The-sides-of-a-triangle-are-of-lengths-m-2-n-2-m-2-n-2-2mn-Show-that-it-is-a-right-angle-

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

$${The}\:{sides}\:{of}\:{a}\:{triangle}\:{are}\:{of} \\ $$$${lengths}\:\sqrt{\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)}\:,{m}^{\mathrm{2}} +{n}^{\mathrm{2}} ,\:\mathrm{2}{mn}. \\ $$$${Show}\:{that}\:{it}\:{is}\:{a}\:{right}\:{angle}\:\Delta. \\ $$

Commented by Tinkutara last updated on 15/Aug/17

Right triangle is not possible with these sides.

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

Commented by NECC last updated on 15/Aug/17

$${please}\:{prove}\:{your}\:{statement} \\ $$$${mathematically}. \\ $$

Commented by Tinkutara last updated on 15/Aug/17

Since m^2 − n^2 + 4m^2 n^2 ≠ m^4 + n^4 + 2m^4 n^4 . Hence right triangle is not possible.

$$\mathrm{Since}\:{m}^{\mathrm{2}} \:−\:{n}^{\mathrm{2}} \:+\:\mathrm{4}{m}^{\mathrm{2}} {n}^{\mathrm{2}} \:\neq\:{m}^{\mathrm{4}} \:+\:{n}^{\mathrm{4}} \:+ \\ $$$$\mathrm{2}{m}^{\mathrm{4}} {n}^{\mathrm{4}} .\:\mathrm{Hence}\:\mathrm{right}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{possible}. \\ $$

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

(m^2 −n^2 )^2 +(2mn)^2 =m^4 −2m^2 n^2 +n^4 +4m^2 n^2 =m^4 +2m^2 n^2 +n^4 =(m^2 +n^2 )^2 Hence rightangle triangle.

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

Commented by Tinkutara last updated on 15/Aug/17

But the side is (√(m^2 − n^2 )). Its square will be m^2 − n^2 only. (OK, it can be considered a typing error in the question)

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

Commented by NECC last updated on 16/Aug/17

$${thanks}….\:{its}\:{clear}\:{now}. \\ $$

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