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Question-179198

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Question Number 179198 by Acem last updated on 26/Oct/22
Commented by Acem last updated on 26/Oct/22
Area_(Square) ?
$${Area}_{{Square}} ? \\ $$
Commented by Acem last updated on 26/Oct/22
All the methods have good ideas, Thanks for all
$${All}\:{the}\:{methods}\:{have}\:{good}\:{ideas},\:{Thanks}\:{for}\:{all} \\ $$
Answered by cortano1 last updated on 26/Oct/22
(5/x) = (x/(20)) ⇒x^2 =100
$$\:\frac{\mathrm{5}}{\mathrm{x}}\:=\:\frac{\mathrm{x}}{\mathrm{20}}\:\Rightarrow\mathrm{x}^{\mathrm{2}} =\mathrm{100}\: \\ $$$$\: \\ $$
Commented by Rasheed.Sindhi last updated on 26/Oct/22
Efficient way!
$$\mathcal{E}{fficient}\:{way}! \\ $$
Commented by Acem last updated on 26/Oct/22
Very well!
$${Very}\:{well}! \\ $$
Answered by Rasheed.Sindhi last updated on 26/Oct/22
Through Areas.... ▲=(1/2)(a+5)(a+20) ▲=(1/2)(a)(20) , ▲=(1/2)(5)(a) ■=▲−▲−▲ =(1/2)(a+5)(a+20)−(1/2)(a)(20)−(1/2)(5)(a) =(1/2)(a^2 +25a+100−20a−5a) 2■=■+100 [∵ a^2 =■] ■=100
$$\boldsymbol{\mathcal{T}{hrough}}\:\boldsymbol{\mathcal{A}{reas}}…. \\ $$$$\blacktriangle=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+\mathrm{5}\right)\left({a}+\mathrm{20}\right) \\ $$$$\blacktriangle=\frac{\mathrm{1}}{\mathrm{2}}\left({a}\right)\left(\mathrm{20}\right)\:\:\:,\:\:\:\blacktriangle=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}\right)\left({a}\right) \\ $$$$\blacksquare=\blacktriangle−\blacktriangle−\blacktriangle \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+\mathrm{5}\right)\left({a}+\mathrm{20}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({a}\right)\left(\mathrm{20}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}\right)\left({a}\right) \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} +\mathrm{25}{a}+\mathrm{100}−\mathrm{20}{a}−\mathrm{5}{a}\right) \\ $$$$\mathrm{2}\blacksquare=\blacksquare+\mathrm{100}\:\:\:\:\left[\because\:{a}^{\mathrm{2}} =\blacksquare\right] \\ $$$$\blacksquare=\mathrm{100} \\ $$
Commented by Acem last updated on 26/Oct/22
Very nice idea too
$${Very}\:{nice}\:{idea}\:{too} \\ $$
Answered by Rasheed.Sindhi last updated on 26/Oct/22
With the help of_(Hypotenuses) H=(√((a+5)^2 +(a+20)^2 )) h_1 =(√(a^2 +5^2 )) , h_2 =(√(a^2 +20^2 )) H=h_1 +h_2 (√((a+5)^2 +(a+20)^2 )) =(√(a^2 +5^2 )) +(√(a^2 +20^2 )) a^2 +10a+25+a^2 +40a+400 =a^2 +25+a^2 +400+2(√(a^2 +25)) (√(a^2 +400)) 2(√(a^2 +25)) (√(a^2 +400)) =50a (a^2 +25)(a^2 +400)=625a^2 (a^2 )^2 −200a^2 +10000=0 (a^2 −100)^2 =0 a^2 =100 ■=100
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\boldsymbol{\mathrm{Hypotenuses}}} {\mathrm{W}\boldsymbol{\mathrm{ith}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{of}}} \\ $$$$\mathrm{H}=\sqrt{\left({a}+\mathrm{5}\right)^{\mathrm{2}} +\left({a}+\mathrm{20}\right)^{\mathrm{2}} }\: \\ $$$$\mathrm{h}_{\mathrm{1}} =\sqrt{{a}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }\:\:\:\:,\:\:\:\mathrm{h}_{\mathrm{2}} =\sqrt{{a}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} }\: \\ $$$$\mathrm{H}=\mathrm{h}_{\mathrm{1}} +\mathrm{h}_{\mathrm{2}} \\ $$$$\sqrt{\left({a}+\mathrm{5}\right)^{\mathrm{2}} +\left({a}+\mathrm{20}\right)^{\mathrm{2}} }\:=\sqrt{{a}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }\:+\sqrt{{a}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} }\: \\ $$$${a}^{\mathrm{2}} +\mathrm{10}{a}+\mathrm{25}+{a}^{\mathrm{2}} +\mathrm{40}{a}+\mathrm{400} \\ $$$$\:\:\:\:\:\:={a}^{\mathrm{2}} +\mathrm{25}+{a}^{\mathrm{2}} +\mathrm{400}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{25}}\:\sqrt{{a}^{\mathrm{2}} +\mathrm{400}} \\ $$$$\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{25}}\:\sqrt{{a}^{\mathrm{2}} +\mathrm{400}}\:=\mathrm{50}{a} \\ $$$$\:\:\:\left({a}^{\mathrm{2}} +\mathrm{25}\right)\left({a}^{\mathrm{2}} +\mathrm{400}\right)=\mathrm{625}{a}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{200}{a}^{\mathrm{2}} +\mathrm{10000}=\mathrm{0} \\ $$$$\left({a}^{\mathrm{2}} −\mathrm{100}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:{a}^{\mathrm{2}} =\mathrm{100} \\ $$$$\:\blacksquare=\mathrm{100} \\ $$
Commented by Acem last updated on 26/Oct/22
You′re accurate! thanks
$${You}'{re}\:{accurate}!\:{thanks} \\ $$
Answered by CElcedricjunior last updated on 26/Oct/22
calculons l′aire du carre d′apres notre celebre frere Thales on a deux droites paralleles donc ((20)/(20+c))=(a/(a+(√(c^2 +25))))=(c/(5+c)) =>((20)/(20+c))=(c/(5+c))=>100+20c=20c+c^2 =>c^2 =100 d′ou^� l′aire du carre est c^2 =100 um .............le celebre]cedric junior.........
$$\boldsymbol{\mathrm{calculons}}\:\boldsymbol{\mathrm{l}}'\boldsymbol{\mathrm{aire}}\:\boldsymbol{\mathrm{du}}\:\boldsymbol{\mathrm{carre}} \\ $$$$\boldsymbol{\mathrm{d}}'\boldsymbol{{ap}\mathrm{res}}\:\boldsymbol{\mathrm{notre}}\:\boldsymbol{\mathrm{celebre}}\:\boldsymbol{\mathrm{frere}}\:\boldsymbol{\mathrm{Thales}} \\ $$$$\boldsymbol{\mathrm{on}}\:\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{deux}}\:\boldsymbol{\mathrm{droites}}\:\boldsymbol{\mathrm{parallele}{s}} \\ $$$$\boldsymbol{\mathrm{donc}} \\ $$$$\frac{\mathrm{20}}{\mathrm{20}+\boldsymbol{\mathrm{c}}}=\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{a}}+\sqrt{\boldsymbol{\mathrm{c}}^{\mathrm{2}} +\mathrm{25}}}=\frac{\boldsymbol{\mathrm{c}}}{\mathrm{5}+\boldsymbol{\mathrm{c}}} \\ $$$$=>\frac{\mathrm{20}}{\mathrm{20}+\boldsymbol{\mathrm{c}}}=\frac{\boldsymbol{\mathrm{c}}}{\mathrm{5}+\boldsymbol{\mathrm{c}}}=>\mathrm{100}+\mathrm{20}\boldsymbol{\mathrm{c}}=\mathrm{20}\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{c}}^{\mathrm{2}} \\ $$$$=>\boldsymbol{\mathrm{c}}^{\mathrm{2}} =\mathrm{100} \\ $$$$\boldsymbol{\mathrm{d}}'\boldsymbol{\mathrm{o}}\grave {\boldsymbol{\mathrm{u}}}\:\boldsymbol{\mathrm{l}}'\boldsymbol{\mathrm{aire}}\:\boldsymbol{\mathrm{du}}\:\boldsymbol{\mathrm{carre}}\:\boldsymbol{\mathrm{est}}\:\boldsymbol{\mathrm{c}}^{\mathrm{2}} =\mathrm{100}\:\boldsymbol{\mathrm{um}} \\ $$$$\: \\ $$$$\left.\:………….{le}\:{celebre}\right]{cedric}\:{junior}……… \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 26/Oct/22
Efficient approach!
$$\mathbb{E}\mathrm{fficient}\:\mathrm{approach}! \\ $$
Commented by Acem last updated on 26/Oct/22
It was wonderful! thanks
$${It}\:{was}\:{wonderful}!\:{thanks} \\ $$
Answered by Rasheed.Sindhi last updated on 29/Oct/22
Trigonometric approach { ((tan θ=(a/(20))⇒θ=tan^(−1) (a/(20)))),((tan θ=(5/a)⇒θ=tan^(−1) (5/a))) :}⇒tan^(−1) (a/(20))=tan^(−1) (5/a) ⇒(a/(20))=(5/a)⇒a^2 =100⇒ ■=100
$$\mathbb{T}\mathrm{rigonometric}\:\mathrm{approach} \\ $$$$\begin{cases}{\mathrm{tan}\:\theta=\frac{{a}}{\mathrm{20}}\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{\mathrm{20}}}\\{\mathrm{tan}\:\theta=\frac{\mathrm{5}}{{a}}\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{5}}{{a}}}\end{cases}\Rightarrow\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{\mathrm{20}}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{5}}{{a}} \\ $$$$\Rightarrow\frac{{a}}{\mathrm{20}}=\frac{\mathrm{5}}{{a}}\Rightarrow{a}^{\mathrm{2}} =\mathrm{100}\Rightarrow\:\blacksquare=\mathrm{100} \\ $$
Commented by Rasheed.Sindhi last updated on 29/Oct/22
Commented by Acem last updated on 29/Oct/22
Genius idea! thank you
$${Genius}\:{idea}!\:{thank}\:{you} \\ $$

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