The Timpson Lyceum · Geometry (2024)

Area & Pythagorean Theorem – Lecture 13 (Geometry)

Overview

This lecture builds the theory of area from the ground up. Defining area precisely (filling a region with arbitrarily small squares) ultimately requires limits and calculus, which this course avoids. Instead, we adopt three simple, intuitive starting points — provable in calculus but here treated as postulates — and use them to derive every area formula we need. From these we prove, in order, the areas of the rectangle, the right triangle, the parallelogram, and the general triangle, and we finish with an elegant area-based proof of the Pythagorean Theorem. The recurring strategy is to enclose an unknown shape inside a figure whose area we already know, then subtract congruent triangular pieces.

Key Concepts

Why area needs a foundation. Measuring area mirrors measuring length: pick a unit square (e.g. one square foot, a $1 \times 1$ region) and count how many tile the region, then refine with tenth-size and hundredth-size squares for the leftover. This works only when the unit (or a power-of-ten fraction of it) tiles the region exactly. For incommensurable sides — say a rectangle that is $1$ by $\sqrt{2}$ — no square ever tiles both sides perfectly, so the process never terminates. A rigorous definition of area therefore depends on a limit, which belongs to calculus. We sidestep this by assuming three results.

The three starting points (treated as postulates).

Area of a square: A square of side length $x$ has area $x^2$ . This is the only one of the three that actually tells us how to compute an area, so every derivation must somehow start from squares.
Congruent triangles, equal areas: If two triangles are congruent, their areas are equal. This also fixes an important idea — area is a number (a real quantity), not just a geometric figure to be compared.
Additivity: If a figure is split into two non-overlapping pieces of areas $A_1$ and $A_2$ , the whole figure has area $A_1 + A_2$ .

The derived theorems (proved from the three above).

Rectangle: A rectangle of length $L$ and width $W$ has area $A = LW$ .
Right triangle: A right triangle with legs $a$ and $b$ has area $A = \tfrac{1}{2}ab$ — duplicate it to form a rectangle of area $ab$ , which is twice the triangle.
Parallelogram: A parallelogram with base $b$ and height $h$ has area $A = bh$ , regardless of how much it slants. Shearing a parallelogram keeps its area fixed.
General triangle: Any triangle has area $A = \tfrac{1}{2}bh$ , where $h$ is the altitude (the perpendicular distance from the chosen base to the opposite vertex). This follows immediately by viewing the triangle as half of a parallelogram.

Pythagorean Theorem as an area statement. For a right triangle with legs $a$ , $b$ and hypotenuse $c$ , the square built on the hypotenuse equals the combined squares built on the two legs: $a^2 + b^2 = c^2$ . The proof constructs one large square of side $a+b$ in two ways and compares areas.

Constructing a triangle equal in area to a quadrilateral. Given any quadrilateral, one can construct a triangle with the same area using only a line drawn parallel to a diagonal. Sliding a vertex along a line parallel to the base leaves a triangle's area unchanged (same base, same height). This classical Greek construction works because the ancients reasoned about areas geometrically, without numbers. Note that "squaring the circle" — constructing a square equal in area to a given circle — is impossible with compass and straightedge, since it would require constructing $\pi$ (or $\sqrt{\pi}$ ).

Worked Example

Pythagorean Theorem by areas. Build a square of side $a+b$ . Place four copies of the right triangle (legs $a$ , $b$ , hypotenuse $c$ ) around the inside; an angle chase shows the inner quadrilateral is a square of side $c$ . Equating the big square's area to its four triangles plus the inner square gives the result.

\begin{aligned} (a+b)^2 &= 4\cdot\left(\tfrac{1}{2}ab\right) + c^2 \\ a^2 + 2ab + b^2 &= 2ab + c^2 \\ a^2 + b^2 &= c^2 \end{aligned}

Area of a rectangle (the same idea in miniature). Enclose the $L \times W$ rectangle in a square of side $L+W$ . The square decomposes into one $L^2$ square, one $W^2$ square, and four congruent right triangles of equal area $A_1$ , where two of those triangles together form the target rectangle (so $4A_1 = 2A$ , the rectangle's area $A$ ).

\begin{aligned} (L+W)^2 &= L^2 + W^2 + 4A_1 \\ L^2 + 2LW + W^2 &= L^2 + W^2 + 2A \\ 2LW &= 2A \\ A &= LW \end{aligned}

Summary

A rigorous definition of area requires limits/calculus; this course instead assumes three provable starting points.
The three postulates: a square of side $x$ has area $x^2$ ; congruent triangles have equal areas (area is a number); areas of non-overlapping pieces add.
Only the square postulate computes an area outright, so every derivation begins by relating the shape to squares or rectangles.
The area chain builds in order: rectangle $LW$ → right triangle $\tfrac{1}{2}ab$ → parallelogram $bh$ → general triangle $\tfrac{1}{2}bh$ (with $h$ the altitude).
The Pythagorean Theorem $a^2+b^2=c^2$ is itself a statement about areas, provable by comparing two dissections of a square of side $a+b$ .
The same "enclose, then subtract congruent triangles" technique converts any quadrilateral into a triangle of equal area; squaring the circle, however, is impossible by compass and straightedge.