Greek Mathematics
αὐτὸς αὑτὸν τέμνει. γίνονται δὲ καὶ τούτων τομαὶ γραμμαί τινες ἰδιάζουσαι. οἱ δὲ τετράγωνοι κρίκοι ἐκπρίσματά εἰσι κυλίνδρων· γίνονται δὲ καὶ ἄλλα τινὰ ποικίλα πρίσματα ἔκ τε σφαιρῶν καὶ ἐκ μικτῶν ἐπιφανειῶν.
(b) Measurement of Areas and Volumes
(i) Area of a Triangle Given the Sides
Heron, Metr. i. 8, ed. H. Schöne (Heron iii.) 18. 12–24. 21
Ἔστι δὲ καθολικὴ μέθοδος ὥστε τριῶν πλευρῶν δοθεισῶν οἱουδηποτοῦν τριγώνου τὸ ἐμβαδὸν εὑρεῖν χωρὶς καθέτου· οἷον ἔστωσαν αἱ τοῦ τριγώνου πλευραὶ μονάδων ζ̄, η̄, θ̄ σύνθες τὰ ζ̄ καὶ τὰ η̄ καὶ τὰ θ̄· γίγνεται κ̅δ̅. τούτων λαβὲ τὸ ἥμισυ· γίγνεται ι̅β̅ ἄφελε τὰς ζ̄ μονάδας· λοιπαὶ ε̄. πάλιν ἄφελε ἀπὸ τῶν ι̅β̅ τὰς η̄· λοιπαὶ δ̄. καὶ ἔτι τὰς θ̄· λοιπαὶ γ̄. ποίησον τὰ ι̅β̅ ἐπὶ τὰ ε̄· γίγνονται ξ̄. ταῦτα ἐπὶ τὸν δ̄· γίγνονται σ̅μ̅ ταῦτα ἐπὶ τὸν γ̄· γίγνεται ψ̅κ̅ τούτων λαβὲ πλευρὰν καὶ ἔσται τὸ ἐμβαδὸν τοῦ τριγώνου. ἐπεὶ οὖν αἱ ψ̅κ̅ ῥητὴν τὴν πλευρὰν οὐκ ἔχουσι, ληψόμεθα μετὰ διαφόρου ἐλαχίστου τὴν πλευρὰν οὕτως· ἐπεὶ ὁ συνεγγίζων τῷ ψκ τετράγωνός ἐστιν ὁ ψ̅κ̅θ̅ καὶ πλευρὰν ἔχει τὸν κ̅ζ̅ μέρισον τὰς ψ̅κ̅ εἰς τὸν κ̅ζ̅· γίγνεται κ̅ς̅ καὶ τρίτα δύο· πρόσθες τὰς κ̅ζ̅ γίγνεται ν̅γ̅ τρίτα δύο. τούτων τὸ ἥμισυ· γίγνεται κ̅ς̅ ∠γʹ ἔσται ἄρα τοῦ ψ̅κ̅ ἡ πλευρὰ ἔγγιστα τὰ κ̅ς̅ ∠γʹ τὰ γὰρ κ̅ς̅ ∠γʹ ἐφ᾿ ἑαυτὰ γίγνεται ψ̅κ̅ λ̅ςʹ̅. ὥστε τὸ διάφορον μονάδος ἐστὶ μόριον λ̅ςʹ̅. ἐὰν δὲ βουλώμεθα
Mensuration: Heron of Alexandria
Certain special curves are generated by sections of these spires. But the square rings are prismatic sections of cylinders; various other kinds of prismatic sections are formed from spheres and mixed surfaces.a
(b) Measurement of Areas and Volumes
(i.) Area of a Triangle Given the Sides
Heron, Metrica i. 8, ed. H. Schöne (Heron iii.) 18. 12–24. 21
There is a general method for finding, without drawing a perpendicular, the area of any triangle whose three sides are given. For example, let the sides of the triangle be 7, 8 and 9. Add together 7, 8 and 9; the result is 24. Take half of this, which gives 12. Take away 7; the remainder is 5. Again, from 12 take away 8; the remainder is 4. And again 9; the remainder is 3. Multiply 12 by 5; the result is 60. Multiply this by 4; the result is 240. Multiply this by 3; the result is 720. Take the square root of this and it will be the area of the triangle. Since 720 has not a rational square root, we shall make a close approximation to the root in this manner. Since the square nearest to 720 is 729, having a root 27, divide 27 into 720; the result is 26 2/3; add 27; the result is 53 2/3. Take half of this; the result is 26 1/2 + 1/3(= 26 5/6). Therefore the square root of 720 will be very nearly 26 5/6. For 26 5/6 multiplied by itself gives 720 1/36; so that the difference is 1/36. If we wish to make the difference less than 1/36,