Heron Metrica is the name given to the three-book compendium of geometric rules and formulas by Heron of Alexandria (1st/2nd Century AD), which covers the calculation of areas and volumes of plane and solid figures. The work includes formulas and methods for calculating areas of various geometric shapes and volumes of solids, such as pyramids, cylinders, and spheres. A key inclusion is Heron's formula for the area of a triangle, which relies on the lengths of its sides. The treatise also describes an iterative method for approximating square roots, a technique with Babylonian origins that is still relevant today. The formula can also be derived using the Newton-Raphson method by assuming the equation x^2 - a = 0.
The three books comprise
Book I This book focuses on methods for finding the area of plane figures, such as triangles and polygons, and the surface areas of common solids, like spheres and cylinders. This book also includes "Hero's formula" for calculating a triangle's area using its side lengths.
Book II It deals with methods for computing the volumes of various solids, including the five regular Platonic solids.
Book III It explores the division of plane and solid figures into parts based on a given ratio.
The book itself had been lost for almost a 1000 years. In 1894, historian Paul Tannery discovered a fragment of the book in Paris. Then, in 1896, R. Schone found a complete copy in Constantinople. The manuscript is called the Constantinopolitanus Palatii Veteris No. 1. and folios 67r-110v contain the Heron manuscript. Some consider (pp190) the copyist to be the monk Ephrem who lived in Constantinople around 950 AD but this is not certain.
The following translation was made using Claude.ai based on the original Greek text. The translation compares favorably with the translation by Henry Mendell and E.M Bruins (1964), Part II and III. The AI made some mistakes which I corrected, fractions weren't always correct, due to the Greek text being slightly ambiguous with 1/2. And sometimes it mistook Delta for an angle symbol.
I have since discovered a Greek rendering at https://scaife.perseus.org although a 1/2 is rendered as 'U2220' unicode which is an angle symbol, again due to ambiguity in the original Greek manuscript. For example, λθU2220η ιϛ, means 39 1/2 1/8 1/6 but the perseus text also misses the accents which are found in the original Greek text.
I will add figures at a later date. This is a work in progress.
Preliminaries
The first geometry, as ancient discourse teaches us, was occupied with measurements and divisions on the earth, whence it was also called "geometry." Since the subject was useful to people, the field was extended further, so that the administration of measurements and divisions also came to include solid bodies. And since the theorems first devised were not sufficient, they required still more thorough investigation, so that even now some of them remain unsolved, although Archimedes and Eudoxus nobly applied themselves to the work. For it was impossible, before Eudoxus's invention, to make a proof (by which) a cylinder is three times the cone that has the same base and equal height as it, and that circles are to one another as the squares on their diameters are to one another. And before Archimedes's understanding, it was incredible to conceive that the surface of a sphere is four times its greatest circle and that its solid is two-thirds of the cylinder containing it, and however many theorems are related to these.
Since the aforementioned study is necessary, we thought it good to collect together both what has been usefully written by those before us and what we ourselves have additionally investigated. Let us begin with plane measurements, including along with the planes other surfaces, whether concave or convex, since every surface is conceived from two <dia>mensions. The comparisons of the aforementioned surfaces are made to some rectilinear and rectangular area—rectilinear because the straight line is unchanging compared to other lines (for every straight line fits onto every straight line, but other lines, concave or convex, do not all fit onto all others). <...> Therefore they made the comparison to something fixed—I mean the straight line—and furthermore to the right angle; for again every right angle fits onto every right angle, but others do not all fit onto all others.
A "cubit area" is so called when a square region has each side of one cubit; similarly a "foot area" is so called when a square region has each side of one foot. Thus the aforementioned surfaces take their comparisons to the aforementioned areas or their parts. Again, solid bodies take their comparisons to a solid region that is rectilinear and rectangular, equal-sided in every direction—this is a cube having each side either of one cubit or one foot, or again to parts of these.
The reason why the comparison is made to the aforementioned areas has been stated. Next let us begin with measurements of surfaces. So that we may not name feet or cubits or their parts for each measurement, we shall set forth the numbers in units; for it is possible to assign them to whatever measure one wishes.
Notes:
The text marked with <..> appears to be missing some words that would complete the thought about why straight lines and right angles are used as standards
The following has be restored <δια>στάσεων (dimensions) where it appears corrupted
Proposition 1: Let there be an oblong plot ΑΒΓΔ having AB of 5 units and ΑΓ of 3 units. To find its area.
Since every rectangular parallelogram is said to be contained by two of the straight lines containing the right angle, and since what is contained by BA, ΑΓ is rectangular, the area of the oblong will be 15 units. For if each of the sides is divided—AB into 5 units and ΑΓ similarly into 3 units—and parallels are drawn through the division points to the sides of the parallelogram, the plot will be divided into 15 plots, each of which will be 1 unit. And if the plot is square, the same reasoning will apply.
Notes: This describes the area of a rectangle and the number of width units times the number of breadth units.
Proposition 2: Let there be a right triangle ABC having a right angle at B, and let AB be 3 units, and BC be 4 units. To find the area of the triangle and the hypotenuse.
Let the rectangle ABCD be completed. Since the area of the rectangle ABCD will be, as shown above, 12 units, and the triangle ABC is half of the rectangle ABCD, therefore the area of triangle ABC will be 6 units.
And since the angle ABC is a right angle, the squares on AB and BC are equal to the square on the hypotenuse AC. The squares on AB and BC together are 25 units, therefore the square on AC will also be 25 units. Therefore AC itself will be 5 units.
The method is this: Multiply 3 by 4 and take half of this result - this gives the area of the triangle. For the hypotenuse: multiply 3 by itself, and similarly multiply 4 by itself, then add them together - this gives 25. Taking the square root of this gives the hypotenuse of the triangle.
Note: This proposition demonstrates the basic application of the Pythagorean theorem to find both the area (using the formula ½ × base × height) and hypotenuse (using a² + b² = c²) of a right triangle with sides 3 and 4 units, yielding the famous 3-4-5 right triangle.
Proposition 3: Let there be an isosceles triangle ABC having AB equal to AC, each of the equal sides being 10 units, and BC being 12 units. To find its area.
Let the perpendicular AD be drawn to BC. Through A let EZ be drawn parallel to BC, and through B and C let BE and CZ be drawn parallel to AD. Therefore the parallelogram BCEZ is double the triangle ABC, for it has the same base and lies between the same parallels.
Since the triangle is isosceles and the perpendicular AD has been drawn, BD is equal to DC. And BC is 12 units, therefore BD is 6 units. AB is 10 units, therefore AD will be 8 units, since the square on AB is equal to the squares on BD and DA. So BE will also be 8 units. BC is 12 units.
Therefore the area of the parallelogram BCEZ is 96 units, so the area of triangle ABC is 48 units.
The method is this: Take half of 12 - this gives 6. Square 10 - this gives 100. Subtract the square of 6, which is 36 - this leaves 64. The square root of this is 8 - this will be the perpendicular AD. Multiply 12 by 8 - this gives 96. Take half of this - this gives 48. This will be the area of the triangle.
Note: This proposition demonstrates how to find the area of an isosceles triangle by first calculating the height using the Pythagorean theorem, then applying the standard area formula (½ × base × height). The triangle has equal sides of 10 units, base of 12 units, height of 8 units, and area of 48 square units.
Proposition 4: For scalene triangles, it is necessary to examine the angles so that we may know whether the perpendiculars drawn from the angles to the sides fall inside the angles or outside. Let there be given a triangle ABC having each side of given measurements, and it is necessary to examine, say, the angle at A, whether it is right or obtuse or acute.
If the square on BC is equal to the squares on BA and AC, it is clear that the angle at A is right. If it is less, the angle is acute. If it is greater, it is clear that the angle at A is obtuse.
Let it be assumed that the square on BC is less than the squares on BA and AC. Therefore the angle at A is acute. For if it were not acute, it would be either right or obtuse. It is not right, for then the square on BC would have to be equal to the squares on CA and AB - but it is not. Therefore the angle at A is not right. Nor is it obtuse, for then the square on BC would have to be greater than the squares on CA and AB - but it is not. Therefore it is not obtuse either. But it was shown that it is not right either. Therefore it is acute.
Similarly we can reason that if the square on BC is greater than the squares on BA and AC, then the angle at A is obtuse.
Note: This proposition establishes the method for determining the nature of angles in scalene triangles using what we now call the Law of Cosines in its classical form. It shows how to classify angles as acute, right, or obtuse by comparing the square of the opposite side with the sum of squares of the adjacent sides - a fundamental principle for understanding triangle geometry that Heron uses in subsequent propositions for calculating areas.
Proposition 5 (epsilon): Let there be an acute triangle ABC having AB of 13 units, BC of 14 units, and AC of 15 units. To find its area.
It is evident that the angle at B is acute, for the square on AC is less than the squares on AB and BC. Let the perpendicular AD be drawn to BC.
Therefore the square on AC together with twice the rectangle under CB and BD is less than the squares on AB and BC, as is demonstrated in the Elements. The squares on AB and BC are 365 units, and the square on AC is 225 units. Therefore the remainder, which is twice the rectangle under CB and BD, is 140 units. Therefore the rectangle under CB and BD once will be 70 units.
Since BC is 14 units, therefore BD will be 5 units. And since the square on AB is equal to the squares on AD and DB, and the square on AB is 169 units, and the square on BD is 25 units, therefore the remainder, the square on AD, will be 144 units. Therefore AD itself will be 12 units.
BC is also 14 units, therefore the rectangle under BC and AD will be 168 units. This is double the triangle ABC, therefore triangle ABC will be 84 units.
The method will be as follows: 13 squared gives 169, and 14 squared gives 196, and 15 squared gives 225. Add 169 and 196 - this gives 365. From this subtract 225 - this leaves 140. Take half of this - this gives 70. Divide this by 14 - this gives 5. Square 13 - this gives 169. From this subtract 5 squared, leaving 144. The square root of this gives 12 - this will be the perpendicular. Multiply this by 14 - this gives 168. Take half of this - 84. This will be the area.
Note: This proposition demonstrates Heron's method for finding the area of an acute triangle when all three sides are known, using the relationship derived from the law of cosines to find the altitude, then applying the standard area formula. The triangle has sides 13, 14, and 15 units with an area of 84 square units.
Proposition 6: Let there be an obtuse triangle ABC having AB of 13 units, BC of 11 units, and AC of 20 units. To find its perpendicular and area.
Let BC be extended and let the perpendicular AD be drawn to it. Therefore the square on AC is greater than the squares on AB and BC by twice the rectangle under CB and BD.
The square on AC is 400 units, the square on BC is 121 units, and the square on BA is 169 units. Therefore twice the rectangle under CB and BD is 110 units. Therefore the rectangle under CB and BD once is 55 units.
Since BC is 11 units, therefore BD will be 5 units. But AB is also 13 units, therefore AD will be 12 units. But BC is also 11 units. Therefore AD multiplied by BC will be 132 units. This is double the triangle ABC, therefore triangle ABC will be 66 units.
The method will be this: 13 squared gives 169, and 11 squared gives 121, and 20 squared gives 400. Add 169 and 121 - this gives 290. Subtract this from 400, leaving 110. Take half of this - this gives 55. Divide by 11 - this gives 5. Square 13 - this gives 169. Subtract 5 squared, leaving 144. The square root of this gives 12. The perpendicular will be 12 units. Multiply this by 11 - this gives 132. Take half of this - 66. This will be the area of the triangle.
Note: This proposition demonstrates how to find the area of an obtuse triangle. The key difference from the acute case is that the perpendicular falls outside the triangle, requiring the base to be extended. The triangle has sides 13, 11, and 20 units, with a perpendicular of 12 units and an area of 66 square units.
Up to this point we have made our calculations following geometric proofs, but from here on we shall make our measurements by analysis through the composition of numbers.
Note: This is a transitional statement in Heron's Metrica where he shifts from demonstrating geometric proofs (as in the previous propositions that showed the theoretical basis for finding areas and perpendiculars) to presenting purely numerical/algebraic methods for practical computation. It marks the move from geometric reasoning to what we might call "cookbook" formulas for calculation.
Proposition 7: If there are two numbers AB and BC, the square root of the product of the square on AB and the square on BC will be the number contained by AB and C [i.e., AB × BC].
For since AB is to BC as the square on AB is to the number contained by AB and C, and as the number contained by AB and C is to the square on BC, therefore as the square on AB is to the number contained by AB and C, so the number contained by AB and C is to the square on BC.
Since three numbers are in proportion, the product of the extremes will be equal to the square of the mean. Therefore the square on AB multiplied by the square on BC will be equal to the square of the number contained by AB and C. Therefore the square root of the product of the square on AB and the square on BC is the number contained by AB and C.
Note: This proposition establishes the mathematical principle that √(a² × b²) = ab, which is fundamental for Heron's later calculations. In modern notation, it's simply stating that √(a² × b²) = ab. This algebraic relationship will be used in subsequent propositions for calculating areas using what becomes known as Heron's formula.
Proposition 8: There is a general method so that when three sides of any triangle are given, the area can be found without a perpendicular. For example, let the sides of the triangle be 7, 8, and 9 units.
Add 7 and 8 and 9 - this gives 24. Take half of this - this gives 12. Subtract the 7 units, leaving 5. Again subtract the 8 from the 12, leaving 4. And also the 9, leaving 3.
Multiply 12 by 5 - this gives 60. Multiply this by 4 - this gives 240. Multiply this by 3 - this gives 720. Take the square root of this, and it will be the area of the triangle.
Since 720 does not have a rational square root, we shall take the square root with the smallest possible difference as follows: Since the square closest to 720 is 729, which has square root 27, divide 720 by 27 - this gives 26⅔. Add 27 - this gives 53⅔. Take half of this - this gives 26⅚. Therefore the square root of 720 is approximately 26⅚.
For 26⅚ squared gives 720 1/36, so the difference is 1/36 of a unit. If we wish the difference to be a fraction smaller than 1/36, instead of 729 we shall use the now-found 720 1/36, and doing this we shall find the difference becoming much smaller than 1/36.
Note: This is the famous Heron's Formula: For a triangle with sides a, b, c, the area equals √[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2. The proposition also demonstrates an early method for approximating square roots of non-perfect squares.
The geometrical proof of this (of the above) is as follows: to find the area of a triangle when the sides are given. It is indeed possible to draw a perpendicular and, having found its magnitude, to find the area of the triangle, but let it be required to obtain the area without the perpendicular.
Geometric Proof of Heron's Formula
Let the given triangle be ABC, and let each of AB, BC, and CA be given. To find the area.
Let a circle DEZ be inscribed in the triangle, with center H, and let AH, BH, CH, DH, EH, and ZH be joined. The rectangle under BC and EH is double the triangle BHC, the rectangle under CA and ZH is double triangle ACH, and the rectangle under AB and DH is double triangle ABH. Therefore the rectangle under the perimeter of triangle ABC and EH (that is, the radius from the center of circle DEZ) is double triangle ABC.
Let CB be extended, and let BΘ be placed equal to AD. Therefore CΘ is half the perimeter of triangle ABC, because AD is equal to AZ, DB is equal to BE, and ZC is equal to CE. Therefore the rectangle under CΘ and EH is equal to triangle ABC.
But the rectangle under CΘ and EH is the square root of the product of the square on CΘ and the square on EH. Therefore the area of triangle ABC multiplied by itself will be equal to the product of the square on ΘC and the square on EH.
Let HΛ be drawn perpendicular to CH, and BΛ perpendicular to CB, and let CΛ be joined. Since each of the angles CHΛ and CBΛ is right, the quadrilateral CHBΛ is inscribed in a circle. Therefore angles CHB and CΛB together equal two right angles. But angles CHB and AHD also equal two right angles, because the angles at H are bisected by AH, BH, CH, and angles CHB and AHD are equal to angles AHC and DHB, and all together equal four right angles. Therefore angle AHD equals angle CΛB. But angle ADH is also right, equal to right angle CBA. Therefore triangle AHD is similar to triangle CBA.
Therefore as BC is to BA, so AD is to DH, that is, BΘ is to EH. By alternation, as CB is to BΘ, so BA is to EH, that is, BK is to KE (because BΛ is parallel to EH). By composition, as CΘ is to BΘ, so BE is to EK. Therefore as the square on CΘ is to the rectangle under CΘ and ΘB, so the rectangle under BEC is to the rectangle under CEK, that is, to the square on EH (for in the right triangle, the perpendicular EH has been drawn from the right angle to the base).
Therefore the product of the square on CΘ and the square on EH, whose square root was the area of triangle ABC, will be equal to the product of the rectangle under CΘB and the rectangle under CEB.
Each of CΘ, ΘB, BE, and CE is given: CΘ is half the perimeter of triangle ABC; ΘB is the excess by which half the perimeter exceeds CB; BE is the excess by which half the perimeter exceeds AC; and EC is the excess by which half the perimeter exceeds AB, since EC equals CZ and BΘ equals AZ, since it is also equal to AD.
Therefore the area of triangle ABC is also given.
Numerical Example
It will be constructed as follows: Let AB be 13 units, BC be 14 units, and AC be 15 units. Add 13 and 14 and 15 - this gives 42. Half of this gives 21. Subtract 13, leaving 8; then 14, leaving 7; and also 15, leaving 6. Multiply 21 by 8, and the result by 7, and the result again by 6 - this gives 7056. The square root of this is 84. This will be the area of the triangle.
Note: This is Heron's complete geometric proof of his area formula, showing that for a triangle with sides a, b, c and semiperimeter s = (a+b+c)/2, the area equals √[s(s-a)(s-b)(s-c)]. The proof uses properties of the inscribed circle and similar triangles to derive the formula rigorously.
(Next is theta. Prop 9)
Proposition 9: Since we have learned to find the area of a triangle when the sides are given and the perpendicular is rational, let it be required to find the area when the perpendicular is not rational.
Let there be a triangle ABC having AB of 8 units, BC of 10 units, and AC of 12 units. Let the perpendicular AΔ be drawn.
Following what was said about the acute triangle, twice the rectangle under CB and BΔ will be 20 units. Therefore BΔ will be 1 unit, and the square on it will therefore be 1 unit. But the square on AB is also 64 units. Therefore the remainder, the square on AΔ, will be 63 units.
But the square on BC is also 100 units. Therefore the product of the square on BC and the square on AΔ will be 6300 units. The square root of this is the rectangle under BC and AΔ multiplied by itself. Therefore the rectangle under BC and AΔ multiplied by itself will be 6300 units.
Therefore half of the rectangle under BC and AΔ multiplied by itself is 1575 units - for when the sides of squares are double each other, the squares are four times those of the halves. But half of the rectangle under BC and AΔ is the area of the triangle. Therefore the area of the triangle is √1575.
It is possible to take the square root of 63 approximately and find the area as if the perpendicular were rational. The square root of 63 is approximately 7 1/2 1/4 1/8 1/16. It will therefore be necessary, having assumed the perpendicular to be this much, to find the area, which is 39 1/2 1/8 1/16.
Note: This proposition shows how to handle cases where the perpendicular calculated using the Pythagorean theorem yields an irrational number (√63), and demonstrates methods for approximating such results to get a practical numerical answer for the area. In Greek arithmetic the fractions are added, therefore Heron's approximation to the square root of 63 is 7 + 1/2 + 1/4 + 1/8 + 1/16 or more easily expressed as 7 + 8/16 + 4/16 + 2/16 + 1/16 = 7 15/16 or 7.9375, when the actual approximate value is 7.93725. In other words Heron was correct to three decimal places.
Proposition 10: Let there be a right trapezoid ABCD having right angles at A and B, and let AD be 6 units, BC be 11 units, and AB be 12 units. To find its area and also CD.
Let CD be bisected at E, and through E let ZEH be drawn parallel to AB, and let AD be extended to Z. Since DE is equal to EC, therefore DZ is also equal to HC. Let the common segments AD and BH be added; therefore AZ together with BH is equal to AD together with BC. But AD together with BC is given, since each of them is given. Therefore AZ together with BH is also given, that is, twice BH. Therefore BH is given. But AB is also given. Therefore the parallelogram ABZH is given.
Since triangle DEZ is equal to triangle EHC, let the common pentagon ABHED be added. Therefore the whole parallelogram ABZH is equal to the whole trapezoid ABCD. But the parallelogram ABZH was shown to be given, therefore the trapezoid ABCD is also given.
CD will be found as follows: Let the perpendicular DΘ be drawn. Since AD is given, therefore BΘ is also given. But BC is also given, and therefore the remainder CΘ is given. But DΘ is also given, for it is equal to AB and the angle at Θ is right. Therefore CD is also given.
It will be constructed according to the analysis as follows: Add 6 and 11 - this gives 17. Take half of this - this gives 8½. Multiply this by 12 - this gives 102. Therefore this is the area.
DC thus: Subtract 6 from 11 - this gives remainder 5. Square this - this gives 25. Square 12 - this gives 144. Add 25 - this gives 169. The square root of this gives 13. DC will be this many units.
Note: This proposition shows how to find the area and diagonal of a right trapezoid using the formula: Area = ½(sum of parallel sides) × height, and the diagonal using the Pythagorean theorem. The Perseus Greek texts appears to be in error. The first sentence refers to ABC rather than ABCD. Teh original Greek text is "Έστω τραπέζιον ορθογώνιον τδ Α Β Γ Δ όρθάς". This highlights another issue, the AI in this case converted ABΓΔ to ABCD I'm not sure what do to the Bruins translation does the same. In some other cases (Prop. 1) Claude has kept the Greek letters. I'll probably go back over the translation and switch t pute Roman lettering for the figures. Henry Mendell in his translation kept the original lettering.