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The trapezoidal example is the classic way to illustrate how Al-Khwārizmī’s algebra solved real-world land measurement problems during the Islamic Golden Age. It shows the systematic process surveyors used to find the area of an irregular field.

Here is the explanation of the trapezoidal example, focusing on the steps:

The Problem: Finding the Area of an Irregular Field

A surveyor needed to find the area of a trapezoidal field where the four side lengths are known, but the crucial perpendicular height ($h$) is unknown. This is vital for calculating land area for tax or inheritance purposes.

Known Sides

• Parallel Side A and Side C (e.g., $20$ and $15$)

• Non-parallel Side B and Side D (e.g., $10$ and $13$)

The goal is to find $h$ to use the area formula: Area=2(A+C)​×h.

Al-Khwārizmī’s Solution (The Four Steps)

The solution relies on breaking the shape down and using algebra to find the unknown height.

1. Geometric Decomposition (The Setup) 📐

The surveyor mentally divides the trapezoid into a central rectangle and two flanking right-angled triangles.

• The difference between the parallel sides ($20-15=5$) is the combined base of the two triangles.

• The surveyor defines one segment of the base as the unknown $x$, making the other segment $(5-x)$.

2. Formulating the Equation (Pythagorean Theorem)

The perpendicular height ($h$) is common to both triangles. The surveyor creates two expressions for h2 using the Pythagorean theorem (a2+b2=c2):

• From Triangle 1 (Side B = 10):

  h2=102−x2

• From Triangle 2 (Side D = 13):

  h2=132−(5−x)2

Setting these two expressions equal creates the complex quadratic equation:

3. Solving with Al-Jabr and Al-Muqābala

The surveyor applies Al-Khwārizmī’s rules to solve for $x$:

• Al-Jabr (Restoration): This is used to eliminate the negative term (−x2) from both sides by adding x2.

  $$100=169-25+10x$$
  $$100=144+10x$$

• Al-Muqābala (Balancing): Terms are simplified to isolate $x$.

  $$-44=10x$$

  This is reduced to a standard form, yielding $x$ (in this specific example, $x=-4.4$, which simply indicates the perpendicular drops outside the base, meaning the trapezoid is obtuse, but the calculation remains correct).

4. Final Area Calculation

With $x$ solved, the height $h$ can be found:

• The value of $x$ is substituted back into the h2 equation to determine the precise value of $h$.

• Finally, $h$ is plugged into the area formula, giving the accurate size of the field.

This procedure demonstrated the power of algebra to solve a geometric problem that was previously intractable, making it a cornerstone of medieval land administration.