00.13.01 · precalc / geometry

Plane geometry (distance, area, pi)

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Anchor (Master): Euclid ~300 BC Elements Books I and XII; Pythagoras ~500 BC; Archimedes ~250 BC Measurement of a Circle

Intuition [Beginner]

The most important fact about right triangles is the Pythagorean theorem: if a right triangle has legs of length $a$ and $b$ and hypotenuse (the long side) of length $c$ , then $a^{2} + b^{2} = c^{2}$ . This formula connects the three sides of a right triangle with a single equation.

The theorem is useful because it lets you compute any side from the other two. A $3$ - $4$ - $5$ triangle satisfies $3^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}$ . A right triangle with legs $5$ and $12$ has hypotenuse $25 + 144 = 169 = 13$ . The formula works for every right triangle, no matter the size.

The area of a triangle is half the base times the height. For a circle of radius $r$ , the area is $π r^{2}$ , where $π$ (pronounced "pie") is the ratio of a circle's circumference to its diameter. Every circle has the same ratio $π \approx 3.14159$ , whether it is a coin or a planet. The Pythagorean theorem and the area formulas are the two load-bearing results of plane geometry.

Visual [Beginner]

A right triangle with legs $a = 3$ and $b = 4$ drawn on a grid, with a square built on each side. The square on leg $a$ has area $9$ , the square on leg $b$ has area $16$ , and the square on the hypotenuse $c$ has area $25 = 9 + 16$ .

The areas of the two small squares add up to the area of the large square. That is the Pythagorean theorem in pictures.

Worked example [Beginner]

A right triangle has legs of length $5$ and $12$ . Find the length of the hypotenuse.

Step 1. Apply the Pythagorean theorem: $c^{2} = a^{2} + b^{2}$ .

Step 2. Compute: $c^{2} = 25 + 144 = 169$ .

Step 3. Take the square root: $c = 13$ .

What this tells us: the hypotenuse is $13$ . The Pythagorean theorem turns a geometry question (how long is the diagonal?) into an arithmetic question (what is the square root of the sum of two squares?).

Check your understanding [Beginner]

Formal definition [Intermediate+]

Area of a triangle. The area of a triangle with base $b$ and corresponding height $h$ is

A_{△} = \frac{1}{2} bh .

The height is the perpendicular distance from the opposite vertex to the line containing the base ^{[Lang — Basic Mathematics Ch. 6]}.

Area of a circle. The area enclosed by a circle of radius $r$ is

A_{\circ} = π r^{2},

where $π$ is the ratio of the circumference $C$ of any circle to its diameter $2 r$ : $π = C / (2 r)$ .

The Pythagorean theorem. In a right triangle with legs $a$ , $b$ and hypotenuse $c$ ,

a^{2} + b^{2} = c^{2} .

Heron's formula. The area of a triangle with side lengths $a$ , $b$ , $c$ and semi-perimeter $s = (a + b + c) /2$ is

A = s (s - a) (s - b) (s - c) .

Counterexamples to common slips

The Pythagorean theorem applies only to right triangles. A triangle with sides $3$ , $4$ , $6$ does not satisfy $3^{2} + 4^{2} = 6^{2}$ ( $9 + 16 \neq = 36$ ) because it is not a right triangle.
The area formula $\frac{1}{2} bh$ requires the height to be perpendicular to the base. The slant height of a triangle is not the height.
The formula $A = π r^{2}$ gives the area, not the circumference. The circumference is $C = 2 π r$ .

Key theorem with proof [Intermediate+]

Theorem (Pythagorean theorem, area-decomposition proof). In a right triangle with legs $a$ , $b$ and hypotenuse $c$ , the identity $a^{2} + b^{2} = c^{2}$ holds.

Proof. Construct a square of side $a + b$ . Inside this square, place four copies of the right triangle, each with legs $a$ and $b$ , arranged so that they form a tilted square of side $c$ in the centre. The area of the large square is $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ .

Each of the four triangles has area $\frac{1}{2} ab$ , so the four triangles together have area $2 ab$ . The tilted square in the centre has side length $c$ (the hypotenuse), so its area is $c^{2}$ .

The large square is decomposed into the four triangles plus the central tilted square:

(a + b)^{2} = 4 \cdot \frac{1}{2} ab + c^{2} .

Expanding: $a^{2} + 2 ab + b^{2} = 2 ab + c^{2}$ . Cancelling $2 ab$ from both sides gives $a^{2} + b^{2} = c^{2}$ . $□$

Bridge. The Pythagorean theorem builds toward the distance formula in 00.09.01 where the theorem is applied to compute the distance between two points in the Cartesian plane, and appears again in 00.10.01 where the focus-directrix condition for conics reduces to a Pythagorean-distance equation. The foundational reason the theorem holds is that the plane has a Euclidean metric, and this is exactly the content that makes the area-decomposition proof work: the area of the large square can be computed two different ways. The central insight is that the identity $a^{2} + b^{2} = c^{2}$ is an area-preserving statement. Putting these together, the bridge is between the algebraic identity (a relation among three squared numbers) and the geometric decomposition (a tiling of a square). The pattern generalises to the law of cosines $c^{2} = a^{2} + b^{2} - 2 ab cos C$ , which reduces to the Pythagorean theorem when $C = 9 0^{\circ}$ .

Exercises [Intermediate+]

Exercise 3 (medium, symbolic).

Prove that the area of a triangle with vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $(x_{3}, y_{3})$ is $\frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$ .

Hint

Enclose the triangle in a bounding rectangle and subtract the three exterior right triangles.

Answer

Let $x_{m i n} = min (x_{1}, x_{2}, x_{3})$ , $x_{m a x} = max (x_{1}, x_{2}, x_{3})$ , and similarly for $y$ . The bounding rectangle has area $(x_{m a x} - x_{m i n}) (y_{m a x} - y_{m i n})$ . The three exterior triangles are right triangles whose total area subtracts from the rectangle to leave the triangle. The signed-area formula $A = \frac{1}{2} (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}))$ computes this by a direct expansion of the determinant $\frac{1}{2} det x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111$ . The absolute value ensures a non-negative area regardless of vertex ordering.

Rubric: full credit for the bounding-rectangle argument or the determinant computation.

Exercise 4 (medium, symbolic).

Prove that $π$ is the same constant for all circles: if two circles have radii $r_{1}$ and $r_{2}$ and circumferences $C_{1}$ and $C_{2}$ , then $C_{1} / (2 r_{1}) = C_{2} / (2 r_{2})$ .

Hint

Use similarity: two circles are similar figures, so the ratio of circumference to diameter is preserved.

Answer

Two circles are similar under the scaling map $x \mapsto (r_{2} / r_{1}) x$ . Under this scaling, the circle of radius $r_{1}$ maps to the circle of radius $r_{2}$ . Every arc of the first circle maps to an arc of the second scaled by the factor $r_{2} / r_{1}$ . The circumference, being the total arc length of the circle, scales by the same factor: $C_{2} = (r_{2} / r_{1}) C_{1}$ . Dividing by $2 r_{2}$ : $C_{2} / (2 r_{2}) = C_{1} / (2 r_{1})$ . Hence the ratio $C / (2 r)$ is independent of the circle.

Rubric: full credit for the similarity argument.

Exercise 6 (medium, symbolic).

Derive Heron's formula from the area formula $A = \frac{1}{2} ab sin C$ and the law of cosines $c^{2} = a^{2} + b^{2} - 2 ab cos C$ .

Hint

Express $sin C$ in terms of $cos C$ via $sin^{2} C = 1 - cos^{2} C$ , then express $cos C$ from the law of cosines.

Answer

From the law of cosines: $cos C = (a^{2} + b^{2} - c^{2}) / (2 ab)$ . Then $sin^{2} C = 1 - ((a^{2} + b^{2} - c^{2}) / (2 ab))^{2}$ . Factoring: $sin^{2} C = (4 a^{2} b^{2} - (a^{2} + b^{2} - c^{2})^{2}) / (4 a^{2} b^{2})$ . The numerator is a difference of squares: $(2 ab - (a^{2} + b^{2} - c^{2})) (2 ab + (a^{2} + b^{2} - c^{2})) = (c^{2} - (a - b)^{2}) ((a + b)^{2} - c^{2}) = (c - a + b) (c + a - b) (a + b - c) (a + b + c)$ . With $s = (a + b + c) /2$ : the four factors are $2 (s - a)$ , $2 (s - b)$ , $2 (s - c)$ , and $2 s$ . So $16 A^{2} = 4 ab \cdot sin C \cdot 4 ab \cdot sin C$ ... actually, $A = \frac{1}{2} ab sin C$ , so $A^{2} = \frac{1}{4} a^{2} b^{2} sin^{2} C = \frac{1}{4} a^{2} b^{2} \cdot \frac{16 s ( s - a ) ( s - b ) ( s - c )}{4 a ^{2} b ^{2}} = s (s - a) (s - b) (s - c)$ .

Rubric: full credit for the algebraic derivation.

Advanced results [Master]

Theorem 1 (law of cosines). In a triangle with sides $a$ , $b$ , $c$ and angle $C$ opposite side $c$ : $c^{2} = a^{2} + b^{2} - 2 ab cos C$ . The Pythagorean theorem is the special case $C = 9 0^{\circ}$ , $cos 9 0^{\circ} = 0$ .

Theorem 2 (Archimedes' approximation of $π$ ). Archimedes (~250 BC) used inscribed and circumscribed regular $96$ -gons to prove $3 + 10/71 < π < 3 + 1/7$ , i.e., $3.1408 < π < 3.1429$ ^{[Archimedes — Measurement of a Circle]}. The method iterates the half-angle formula to double the number of sides at each step.

Theorem 3 (irrationality of $π$ , statement). The constant $π$ is irrational: there are no integers $p$ , $q$ with $q \neq = 0$ such that $π = p / q$ . First proved by Lambert 1761 using continued fractions; a simpler proof is due to Niven 1947.

Theorem 4 (Euclid's parallel postulate). Euclid's fifth postulate states: given a line $ℓ$ and a point $P$ not on $ℓ$ , there is exactly one line through $P$ parallel to $ℓ$ . This postulate is independent of the other four: denying it produces consistent non-Euclidean geometries (hyperbolic geometry with infinitely many parallels, elliptic geometry with none).

Theorem 5 (area of a regular polygon). A regular $n$ -gon with circumradius $R$ has area $A = \frac{1}{2} n R^{2} sin (2 π / n)$ . As $n \to \infty$ this approaches $π R^{2}$ , the area of the circumscribed circle.

Theorem 6 (isosceles-triangle theorem). In a triangle, equal sides subtend equal angles, and conversely. This is Proposition 5 of Euclid's Elements Book I (the "pons asinorum"), and its proof is the first non-evident argument in the Elements.

Synthesis. The foundational reason plane geometry coheres is the Pythagorean theorem, which identifies the Euclidean metric on $R^{2}$ with the sum-of-squares norm. The central insight is that area is the correct invariant for proving metric relations: the area-decomposition proof converts a statement about lengths into a statement about areas, which is a bilinear computation. The bridge is between the synthetic-geometric framework of Euclid (constructions, congruences, areas) and the analytic framework of Descartes (coordinates, distances, algebraic identities). Putting these together, the Pythagorean theorem generalises to the law of cosines (adding the angular correction $- 2 ab cos C$ ), the area formulas generalise from triangles to arbitrary polygons by triangulation, and the approximation of $π$ by inscribed polygons generalises to the Riemann integral that computes $π r^{2}$ as the limit of polygonal areas. The pattern recurs throughout analysis: the limit $n \to \infty$ of regular $n$ -gon areas is the prototype of the limiting process that defines the integral, and the irrationality of $π$ is the prototype of the transcendence results that separate geometric constants from algebraic ones.

Full proof set [Master]

Proposition 1 (law of cosines). In a triangle with sides $a$ , $b$ , $c$ and angle $C$ opposite $c$ : $c^{2} = a^{2} + b^{2} - 2 ab cos C$ .

Proof. Place $C$ at the origin with side $a$ along the positive $x$ -axis. The vertices are $A = (b cos C, b sin C)$ , $B = (a, 0)$ , and $C = (0, 0)$ . By the distance formula, $c^{2} = (b cos C - a)^{2} + (b sin C)^{2} = b^{2} cos^{2} C - 2 ab cos C + a^{2} + b^{2} sin^{2} C = a^{2} + b^{2} (cos^{2} C + sin^{2} C) - 2 ab cos C = a^{2} + b^{2} - 2 ab cos C$ . $□$

Proposition 2 (Archimedes' iteration step). Let $p_{n}$ be the perimeter of a regular $n$ -gon inscribed in a circle of radius $1$ . Then $p_{2 n} = 2 n 2 - 2 1 - (p_{n} / n)^{2}$ .

Proof. Each side of the $n$ -gon has length $s_{n} = p_{n} / n = 2 sin (π / n)$ and subtends angle $2 π / n$ at the centre. Bisecting each central angle produces $2 n$ isosceles triangles with apex angle $π / n$ and equal sides of length $1$ . The base of each triangle is $s_{2 n} = 2 sin (π / (2 n)) = 2 (1 - cos (π / n)) /2$ by the half-angle identity. Since $cos (π / n) = 1 - sin^{2} (π / n) = 1 - (s_{n} /2)^{2}$ , we get $s_{2 n} = 2 (1 - 1 - (s_{n} /2)^{2}) /2$ . The perimeter $p_{2 n} = 2 n \cdot s_{2 n}$ follows. $□$

Proposition 3 (converse of the Pythagorean theorem). If $a^{2} + b^{2} = c^{2}$ in a triangle, then the angle opposite $c$ is a right angle.

Proof. Given in Exercise 7 solution. Construct a right triangle with the same legs and apply SSS congruence. $□$

Connections [Master]

Real numbers, integers, rationals 00.01.01. The irrationality of $π$ extends the number-system hierarchy from that unit: $π$ is a real number that is neither rational nor algebraic (it is transcendental, proved by Lindemann 1882). The Archimedean property of $R$ (every real number can be approximated by rationals) is what makes the polygon-perimeter approximation of $π$ converge.
Cartesian coordinates and distance 00.09.01. The distance formula $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ from that unit is the Pythagorean theorem in coordinate disguise. The coordinate framework turns every geometric-length question into a computation with number pairs, and the Pythagorean theorem is the engine that makes the computation work.
Conic sections 00.10.01. The focus-directrix definition of conics relies on the distance formula, which is the Pythagorean theorem. The area of an ellipse ( $π ab$ ) generalises the area of a circle ( $π r^{2}$ ) by the stretch factor $b / a$ , and the reflection properties of conics follow from metric computations using the Pythagorean theorem.

Historical & philosophical context [Master]

Pythagoras of Samos (~570–495 BC) and his school are credited with the first proof of the theorem that bears his name, though the result was known to Babylonian mathematicians circa 1800 BCE (tablet Plimpton 322 lists Pythagorean triples) and to Chinese mathematicians (the Zhou Bi Suan Jing, ~1046 BCE). The Pythagorean school's contribution was the proof: the transition from empirical observation to deductive argument ^{[Euclid — Elements]}.

Euclid of Alexandria ~300 BC wrote the Elements ^{[Euclid — Elements]}, the most influential mathematical text ever written. Book I contains 48 propositions building plane geometry from five postulates, including the Pythagorean theorem (Proposition I.47) and its converse (Proposition I.48). Book XII develops the method of exhaustion (Eudoxus ~370 BC), the ancient precursor of integration, and uses it to prove that the ratio of the areas of two circles equals the ratio of the squares of their diameters. Archimedes of Syracuse ~250 BC Measurement of a Circle ^{[Archimedes — Measurement of a Circle]} applied the method of exhaustion to inscribe and circumscribe regular polygons around a circle, bounding $π$ between $3 + 10/71$ and $3 + 1/7$ . Lambert 1761 proved the irrationality of $π$ , and Lindemann 1882 proved the transcendence of $π$ (resolving the ancient problem of squaring the circle in the negative).

Bibliography [Master]

@book{EuclidElements,
  author = {Euclid},
  title = {Elements},
  note = {Books I--XIII. English translation by T. L. Heath, Dover 1956},
  year = {-300}
}

@book{ArchimedesCircle,
  author = {Archimedes},
  title = {Measurement of a Circle},
  note = {In {\em The Works of Archimedes}, ed. T. L. Heath, Cambridge University Press 1897},
  year = {-250}
}

@book{Lang1988,
  author = {Lang, Serge},
  title = {Basic Mathematics},
  publisher = {Springer},
  year = {1988}
}

@book{Apostol1967,
  author = {Apostol, Tom M.},
  title = {Calculus, Volume 1},
  publisher = {Wiley},
  edition = {2nd},
  year = {1967}
}