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