The Ratio Every Circle Shares Measuring Space: Perimeter and Area

Watch the start of a 4 x 100 metre relay race and something looks unfair at first glance: the runners in the outer lanes start noticeably ahead of the runners in the inner lanes, all lined up on a diagonal rather than a straight row. Is this actually fair? Answering that question properly needs exactly the tool this chapter builds: a way to measure the length of a curved path, not just a straight one. A running track is built from straight sections joined by curved, semicircular ends, and a runner in an outer lane covers a longer curve at the bend than a runner in an inner lane, even though both run 'the same' one lap. The stagger exists precisely to cancel out that extra curved distance, so that every runner covers the same total length by the time the lap finishes. This chapter returns to this exact scenario with real numbers once the tools for measuring curves are built.

Measure the distance around any circle (its circumference) and the distance straight across it through the centre (its diameter), then divide the first by the second. Do this for a coin, a dinner plate, and a running track, and remarkably, the same number comes out every single time. This constant ratio, circumference divided by diameter, is given its own name: pi. Civilisations across the world independently landed on increasingly precise estimates of this same number, long before they could prove it was exactly the same constant for every circle. Ancient Mesopotamian tablets used 3 and 1/8. Archimedes, using a clever method of trapping a circle between two many-sided polygons whose perimeters he COULD compute exactly, proved pi was trapped between 3 + 10/71 and 3 + 1/7. In China, Liu Hui and later Zu Chongzhi refined this further, reaching the fraction 355/113, astonishingly accurate. In India, Aryabhata gave the approximation 62832/20000 = 3.1416, explicitly calling it 'asanna', meaning approximate, showing he understood it wasn't exact. Centuries later, the mathematician-astronomer Madhava of Kerala found something remarkable: an exact infinite sum that equals pi/4 exactly: 1 - 1/3 + 1/5 - 1/7 + 1/9 - ..., continuing forever. Today the value used in this course is almost always the simple approximation 22/7, or sometimes 3.14, chosen for how cleanly it divides into typical exam numbers, not because either is exactly correct.

Pi is irrational, meaning it can never be written exactly as one whole number divided by another, no matter how large the numbers used; its decimal expansion goes on forever without ever settling into a repeating pattern. Ancient mathematicians like Aryabhata and Zu Chongzhi seem to have suspected this was true, from how carefully they hedged their approximations as approximations, but a rigorous proof did not arrive until 1761, when Johann Lambert proved it conclusively, over a thousand years later. With circumference C = 2 pi r established, the length of just part of a circle, an arc, follows directly: since the full circumference corresponds to the full 360 degrees around the centre, an arc spanning only theta degrees is that same fraction of the whole circumference, giving arc length l = 2 pi r x (theta/360). Now the relay race stagger can finally be checked with real numbers. A standard 400 metre track has straight sections of 84.39 metres and a curved section with an inner radius of 36.5 metres; assuming a runner's actual running line sits about 0.3 metres in from the innermost border, an effective radius of roughly 36.8 metres, the full curved distance covered comes to approximately 2 x pi x 36.8, about 231.2 metres for the innermost lane's semicircular ends combined with the straights. The stagger for each outer lane is calculated to add back exactly the extra arc length that lane's larger radius would otherwise cost it, so that every lane's total lap distance comes out equal.

Grade 8 already established area = ab for a rectangle with sides a and b, and area = base x height for a parallelogram (found by cutting a triangular sliver off one end and reattaching it to the other, turning the parallelogram into a rectangle of the same base and height without changing its area). A triangle's area, half of base times height, comes from noticing any triangle is exactly half of a parallelogram sharing the same base and height, formed by duplicating the triangle and rotating the copy 180 degrees to fit alongside the original. One extra fact worth stating clearly: a median (the line from a triangle's vertex to the midpoint of the opposite side) always splits a triangle into two smaller triangles of exactly equal area, since both share the same height from that vertex, and their bases are equal by definition of the midpoint. This fact reappears constantly in area-based proofs.

Half base times height needs a height, and a height is not always easy to measure directly, especially for a scalene triangle with no obviously perpendicular side. Heron's Formula solves exactly this: given only the three side lengths a, b, c, first compute the semi-perimeter s = (a+b+c)/2, then the area is the square root of s(s-a)(s-b)(s-c). This book states the result and checks it against special cases you can verify another way, but does not prove it, explicitly setting that proof aside for a later class. It is worth seeing where it actually comes from, since the proof uses nothing beyond the Pythagoras theorem and the difference-of-two-squares identity, both already familiar. Drop a perpendicular from vertex A to side BC (length a), meeting it at point D, splitting BC into segments of length x and (a-x), with height h. From the two right triangles this creates: h squared = c squared - x squared, and also h squared = b squared - (a-x) squared. Setting these equal and expanding gives 2ax = a squared + c squared - b squared, so x = (a squared + c squared - b squared)/(2a). Substituting back, h squared = c squared - x squared factors, using difference of squares, as (c-x)(c+x), and after replacing x and simplifying every bracket using s = (a+b+c)/2, each factor turns out to be a clean multiple of (s-a), (s-b), (s-c), or s itself. The result, after collecting everything together: h squared = 4s(s-a)(s-b)(s-c)/(a squared). Since the triangle's area is (1/2)(a)(h), squaring both sides gives Area squared = (1/4)(a squared)(h squared) = s(s-a)(s-b)(s-c) exactly, and taking the square root recovers Heron's Formula in full, built from nothing more exotic than Pythagoras and algebra.

Heron's Formula handles triangles. Brahmagupta's Formula extends the same idea to any cyclic quadrilateral, a four-sided figure whose corners all sit on one circle: with sides a, b, c, d and semi-perimeter s = (a+b+c+d)/2, its area is the square root of (s-a)(s-b)(s-c)(s-d). Once again the book states this and checks it against special cases (a rectangle, an isosceles trapezium) rather than proving it fully. What is genuinely elegant is the connection between the two formulas: shrink side d down to zero, so the quadrilateral collapses into a triangle (two of its four corners now sitting on top of each other). Substitute d=0 into Brahmagupta's Formula, and s becomes exactly (a+b+c)/2, matching a triangle's own semi-perimeter, and the formula collapses into precisely Heron's Formula. A triangle is, in this precise sense, a cyclic quadrilateral that has been flattened until one side disappeared, and Heron's Formula is simply the special case of Brahmagupta's where that happens.

Just as pi connects a circle's circumference to its diameter, it also connects a circle's area to its radius: Area = pi r squared. Different civilisations reached this by different, equally clever routes. Babylonian tablets used a rough rule relating circumference squared to area. Egyptian and early Indian (Baudhayana) approximations used a formula close to (8d/9) squared. Archimedes gave an actual proof: he showed a circle's area always equals exactly half its circumference times its radius, (1/2)(2 pi r)(r) = pi r squared, by comparing the circle to many-sided polygons whose areas he could compute directly, the same style of argument he used for the circumference bounds. Centuries later, the Kerala mathematician Nilakantha gave a beautifully visual proof: slice a circle into many thin pie-shaped wedges, then rearrange these wedges, alternating point-up and point-down, into a shape that looks almost exactly like a rectangle of height r and length half the circumference (pi r); as the slices get thinner, the shape becomes exactly that rectangle, whose area is directly pi r times r, giving pi r squared. A sector, a pie-slice cut off by two radii and an arc, is to the whole circle's area exactly what an arc is to circumference: a fraction theta/360 of the whole, giving sector area = pi r squared x (theta/360).