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Question-49384




Question Number 49384 by Pk1167156@gmail.com last updated on 06/Dec/18
Commented by Kunal12588 last updated on 06/Dec/18
for △ABD  AD=39  pythagorean triples with side 39  1. 39,52,65           ×  2.39,80,89     3.39,252,255      ×  4.39,760,761      ×  5.15,36,39            ×  in △ABD   AD <AB(hypotaneous)  so 5^(th)  triple is not for here  also, ar(△ABD)<ar(△ABC)=4920  here BD=52,80,252,760  BD→52  (1/2)×39×52<4920  BD→80  (1/2)×39×80<4920  BD→252  (1/2)252×39<4920  BD→760  (1/2)×39×760>4920 so 4^(th)  triple is not for here  case 1 . BD=52  ar(△ABC)=4920  (1/2)×(39+DC)×52=4920  39+DC=((4920)/(26))=((2460)/(13)) but DC is integer   BD≠52  case 2. BD=80  (1/2)×(39+DC)×80=4920  39+DC=((492)/4)=123  so,DC=123−39=84, AC=123  AB=(√(39^2 +BD^2 ))=(√(39^2 +80^2 ))=(√(1521+6400))  AB=(√(7921))=89  BC=(√(BD^2 +DC^2 ))=(√(80^2 +84^2 ))=(√(6400+7056))  BC=(√(13456))=116  question is without units as answer
$${for}\:\bigtriangleup{ABD} \\ $$$${AD}=\mathrm{39} \\ $$$${pythagorean}\:{triples}\:{with}\:{side}\:\mathrm{39} \\ $$$$\mathrm{1}.\:\mathrm{39},\mathrm{52},\mathrm{65}\:\:\:\:\:\:\:\:\:\:\:× \\ $$$$\mathrm{2}.\mathrm{39},\mathrm{80},\mathrm{89}\:\:\: \\ $$$$\mathrm{3}.\mathrm{39},\mathrm{252},\mathrm{255}\:\:\:\:\:\:× \\ $$$$\mathrm{4}.\mathrm{39},\mathrm{760},\mathrm{761}\:\:\:\:\:\:× \\ $$$$\mathrm{5}.\mathrm{15},\mathrm{36},\mathrm{39}\:\:\:\:\:\:\:\:\:\:\:\:× \\ $$$${in}\:\bigtriangleup{ABD}\: \\ $$$${AD}\:<{AB}\left({hypotaneous}\right) \\ $$$${so}\:\mathrm{5}^{{th}} \:{triple}\:{is}\:{not}\:{for}\:{here} \\ $$$${also},\:{ar}\left(\bigtriangleup{ABD}\right)<{ar}\left(\bigtriangleup{ABC}\right)=\mathrm{4920} \\ $$$${here}\:{BD}=\mathrm{52},\mathrm{80},\mathrm{252},\mathrm{760} \\ $$$${BD}\rightarrow\mathrm{52} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{39}×\mathrm{52}<\mathrm{4920} \\ $$$${BD}\rightarrow\mathrm{80} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{39}×\mathrm{80}<\mathrm{4920} \\ $$$${BD}\rightarrow\mathrm{252} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{252}×\mathrm{39}<\mathrm{4920} \\ $$$${BD}\rightarrow\mathrm{760} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{39}×\mathrm{760}>\mathrm{4920}\:{so}\:\mathrm{4}^{{th}} \:{triple}\:{is}\:{not}\:{for}\:{here} \\ $$$${case}\:\mathrm{1}\:.\:{BD}=\mathrm{52} \\ $$$${ar}\left(\bigtriangleup{ABC}\right)=\mathrm{4920} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{39}+{DC}\right)×\mathrm{52}=\mathrm{4920} \\ $$$$\mathrm{39}+{DC}=\frac{\mathrm{4920}}{\mathrm{26}}=\frac{\mathrm{2460}}{\mathrm{13}}\:{but}\:{DC}\:{is}\:{integer}\: \\ $$$${BD}\neq\mathrm{52} \\ $$$${case}\:\mathrm{2}.\:{BD}=\mathrm{80} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{39}+{DC}\right)×\mathrm{80}=\mathrm{4920} \\ $$$$\mathrm{39}+{DC}=\frac{\mathrm{492}}{\mathrm{4}}=\mathrm{123} \\ $$$${so},{DC}=\mathrm{123}−\mathrm{39}=\mathrm{84},\:{AC}=\mathrm{123} \\ $$$${AB}=\sqrt{\mathrm{39}^{\mathrm{2}} +{BD}^{\mathrm{2}} }=\sqrt{\mathrm{39}^{\mathrm{2}} +\mathrm{80}^{\mathrm{2}} }=\sqrt{\mathrm{1521}+\mathrm{6400}} \\ $$$${AB}=\sqrt{\mathrm{7921}}=\mathrm{89} \\ $$$${BC}=\sqrt{{BD}^{\mathrm{2}} +{DC}^{\mathrm{2}} }=\sqrt{\mathrm{80}^{\mathrm{2}} +\mathrm{84}^{\mathrm{2}} }=\sqrt{\mathrm{6400}+\mathrm{7056}} \\ $$$${BC}=\sqrt{\mathrm{13456}}=\mathrm{116} \\ $$$${question}\:{is}\:{without}\:{units}\:{as}\:{answer} \\ $$
Answered by Kunal12588 last updated on 06/Dec/18
AB=89  BC=116  AC=123
$$\mathrm{AB}=\mathrm{89} \\ $$$$\mathrm{BC}=\mathrm{116} \\ $$$$\mathrm{AC}=\mathrm{123} \\ $$
Commented by Kunal12588 last updated on 06/Dec/18
thank you sir who gave me idea of pythagorean triples
$${thank}\:{you}\:{sir}\:{who}\:{gave}\:{me}\:{idea}\:{of}\:{pythagorean}\:{triples} \\ $$
Commented by Pk1167156@gmail.com last updated on 06/Dec/18
Nice sir
Commented by MJS last updated on 06/Dec/18
I guess it was me?  you′re welcome...
$$\mathrm{I}\:\mathrm{guess}\:\mathrm{it}\:\mathrm{was}\:\mathrm{me}? \\ $$$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}… \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
yes sir thanks to you I learned about pythagorean  triple and how to find them.
$${yes}\:{sir}\:{thanks}\:{to}\:{you}\:{I}\:{learned}\:{about}\:{pythagorean} \\ $$$${triple}\:{and}\:{how}\:{to}\:{find}\:{them}. \\ $$

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