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