UK MA 330 - Heron’s Formula for Triangular Area

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1Heron’s Formula for Triangular Areaby Christy Williams, Crystal Holcomb, and Kayla GiffordHeron of Alexandrian Physicist, mathematician, and engineern Taught at the museum in Alexandrian Interests were more practical (mechanics, engineering, measurement) than theoreticaln He is placed somewhere around 75 A.D. (±150)2Heron’s Worksn Automatan Mechanican Dioptran Metrican Pneumatican Catoptrican Belopoecian Geometrican Stereometrican Mensuraen CheirobalistraThe AeolipileHeron’s Aeolipile was the first recorded steam engine. It was taken as being a toy but could have possibly caused an industrial revolution 2000 years before the original.3Metrican Mathematicians knew of its existence for years but no traces of it existedn In 1894 mathematical historian Paul Tannery found a fragment of it in a 13thcentury Parisian manuscriptn In 1896 R. Schöne found the complete manuscript in Constantinople.n Proposition I.8 of Metrica gives the proof of his formula for the area of a triangle251726How would you find the area of the given triangle using the most common area formula?bhA21=Since no height is given, it becomes quite difficult…How is Heron’s formula helpful?4Heron’s formula allows us to find the area of a triangle when only the lengths of the three sides are given. His formula states:()()()csbsassK −−−=Where a, b, and c, are the lengths of the sides and s is the semiperimeter of the triangle.Heron’s FormulaThe Preliminaries…5Proposition 1Proposition IV.4 of Euclid’s Elements.The bisectors of the angles of a triangle meet at a point that is the center of the triangles inscribed circle. (Note: this is called the incenter)Proposition 2Proposition VI.8 of Euclid’s Elements.In a right-angled triangle, if a perpendicular is drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle and to one another.6Proposition 3In a right triangle, the midpoint of the hypotenuse is equidistant from the three vertices.Proposition 4If AHBO is a quadrilateral with diagonals AB and OH, then if and are right angles (as shown), then a circle can be drawn passing through the vertices A, O, B, and H.HOB∠HAB∠7Proposition 5Proposition III.22 of Euclid’s Elements.The opposite angles of a cyclic quadrilateral sum to two right angles.SemiperimeterThe semiperimeter, s, of a triangle with sides a, b, and c, is2cbas++=8Heron’s Proof…Heron’s Proofn The proof for this theorem is broken into three parts.n Part A inscribes a circle within a triangle to get a relationship between the triangle’s area and semiperimeter. n Part B uses the same circle inscribed within a triangle in Part A to find the terms s-a, s-b, and s-c in the diagram.n Part C uses the same diagram with a quadrilateral and the results from Parts A and B to prove Heron’s theorem.9Restatement of Heron’s FormulaFor a triangle having sides of length a, b, and c and area K, we have where s is the triangle’s semiperimeter.()()()csbsassK −−−=Heron’s Proof: Part ALet ABC be an arbitrary triangle such that side AB is at least as long as the other two sides. Inscribe a circle with center O and radius r inside of the triangle.Therefore, . OFOEOD==10Heron’s Proof: Part A (cont.)Now, the area for the three triangles ?AOB, ?BOC, and ?COA is found using the formula ½(base)(height).Area ?AOB = Area ?BOC = Area ?COA =crODAB2121))(( =arOEBC2121))(( =brOFAC2121))(( =Heron’s Proof: Part A (cont.)We know the area of triangle ABC is K. Therefore()()()()COAAreaBOCAreaAOBAreaABCAreaK∆+∆+∆=∆=If the areas calculated for the triangles ?AOB, ?BOC, and ?COA found in the previous slides are substituted into this equation, then K isrscbarbrarcrK =++=++=221212111Heron’s Proof: Part BWhen inscribing the circle inside the triangle ABC, three pairs of congruent triangles are formed (by Euclid’s Prop. I.26 AAS).COFCOEBOEBODAOFAOD∆≅∆∆≅∆∆≅∆Heron’s Proof: Part B (cont.)n Using corresponding parts of similar triangles, the following relationships were found:COFCOEBOEBODAOFAOD∠=∠∠=∠∠=∠CFCEBEBDAFAD===12Heron’s Proof: Part B (cont.)n The base of the triangle was extended to point G where AG = CE. Therefore, using construction and congruence of a triangle:CEADBDAGADBDBG ++=++=()CEADBDBG 22221++=()()()[]CFCEAFADBEBDBG +++++=21()ACBCABBG ++=21()()()[]CFAFCEBEADBDBG +++++=21()sbacBG =++=21Heron’s Proof: Part B (cont.)n Since , the semi-perimeter of the triangle is the long segment straighten out. Now, s-c, s-b, and s-a can be found.sBG =AGABBGcs =−=−()()CFAFAGADBDACBGbs +−++=−=−()()CEADCEADBD +−++=BD=Since AD = AF and AG = CE = CF,13Heron’s Proof: Part B (cont.)()()CEBEAGADBDBCBGas +−++=−=−()()CEBDCEADBD +−++=AD=Since BD = BF and AG = CE,Heron’s Proof: Part B (cont.)n In Summary, the important things found from this section of the proof.()sbacBG =++=21AGcs =−BDbs =−ADas =−14Heron’s Proof: Part Cn The same circle inscribed within a triangle is used except three lines are now extended from the diagram.n The segment OL is drawn perpendicular to OB and cuts AB at point K.n The segment AM is drawn from point A perpendicular to AB and intersects OL at point H. n The last segment drawn is BH.n The quadrilateral AHBO is formed.Heron’s Proof: Part C (cont.)n Proposition 4 says the quadrilateral AHBO is cyclic while Proposition 5 by Euclid says the sum of its opposite angles equals two right angles.anglesright 2=∠+∠AOBAHB15Heron’s Proof: Part C (cont.)n By congruence, the angles around the center O reduce to three pairs of equal angles to give:Therefore,anglesrt 4222=++γβαanglesrt 2=++γβαHeron’s Proof: Part C (cont.)n Since , andTherefore, .AOB∠=+αβAOBAHBAOB∠+∠==∠+ anglesrt 2αanglesrt 2=++γβαAHB∠=α16Heron’s Proof: Part C (cont.)n Since and both angles CFO and BAH are right angles, then the two triangles ?COF and ?BHA are similar. n This leads to the following proportion using from Part B that and:which is equivalent to the proportion(*)AHB∠=αrAGOFCFAHAB==CF AG =r OH =rAHAGAB=Heron’s Proof: Part C (cont.)n Since both angles KAH and KDO are right angles and vertical angles AKH and DKO are equal, the two triangles ?KAH and ?KDO are similar. n This leads to the proportion:Which simplifies to(**)KDrKDODAKAH==KDAKrAH=17Heron’s Proof: Part C (cont.)n The two equations(*) and (**)are combined to form the key


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UK MA 330 - Heron’s Formula for Triangular Area

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