Heron s Formula for Triangular Area


 Jeffrey Camron Hancock
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1 Heron s Formul for Tringulr Are y Christy Willims, Crystl Holom, nd Kyl Gifford Heron of Alexndri Physiist, mthemtiin, nd engineer Tught t the museum in Alexndri Interests were more prtil (mehnis, engineering, mesurement) thn theoretil He is pled somewhere round 75 A.D. (±150) 1
2 Heron s Works Automt Mehni Dioptr Metri Pneumti Ctoptri Belopoei Geometri Stereometri Mensure Cheirolistr The Aeolipile Heron s Aeolipile ws the first reorded stem engine. It ws tken s eing toy ut ould hve possily used n industril revolution 000 yers efore the originl.
3 Metri Mthemtiins knew of its existene for yers ut no tres of it existed In 189 mthemtil historin Pul Tnnery found frgment of it in 13 th entury Prisin mnusript In 1896 R. Shöne found the omplete mnusript in Constntinople. Proposition I.8 of Metri gives the proof of his formul for the re of tringle How is Heron s formul helpful? How would you find the re of the given tringle using the most ommon re formul? 1 A h Sine no height is given, it eomes quite diffiult
4 Heron s Formul Heron s formul llows us to find the re of tringle when only the lengths of the three sides re given. His formul sttes: K s ( s )( s )( s ) Where,, nd, re the lengths of the sides nd s is the semiperimeter of the tringle. The Preliminries
5 Proposition 1 Proposition IV. of Eulid s Elements. The isetors of the ngles of tringle meet t point tht is the enter of the tringles insried irle. (Note: this is lled the inenter) Proposition Proposition VI.8 of Eulid s Elements. In rightngled tringle, if perpendiulr is drwn from the right ngle to the se, the tringles on eh side of it re similr to the whole tringle nd to one nother. 5
6 Proposition 3 In right tringle, the midpoint of the hypotenuse is equidistnt from the three verties. Proposition If AHBO is qudrilterl with digonls AB nd OH, then if HOB nd HAB re right ngles (s shown), then irle n e drwn pssing through the verties A, O, B, nd H. 6
7 Proposition 5 Proposition III. of Eulid s Elements. The opposite ngles of yli qudrilterl sum to two right ngles. Semiperimeter The semiperimeter, s, of tringle with sides,, nd, is s 7
8 Heron s Proof Heron s Proof The proof for this theorem is roken into three prts. Prt A insries irle within tringle to get reltionship etween the tringle s re nd semiperimeter. Prt B uses the sme irle insried within tringle in Prt A to find the terms s, s, nd s in the digrm. Prt C uses the sme digrm with qudrilterl nd the results from Prts A nd B to prove Heron s theorem. 8
9 Resttement of Heron s Formul For tringle hving sides of length,, nd nd re K, we hve K s ( s )( s )( s ) where s is the tringle s semiperimeter. Heron s Proof: Prt A Let ABC e n ritrry tringle suh tht side AB is t lest s long s the other two sides. Insrie irle with enter O nd rdius r inside of the tringle. Therefore, OD OE OF. 9
10 Heron s Proof: Prt A (ont.) Now, the re for the three tringles?aob,?boc, nd?coa is found using the formul Are?AOB Are?BOC Are?COA ½(se)(height). ( AB)( OD 1 r 1 ) ( BC)( OE 1 r 1 ) ( AC)( OF 1 r 1 ) Heron s Proof: Prt A (ont.) K Are We know the re of tringle ABC is K. Therefore ( ABC ) Are( AOB) Are( BOC ) Are( COA) If the res lulted for the tringles?aob,?boc, nd?coa found in the previous slides re sustituted into this eqution, then K is K r r r r rs 10
11 Heron s Proof: Prt B When insriing the irle inside the tringle ABC, three pirs of ongruent tringles re formed (y Eulid s Prop. I.6 AAS). AOD BOD AOF BOE COE COF Heron s Proof: Prt B (ont.) Using orresponding prts of similr tringles, the following reltionships were found: AD AF BD BE CE CF AOD AOF BOD BOE COE COF 11
12 Heron s Proof: Prt B (ont.) The se of the tringle ws extended to point G where AG CE. Therefore, using onstrution nd ongruene of tringle: BG BD AD AG BD AD CE ( BD AD CE) 1 1 [( BD BE ) ( AD AF) ( CE CF )] 1 [( BD AD) ( BE CE) ( AF CF )] 1 ( AB BC AC) BG BG BG BG ( ) s BG 1 Heron s Proof: Prt B (ont.) Sine BG s, the semiperimeter of the tringle is the long segment strighten out. Now, s, s, nd s n e found. s BG AB AG Sine AD AF nd AG CE CF, BG AC ( BD AD AG) ( AF CF ) ( BD AD CE) ( AD CE) s BD 1
13 Heron s Proof: Prt B (ont.) Sine BD BF nd AG CE, BG BC ( BD AD AG) ( BE CE) ( BD AD CE) ( BD CE) s AD Heron s Proof: Prt B (ont.) In Summry, the importnt things found from this setion of the proof. ( ) s BG 1 s AG s s BD AD 13
14 Heron s Proof: Prt C The sme irle insried within tringle is used exept three lines re now extended from the digrm. The segment OL is drwn perpendiulr to OB nd uts AB t point K. The segment AM is drwn from point A perpendiulr to AB nd intersets OL t point H. The lst segment drwn is BH. The qudrilterl AHBO is formed. Heron s Proof: Prt C (ont.) Proposition sys the qudrilterl AHBO is yli while Proposition 5 y Eulid sys the sum of its opposite ngles equls two right ngles. AHB AOB right ngles 1
15 Heron s Proof: Prt C (ont.) By ongruene, the ngles round the enter O redue to three pirs of equl ngles to give: α β γ rt ngles Therefore, α β γ rt ngles α Sine Heron s Proof: Prt C (ont.) β γ β α AOB rt ngles Therefore,., nd α AOB rt ngles AHB AOB α AHB 15
16 Heron s Proof: Prt C (ont.) Sine α AHB nd oth ngles CFO nd BAH re right ngles, then the two tringles?cof nd?bha re similr. This leds to the following proportion using from Prt B tht AG CF nd OH r : AB CF AG AH OF r whih is equivlent to the proportion AB AG AH r (*) Heron s Proof: Prt C (ont.) Sine oth ngles KAH nd KDO re right ngles nd vertil ngles AKH nd DKO re equl, the two tringles?kah nd?kdo re similr. This leds to the proportion: AH AK Whih simplifies to AH r OD KD AK KD r KD (**) 16
17 Heron s Proof: Prt C (ont.) The two equtions AB AG AH (*) nd AH AK (**) r r KD re omined to form the key eqution: AB AG AK KD (***) Heron s Proof: Prt C (ont.) By Proposition,?KDO is similr to?odb where?bok hs ltitude ODr. This gives the eqution: KD r r BD whih simplifies ( )( ) to KD BD r (****) (r is the men proportionl etween mgnitudes KD nd BD) 17
18 Heron s Proof: Prt C (ont.) One is dded to eqution (***), the eqution is simplified, then BG/BG is multiplied on the right nd BD/BD is multiplied on the left, then simplified. AB AG AK KD BG BG AD BD BG AG KD BD AB AK 1 1 AG KD AB AG AK KD AG KD BG AD AG KD Using the eqution KD BD r (****) this simplifies to: ( BG) ( AG)( BG) ( )( ) ( AD)( BD) r Heron s Proof: Prt C (ont.) Crossmultiplition of ( BG ) ( AG)( BG) ( BG) ( AG)( BG)( AD)( BD) r Prt B re needed. These re: ( AD)( BD) r produed. Next, the results from BG s s AG s s BD AD 18
19 Heron s Proof: Prt C (ont.) The results from Prt B re sustituted into the eqution: ( BG) ( AG)( BG)( AD)( BD) r r s ( s )( s)( s )( s ) We know rememer from Prt A tht Krs, so the eqution eomes: K s ( s )( s )( s ) Thus proving Heron s Theorem of Tringulr Are Applition of Heron s Formul We n now use Heron s Formul to find the re of the previously given tringle 5 17 s ( ) K ( 3 17)( 3 5)( 3 6)
20 Euler s Proof of Heron s Formul Leonhrd Euler provided proof of Heron s Formul in 178 pper entitled Vrie demonstrtiones geometrie His proof is s follows Euler s Proof (Piture) For referene, this is piture of the proof y Euler. 0
21 Euler s Proof (ont.) ABC Begin with hving sides,, nd nd ngles, β nd γ Insrie irle within the tringle Let O e the enter of the insried irle with rdius r OS OU From the onstrution of the inenter, we know tht segments OA, OB, nd OC iset the ngles of ABC with OAB α β, OBA, γ nd OCA Euler s Proof (ont.) Extend BO nd onstrut perpendiulr from A interseting this extended line t V Denote y N the intersetion of the extensions of segment AV nd rdius OS Beuse AOV is n exterior ngle of AOB, oserve tht AOV OAB OBA α β 1
22 Euler s Proof (ont.) Beuse AOV is right, we know tht nd OAV re omplementry Thus, But α Therefore, α β OAV 90 β γ 90 s well γ OAV OCU AOV Euler s Proof (ont.) Right tringles nd re similr so we get AV / VO CU / OU z / r Also dedue tht nd re similr, s re nd, s well s nd Hene NAS This results in So, OAV NOV BAV AV / AB OV / ON z r AB ON OCU NAS NOV x y SN r ( SN ) r( x y z) rs z BAV
23 Euler s Proof (ont.) Beuse they re vertil ngles, BOS nd VON re ongruent, so OBS 90 BOS 90 VON ANS NAS Hene, nd This results in BOS re similr SN / AS BS / OS SN / x SN y / r ( xy) / r Euler s Proof (ont.) Lstly, Euler onluded tht Are z ( ABC ) rs rs( rs) z( SN )( rs) xy ( ) rs sxyz s( s )( s )( s ) r 3
24 Pythgoren Theorem Heron s Formul n e used s proof of the Pythgoren Theorem Pythgoren Theorem from Heron s Formul Suppose we hve right tringle with hypotenuse of length, nd legs of length nd The semiperimeter is: s
25 5 Pythgoren Thm. from Heron s Formul (ont.) s Similrly s nd s After pplying lger, we get Pythgoren Thm. from Heron s Formul (ont.) ( )( )( )( ) ( ) [ ]( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) ( ) ( ) ( ) ( )
26 6 Pythgoren Thm. from Heron s Formul (ont.) Returning to Heron s Formul, we get the re of the tringle to e ( )( )( ) ( )( )( )( ) ( ) 16 s s s s K Pythgoren Thm. from Heron s Formul (ont.) Beuse we know the height of this tringle is, we n equte our expression to the expression h K 1 1 ( ) 16 ( ) Equting oth expressions of K nd squring oth sides, we get Crossmultiplition gives us
27 7 Pythgoren Thm. from Heron s Formul (ont.) Tking ll terms to the left side, we hve ( ) ( ) ( ) ( ) [ ] ( ) Thus, Heron s formul provides us with nother proof of the Pythgoren Theorem
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