explain why even the narrowest displayed gap does not give an exact circumference.

Tighten the circumference bounds

Sides

Change both regular polygons together. The inner perimeter stays below the circumference and the tangent polygon perimeter stays above it.

Exhaustion closes the gap

The method of exhaustion traps an unknown quantity between valid bounds whose gap can be made smaller than any chosen positive amount. It is an informal limiting argument: there is no final polygon that becomes a circle. Every requested accuracy is achieved at some finite refinement.

Worked example 1 What a narrow enclosure certifies

Problem. Polygon measurements establish that a circumference, in centimeters, satisfies the following bounds. Find its value rounded to the nearest hundredth, and give a guaranteed error bound for that rounded estimate.

\begin{aligned} 43.276&<C,\\ C&<43.284. \end{aligned}

Solution. Every number in this interval rounds to $43.28$ : the interval lies strictly between the rounding boundaries $43.275$ and $43.285$ .

The estimate is the midpoint of the given bounds. Its distance to either end is

43.284-43.28=0.004.

Therefore the error is less than $0.004$ cm. This certifies an approximation, not the exact equality $C=43.28$ cm. Further bounds would be needed for more digits.

Check your understanding

A circumference estimate has a guaranteed error bound that halves at every refinement. Why can this process reach an error smaller than any chosen positive tolerance?

A Repeated halving can make the positive bound smaller than any prescribed positive amount. B The first sufficiently smooth-looking polygon is exactly the circle. C The error becomes zero after a fixed number of halvings. D Agreement of three rounded values proves the exact answer.

For circle polygons, arc bisection supplies that refinement. The next argument explains why, using the earlier chord-bisection theorem and Pythagoras, rather than an unintroduced trigonometric formula.

Bisect the arc from A to B at T. The radius through chord midpoint M locates T, and AT is one of the two shorter chords.

Why the polygon bounds can be arbitrarily close#

For a radius- $2$ circle, an inscribed regular hexagon has side $2$ and apothem $\sqrt3$ . Half a side and the apothem form a right triangle. Bisecting the arc creates a new chord with squared length

\begin{aligned} b^2&=1+(2-\sqrt3)^2,\\ b^2&=8-4\sqrt3. \end{aligned}

This is about $1.072$ , less than half the old squared side length, which is $4$ .

Now take any nondegenerate regular inscribed polygon. Its radius $r$ and side $s$ are positive, and its apothem satisfies $0<h<r$ . The old half-side and the radial gap give the new chord $b$ by Pythagoras:

\begin{aligned} h^2+(s/2)^2&=r^2,\\ b^2&=(s/2)^2+(r-h)^2,\\ b^2&=2r(r-h). \end{aligned}

Also $s^2=4(r-h)(r+h)$ . Dividing shows

\begin{aligned}\frac{b^2}{s^2}&=\frac{r}{2(r+h)},\\\frac{b^2}{s^2}&<\frac12.\end{aligned}

Each bisection therefore more than halves the squared side length. Repeating it makes $s$ as small as needed. The same right triangle gives

r-h=\frac{s^2}{4(r+h)}.

Its denominator is at least $4r$ , so the radial gap also becomes as small as needed. The apothem approaches $r$ .

The outer polygon is the $r/h$ dilation of the inner one. Thus

\begin{aligned} Q&=\frac{r}{h}P,\\ Q-P&=P\frac{r-h}{h}. \end{aligned}

The inner perimeters stay below the first outer perimeter, and the apothems stay above the first positive apothem. These fixed bounds prevent $P/h$ from growing without limit. As $r-h$ shrinks, the perimeter gap can therefore be made smaller than any chosen positive amount. Since $C$ stays between them, both perimeters approach $C$ . Nothing depended on choosing radius $2$ .

Why every circle uses the same pi

Earlier, circle similarity showed that a translation and positive dilation map any circle onto another. A dilation multiplies every polygon side by its scale factor. It therefore multiplies every inner and outer perimeter by that factor. Their arbitrarily tight enclosure shows that the curved circumference scales by the same factor too.

The circumference-to-diameter ratio is constant#

A diameter- $6$ circle dilated by $5/2$ has diameter $15$ . If its circumference was $C$ , the new circumference is $5C/2$ . Dividing the new circumference by the new diameter gives

\frac{5C/2}{15}=\frac{C}{6}.

For any diameter $d>0$ and positive scale factor $k$ , the same cancellation gives

\frac{kC}{kd}=\frac{C}{d}.

Any positive-radius circle can be scaled to any other, so this ratio is shared by all circles. The particular diameter $6$ and factor $5/2$ were not essential.

That common, dimensionless number is pi, written $\pi$ . Use the same length unit in both measurements before dividing.

\pi=\frac{C}{d}.

Multiplying by the diameter gives the circumference formulas:

\begin{aligned} C&=\pi d,\\ C&=2\pi r. \end{aligned}

The familiar decimal starts $3.14159265\ldots$ . Keeping $\pi$ as a symbol is exact; replacing it by a finite decimal is an approximation.

Check your understanding

A diameter- $8$ cm circle is enlarged to diameter $20$ cm. What happens to its circumference-to-diameter ratio?

A It is multiplied by

5/2

. B It is multiplied by

25/4

. C It stays equal to

\pi

. D It is divided by

5/2

Worked example 2 A finite hexagon enclosure

Problem. A circle has radius $6$ cm. A regular hexagon is inscribed in it. A second regular hexagon has parallel sides tangent to the circle. Find the two polygon perimeters and use them to bound the circumference and $\pi$ .

The inscribed and tangent regular hexagons around the same radius-6 circle. No computed perimeter is marked.

Solution. Each of the six central triangles of the inner hexagon is equilateral, so its side length is $6$ cm. The inner perimeter is

\begin{aligned}P&=6(6)\\P&=36\text{ cm}.\end{aligned}

Pythagoras gives its apothem, $h=3\sqrt3$ cm. The outer hexagon is enlarged by $r/h$ , so its perimeter is

\begin{aligned} Q&=36\frac{6}{3\sqrt3},\\ Q&=24\sqrt3\text{ cm}. \end{aligned}

Therefore $36<C<24\sqrt3$ , with lengths in centimeters. The diameter is $12$ cm. Dividing both bounds by $12$ gives

3<\pi<2\sqrt3.

This interval is valid but wide. It does not identify an exact decimal for $\pi$ .

From polygon areas to disk area

Adding the center-triangle areas of a regular polygon gives half its perimeter times its apothem, as in regular-polygon area. For the outer tangent polygon, the apothem is the circle radius itself.

For example, a radius- $4$ circle lies between a hexagon of area $24\sqrt3$ and a tangent hexagon of area $32\sqrt3$ , in square units. These come from perimeters $24$ and $16\sqrt3$ , and apothems $2\sqrt3$ and $4$ , respectively. The two polygon areas are about $41.57$ and $55.43$ square units, respectively. Refined polygons tighten the exact enclosure.

Why the area of a disk is half its circumference times its radius#

Keep radius $4$ . The circumference is $8\pi$ . On refinement, the inner perimeter approaches $8\pi$ and its apothem approaches $4$ . Its area approaches

\frac12(8\pi)(4)=16\pi.

The outer perimeter approaches the same circumference, and its apothem is already $4$ , so its area approaches the same number. Both the disk area and $16\pi$ lie between the two polygon areas at every refinement. Their difference must be smaller than every chosen positive tolerance, so it is zero.

For a general positive radius $r$ , let $P$ and $Q$ be the inner and outer perimeters, and $h$ the inner apothem. The two polygon areas are

\begin{aligned} I&=\frac12 Ph,\\ U&=\frac12 Qr. \end{aligned}

Both approach $Cr/2$ , because the perimeters approach $C$ and $h$ approaches $r$ . More explicitly, their gap is

\begin{aligned}E_p&=\frac{r(Q-P)}2,\\E_h&=\frac{P(r-h)}2,\\U-I&=E_p+E_h.\end{aligned}

The first contribution uses the shrinking perimeter gap; the second uses the shrinking radial gap while $P$ stays bounded. Both contributions can therefore be made small enough for any chosen total tolerance. The disk and $Cr/2$ remain inside the same shrinking enclosure. If their difference were any fixed positive amount, a smaller enclosure would rule it out. Therefore

A=\frac12 Cr.

Nothing depended on radius $4$ . Substituting the circumference formula gives

\begin{aligned} A&=\frac12(2\pi r)r,\\ A&=\pi r^2. \end{aligned}

You may also recognize the thin-wedge picture from Pre-Algebra. A sector is a wedge bounded by two radii and the joining arc. Cutting a disk into equal sectors and alternating them in a row preserves area exactly. A finite row still has curved, scalloped edges; it is not a perfect parallelogram.

Eight curved sectors before and after rearrangement. This finite approximation preserves the actual pieces and their area.

Sixteen curved sectors. Finer pieces reduce the scalloping, but the finite rearrangement still has curved edges.

Half the original curved boundary runs along the lower side of the row. As the pieces become arbitrarily thin, the corresponding straight base approaches $C/2$ and the perpendicular height approaches $r$ . The limiting area is therefore $(C/2)r$ , consistent with the polygon enclosure proof. The exact conclusion comes from controlling the remaining discrepancy, not from declaring one finite picture rectangular.

Check your understanding

A regular inscribed square has perimeter $32\sqrt2$ units and apothem $4\sqrt2$ units. What is its area?

64\sqrt2

square units B

128

square units C

128\sqrt2

square units D

256

square units

Worked example 3 Keep the exact area before rounding

Problem. A circle has diameter $14/5$ cm. Find its exact area and its area to the nearest hundredth of a square centimeter, using a calculator’s $\pi$ key.

The given diameter crosses the center. The radius and area are unknown.

Solution. Halve the diameter before squaring:

r=\frac75\text{ cm}.

The exact area is

\begin{aligned} A&=\pi\left(\frac75\right)^2,\\ A&=\frac{49\pi}{25}\text{ cm}^2. \end{aligned}

Using the $\pi$ key only at the end gives about $6.16$ square centimeters. The exact answer retains $\pi$ ; the rounded answer uses $\approx$ .

Recover a missing length

A positive radius can be recovered from either measurement. Undo the multiplication in the circumference formula, or undo the multiplication and then the square in the area formula.

Worked example 4 Recover lengths from an exact area

Problem. A disk has area $72\pi$ square centimeters. Find its radius, diameter, and circumference exactly.

Only the disk area is given; the drawn radius has no supplied length.

Solution. Divide the area by $\pi$ :

r^2=72\text{ cm}^2.

A radius is positive, so take the positive square root:

r=6\sqrt2\text{ cm}.

Then

\begin{aligned} d&=12\sqrt2\text{ cm},\\ C&=12\pi\sqrt2\text{ cm}. \end{aligned}

The negative square root does not describe a radius. No decimal approximation was needed.

In general, for exact positive circumference or area,

\begin{aligned} r&=\frac{C}{2\pi},\\ d&=\frac{C}{\pi},\\ r&=\sqrt{\frac{A}{\pi}}. \end{aligned}

An approximate input gives an approximate recovered length; extra calculator digits cannot restore information that was never measured.

Check your understanding

A circle has circumference $29\pi$ cm. What is its exact area?

841\pi

square cm B

29\pi/2

square cm C

841\pi/2

square cm D

841\pi/4

square cm

Compare changes in size

A dilation multiplies circumference by its length factor and disk area by its square. Compare lengths with lengths and areas with areas.

Worked example 5 Use a circumference comparison to find an area

Problem. A larger circle has circumference $5/3$ times that of a smaller circle. The smaller disk has area $54\pi$ square centimeters. Find the larger disk’s area exactly.

The circumference ratio is 5 to 3; only the smaller disk area is supplied.

Solution. Circumference has the same scale factor as radius, so the radius factor is $5/3$ . Area has the square of that factor:

\begin{aligned} A_{\text{large}}&=54\pi\left(\frac53\right)^2,\\ A_{\text{large}}&=150\pi\text{ cm}^2. \end{aligned}

Multiplying area by $5/3$ would treat square units as lengths.

Check your understanding

Which reasoning justifies an exact disk-area formula from finite polygon approximations?

A One drawing looks indistinguishable from a circle. B Three approximate area values agree after rounding. C The disk area and the proposed value stay between bounds whose gap can be made smaller than any positive tolerance. D A finite number of wedges becomes a rectangle once the pieces are small enough.

Circumference and Circle Area

Learning goals

Straight sides give bounds

Exhaustion closes the gap

Why every circle uses the same pi

From polygon areas to disk area

Recover a missing length

Compare changes in size

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress