Circles: Free Response

Sort the five by shape first. Three are straight (the hand, the crack, the scratch) and two are curved (the gold band and the stretch of it between two marks).

The minute hand runs from the middle of the face out to the rim, which is exactly what a radius is: a segment from the center to a point on the circle. Call its length $r$.

The crack is straight, joins two points of the rim, and passes through the middle, so it is a diameter. It is two radii laid end to end, so its length is fixed by the face:

$d = 2r.$

The scratch is straight and joins two points of the rim as well, but it misses the middle, so it is a chord that is not a diameter. Its length is not settled by the size of the face at all: a chord drawn near the rim is short, and one drawn near the middle is long, and both sit on the same face.

The gold band runs the whole way round the boundary, which is the circumference. The stretch of it between the 12 mark and the 3 mark is part of that boundary between two points, which is an arc.

The minute hand reaches from the middle out to the rim, so its length is the radius: $r = 13$ cm. The crack runs from rim to rim through the middle, so it is a diameter, which is two radii end to end:

$d = 2r = 2 \times 13 = 26 \text{ cm}.$

The second clock is described the other way round. The 44 cm measured straight across is its diameter, and its minute hand is a radius, so halve it:

$r = \frac{d}{2} = \frac{44}{2} = 22 \text{ cm}.$

Both hands are radii of their own faces, so they compare directly:

$22 - 13 = 9 \text{ cm}.$

The second hand is 9 cm longer. Notice that the two conversions ran in opposite directions: doubling took the first clock from a radius to a diameter, and halving took the second from a diameter back to a radius. Which one a problem needs is decided by what it hands you, not by which is easier.

Try to break the claim first. Draw one radius to the 12 mark and a second to the 3 mark. Each is 13 cm long, so the two together measure 26 cm, which is exactly the length of a diameter. But they do not make one. They meet at the center and turn a corner there, so what is drawn is a bent path from rim to rim, and a diameter has to be a single straight segment. The claim fails as it stands: total length is not the only thing a diameter has to have.

Now find the case where it does hold. Draw the second radius to the 6 mark instead, straight opposite the 12. The two radii now leave the center in exactly opposite directions, the corner between them opens right out, and the two lie along one straight line. That line runs from rim to rim through the center, which is a diameter. So two radii make a diameter only when they point in opposite directions, and whenever they do point in opposite directions they do make one.

That second picture also shows where the doubling comes from. Start from a diameter rather than from the radii: it is straight, it runs from rim to rim, and it passes through the center, so the center cuts it into two pieces. Each piece runs from the center out to the rim, which makes each piece a radius. A diameter is therefore exactly two radii end to end:

$d = r + r = 2r.$

The doubling is not a separate rule to memorise. It is a record of the center sitting halfway along every diameter.

Take the tabletop first. Dividing 345.4 by 110 asks how many times 110 fits into 345.4. Three copies of 110 make 330, which leaves 15.4, and 15.4 is 0.14 of 110:

$\frac{345.4}{110} = 3.14.$

The dustbin lid next. Three copies of 55 make 165, which leaves 7.7, and 7.7 is 0.14 of 55:

$\frac{172.7}{55} = 3.14.$

The biscuit tin lid last. Three copies of 35 make 105, which leaves 4.9, and 4.9 is 0.14 of 35:

$\frac{109.9}{35} = 3.14.$

Three objects of quite different sizes, one quotient. That quotient is the distance around divided by the distance across, which is the ratio the number pi is defined to be.

It carries no unit because each division is a length in centimetres divided by a length in centimetres, so the centimetres cancel exactly as a common factor does. Pi is not 3.14 cm; it is a bare number, and these measurements put it at about 3.14. It would come out the same had the class worked in inches throughout.

An enlargement multiplies every length in the picture by the same factor, so multiply the two measured lengths by 4:

$4 \times 35 = 140 \text{ cm}, \qquad 4 \times 109.9 = 439.6 \text{ cm}.$

The quotient needs no fresh division. Both numbers carry the same factor of 4, and a factor shared by the top and the bottom of a fraction cancels:

$\frac{4 \times 109.9}{4 \times 35} = \frac{109.9}{35} = 3.14.$

So the enlarged picture reports exactly the quotient the original did, without a ruler going near it.

Now rearrange. Write $C$ for the distance around, $d$ for the distance across, and $\pi$ for the quotient they always give. The definition says

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

Multiplying both sides by $d$ undoes the division and leaves $C$ by itself:

$C = \pi d.$

That is the circumference rule, and it is nothing more than the definition of pi with the division undone. It checks against the table: $3.14 \times 110 = 345.4$ cm, which is what the class measured around the tabletop.

Start from the one thing all circles share, which is shape. A circle is fixed completely by a single number, the distance from its center out to the rim, so any circle can be turned into any other by enlarging or shrinking it. Nothing else about a circle can vary: there is no long circle and no thin circle the way there are long and thin rectangles.

Now follow an enlargement through. Take a circle whose distance around is $C$ and whose distance across is $d$, and enlarge it so that every length becomes $k$ times as long. The distance across becomes $k \times d$. The boundary is stretched by that same factor everywhere along it, so the distance around becomes $k \times C$. Divide the one by the other:

$\frac{k \times C}{k \times d} = \frac{C}{d}.$

The factor $k$ appears once on the top and once on the bottom and cancels, so the enlarged circle reports exactly the quotient the original did, whatever $k$ was. Since every circle can be reached from every other by such an enlargement, every circle reports the same quotient. That is what makes it sensible to name one number pi and use it for circles of every size.

The class's table cannot do that work, for two separate reasons. First, it lists three circles, and a longer table would list a thousand; either way it lists finitely many, while there are infinitely many circles, so some circle is always left untested. A pattern in the rows you happened to measure is evidence for a claim about all of them, never a proof of it.

Second, every entry in it is a measurement, and a measurement is rounded: a string is laid against a ruler and read to the nearest tenth of a centimetre. A quotient built from two rounded numbers is itself only approximate, so the rows that are there are equally consistent with a true quotient a little above or a little below 3.14. The measurements are what suggested the constant and what tell us roughly how big it is. The enlargement argument is what makes it certain that one number serves them all.

The height first. Each sector's two straight edges were radii of the disk, so each is $r$ long, and the strip stands on those edges. As the sectors get thinner those edges lean less and less, so the perpendicular height of the strip closes in on the full radius:

$\text{height} = r.$

The base next. The base is built out of the curved edges, and every sector contributes exactly one of them. The sectors alternate, so half of those curved edges lie along the bottom of the strip and the other half along the top. All of them together make up the disk's whole boundary, $2\pi r$, so the bottom carries half of it:

$\text{base} = \frac{1}{2}(2\pi r) = \pi r.$

A parallelogram's area is its base times its perpendicular height, so

$A = (\pi r)(r) = \pi r^2.$

One factor of $r$ came from the base and the other from the height. That, and nothing else, is why the radius ends up squared.

Substitute $r = 13$ into the two lengths from part A. The base is $\pi r$:

$\text{base} \approx 3.14 \times 13 = 40.82 \text{ cm}.$

The height is the radius itself, 13 cm. Multiply the two:

$A \approx 40.82 \times 13 = 530.66 \text{ cm}^2.$

Now go straight to the rule instead, squaring the radius first and multiplying by $\pi$ afterwards:

$A \approx 3.14 \times 13^2 = 3.14 \times 169 = 530.66 \text{ cm}^2.$

The two agree, and they had to: the second is the first with the same three factors multiplied in a different order. Both are approximate, because 3.14 is a rounded value of $\pi$, which is why each is written with $\approx$ and not with $=$.

The unit is worth a sentence of its own. The base and the height are both lengths in centimetres, and multiplying two lengths gives an area, counted in square centimetres. The base alone, 40.82 cm, is a length and not an area, however similar the two figures look on the page.

Take the base first, and count rather than look. Cutting the disk into $n$ equal sectors produces $n$ sectors, each with exactly one curved edge, and those $n$ curved edges together are the disk's whole boundary cut into $n$ equal pieces. Laying the sectors alternately point-up and point-down sends every second one to the top of the strip and the rest to the bottom, so the bottom edge is built from half of the pieces and the top edge from the other half:

$\text{base} = \frac{1}{2}(2\pi r) = \pi r.$

It is worth seeing what the other choice would cost. Using the whole boundary as the base counts every curved edge twice, once along the bottom and once along the top, and gives $(2\pi r)(r) = 2\pi r^2$, which is twice the disk's area. The half is not a fudge factor; it is a count.

The second step is the more delicate one, and it is easy to state too weakly. The strip is not an approximation of the disk. It is the disk, cut up and rearranged: no piece was stretched, none was thrown away, and none overlaps another. Moving pieces about cannot change how much area they cover between them, so for every number of sectors, however small, the strip covers exactly the area the disk did.

What is only approximate is the strip's shape. With twelve sectors the base is visibly bumpy and the ends lean, so measuring the strip as a parallelogram gives a slightly wrong figure. With a hundred sectors the bumps flatten and the ends stand nearer upright; with more still, nearer again. So the area stays exact throughout while the shape improves, and the parallelogram measurement $(\pi r)(r)$ closes in on the area the strip had all along. That is the whole content of the derivation: an area that cannot be measured directly is pinned down by shapes that can be, and the rearrangement is what guarantees nothing was lost on the way.

The boundary of a half-disk comes in two pieces, and the ribbon has to travel along both to get back where it started.

The curved piece is half of the whole circle's boundary. Half of $2\pi r$ is $\pi r$, and with $r = 21$ the 7 in the fraction cancels into the 21 before any multiplying is done:

$\pi r \approx \frac{22}{7} \times 21 = 22 \times 3 = 66 \text{ cm}.$

The straight piece is the edge the cut left behind. It runs from one end of the curve to the other through the pivot, so it is a diameter:

$d = 2r = 2 \times 21 = 42 \text{ cm}.$

Add the two pieces:

$P \approx 66 + 42 = 108 \text{ cm}.$

The shop needs about 108 cm of ribbon. The answer is approximate because $\frac{22}{7}$ is a rounded value of $\pi$, so it is written with $\approx$ rather than with $=$.

The paper is half a disk, so find what a whole disk of this radius would cover and take half of that. Square the radius first:

$r^2 = 21^2 = 441.$

A whole disk covers $\pi r^2$, and 441 is a whole number of sevens, so the fraction cancels cleanly again:

$\pi r^2 \approx \frac{22}{7} \times 441 = 22 \times 63 = 1386 \text{ cm}^2.$

The fan is half of that:

$A \approx \frac{1386}{2} = 693 \text{ cm}^2.$

So the fan holds about 693 square centimetres of paper. The unit is square centimetres because the radius was squared: an area counts unit squares, whereas the ribbon in part A was a length and was counted in centimetres.

Work out what the supplier's figure is. Half of the whole circle's boundary is

$\frac{1}{2}(2\pi r) = \pi r \approx 66 \text{ cm},$

which is exactly the curved edge from part A. So their arithmetic is right and their reasoning is not: the quote buys ribbon for the curve and nothing at all for the straight edge along the bottom. That edge is a diameter, 42 cm long, so the order falls short by

$108 - 66 = 42 \text{ cm}.$

Now the general point, which is what the supplier really got wrong. Compare each of the two measurements against the whole circle's.

The whole disk covers about $1386 \text{ cm}^2$ and the fan covers about $693 \text{ cm}^2$, which is exactly half. Cutting a disk along a diameter divides the paper into two matching halves and creates no new paper, so each half holds half the area. Halving is the right instinct here.

The whole circle's boundary is $2\pi r \approx 132 \text{ cm}$, and half of that is 66 cm, but the fan's boundary came to 108 cm, which is more. The cut is the reason. It leaves each half with an edge the whole circle never had, and that edge is a diameter, 42 cm of new boundary:

$66 + 42 = 108 \text{ cm}.$

So cutting a disk in half divides what is inside and adds to what is around. Area is what the cut splits; boundary is partly what the cut creates. Any figure made by cutting a shape needs its boundary retraced piece by piece rather than scaled down from the original.

Every corner of a square is a right angle, which is a quarter of a full turn, so the glaze at one corner is exactly one fourth of a disk of radius 11 cm. There are four corners, so the four pieces together amount to four fourths, which is one whole disk:

$4 \times \frac{1}{4}\pi r^2 = \pi r^2.$

That is worth spotting before any arithmetic, because it turns four area calculations into one. Square the radius first:

$r^2 = 11^2 = 121.$

Then multiply by $\pi$:

$A \approx 3.14 \times 121 = 379.94 \text{ cm}^2.$

So about $379.94 \text{ cm}^2$ of the tile is glazed. The long way round agrees, as it must: each corner covers $\frac{1}{4} \times 3.14 \times 121$, and adding four of those fourths puts the whole disk back together.

The glazed corners and the plain region do not overlap, and between them they cover the tile exactly, so their areas add to the tile's area. That makes the plain area what is left when the glaze is taken away.

The tile is a square of side 40 cm:

$A_{\text{tile}} = 40 \times 40 = 1600 \text{ cm}^2.$

Subtract the glazed area found in part A:

$A_{\text{plain}} \approx 1600 - 379.94 = 1220.06 \text{ cm}^2.$

So well over half the tile is left plain, which fits the picture: the four corner pieces are small against the tile they sit on. The subtraction is legitimate only because the pieces do not overlap, which is why it was worth saying so before doing it.

Look back at where the first tile's side length actually entered the work. It entered once, at the very last step, when the plain area was found. The glazed area rested on two facts only: each corner is a right angle, and each quarter has radius 11 cm.

Both facts hold on the rectangle. A rectangle's corners are right angles exactly as a square's are, and a right angle is a quarter of a full turn, so the glaze at each corner is again one fourth of a disk of radius 11 cm. Four of them make four fourths:

$4 \times \frac{1}{4}\pi r^2 = \pi r^2 \approx 379.94 \text{ cm}^2.$

So the glazed area is unchanged. Neither the shape of the tile nor its size came into the argument at all.

The argument does carry a condition, and it is worth stating rather than assuming. Every quarter must fit inside the tile, and no two quarters may overlap; if two overlapped, part of the surface would be counted twice when their areas were added. On this rectangle the two quarters along a 40 cm side reach 11 cm each from their corners, and $11 + 11 = 22$ falls short of 40, so a gap is left; along a 60 cm side the gap is wider still. A tile only 20 cm across would fail the test, since two quarters of radius 11 cm would need 22 cm between them and would run into each other.

The plain area does change, because it is the one part of the work that used the tile's own dimensions. The rectangle covers

$60 \times 40 = 2400 \text{ cm}^2,$

so the plain part is

$A_{\text{plain}} \approx 2400 - 379.94 = 2020.06 \text{ cm}^2.$

That is $800 \text{ cm}^2$ more plain tile than the square had, which is exactly the extra area of the rectangle. Every square centimetre added to the tile is added to the plain part, because the glaze was settled before the tile's dimensions were ever used.