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Question Number 19783 by NECC last updated on 15/Aug/17

$$Thesidesofatriangleareof$$$$lengths\sqrt{(m^2-n^2)},m^2+n^2,2mn.$$$$Showthatitisarightangle\mathrm{\Delta }.$$

Commented by Tinkutara last updated on 15/Aug/17

$$\mathrm{Right}\mathrm{triangle}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{with}\mathrm{these}$$$$\mathrm{sides}.$$

Commented by NECC last updated on 15/Aug/17

$$pleaseproveyourstatement$$$$mathematically.$$

Commented by Tinkutara last updated on 15/Aug/17

$$\mathrm{Since}{m}^2-n^2+4m^2{n}^2\ne m^4+n^4+$$$$2m^4{n}^4.\mathrm{Hence}\mathrm{right}\mathrm{triangle}\mathrm{is}\mathrm{not}$$$$\mathrm{possible}.$$

Answered by Rasheed.Sindhi last updated on 15/Aug/17

$${({\mathrm{m}}^2-{\mathrm{n}}^2)}^2+{\left(2\mathrm{mn}\right)}^2$$$$={\mathrm{m}}^4-{2\mathrm{m}}^2{\mathrm{n}}^2+{\mathrm{n}}^4+{4\mathrm{m}}^2{\mathrm{n}}^2$$$$={\mathrm{m}}^4+{2\mathrm{m}}^2{\mathrm{n}}^2+{\mathrm{n}}^4$$$$={({\mathrm{m}}^2+{\mathrm{n}}^2)}^2$$$$\mathrm{Hence}\mathrm{rightangle}\mathrm{triangle}.$$

Commented by Tinkutara last updated on 15/Aug/17

$$\mathrm{But}\mathrm{the}\mathrm{side}\mathrm{is}\sqrt{m^2-n^2}.\mathrm{Its}\mathrm{square}$$$$\mathrm{will}\mathrm{be}{m}^2-n^2\mathrm{only}.(\mathrm{OK},\mathrm{it}\mathrm{can}\mathrm{be}$$$$\mathrm{considered}\mathrm{a}\mathrm{typing}\mathrm{error}\mathrm{in}\mathrm{the}$$$$\mathrm{question})$$

Commented by NECC last updated on 16/Aug/17

$$thanks....itsclearnow.$$

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