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Question Number 49384 by Pk1167156@gmail.com last updated on 06/Dec/18

Commented by Kunal12588 last updated on 06/Dec/18

$$for△ABD$$$$AD=39$$$$pythagoreantripleswithside39$$$$1.39,52,65\times$$$$2.39,80,89$$$$3.39,252,255\times$$$$4.39,760,761\times$$$$5.15,36,39\times$$$$in△ABD$$$$AD<AB\left(hypotaneous\right)$$$$so{5}^{th}tripleisnotforhere$$$$also,ar(△ABD)<ar(△ABC)=4920$$$$hereBD=52,80,252,760$$$$BD\to 52$$$$\frac{1}{2}\times 39\times 52<4920$$$$BD\to 80$$$$\frac{1}{2}\times 39\times 80<4920$$$$BD\to 252$$$$\frac{1}{2}252\times 39<4920$$$$BD\to 760$$$$\frac{1}{2}\times 39\times 760>4920so{4}^{th}tripleisnotforhere$$$$case1.BD=52$$$$ar(△ABC)=4920$$$$\frac{1}{2}\times (39+DC)\times 52=4920$$$$39+DC=\frac{4920}{26}=\frac{2460}{13}butDCisinteger$$$$BD\ne 52$$$$case2.BD=80$$$$\frac{1}{2}\times (39+DC)\times 80=4920$$$$39+DC=\frac{492}{4}=123$$$$so,DC=123-39=84,AC=123$$$$AB=\sqrt{{39}^2+{BD}^2}=\sqrt{{39}^2+{80}^2}=\sqrt{1521+6400}$$$$AB=\sqrt{7921}=89$$$$BC=\sqrt{{BD}^2+{DC}^2}=\sqrt{{80}^2+{84}^2}=\sqrt{6400+7056}$$$$BC=\sqrt{13456}=116$$$$questioniswithoutunitsasanswer$$

Answered by Kunal12588 last updated on 06/Dec/18

$$\mathrm{AB}=89$$$$\mathrm{BC}=116$$$$\mathrm{AC}=123$$

Commented by Kunal12588 last updated on 06/Dec/18

$$thankyousirwhogavemeideaofpythagoreantriples$$

Commented by Pk1167156@gmail.com last updated on 06/Dec/18

Nice sir

Commented by MJS last updated on 06/Dec/18

$$\mathrm{I}\mathrm{guess}\mathrm{it}\mathrm{was}\mathrm{me}?$$$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}...$$

Commented by Kunal12588 last updated on 07/Dec/18

$$yessirthankstoyouIlearnedaboutpythagorean$$$$tripleandhowtofindthem.$$

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