Question Number 187790 by cherokeesay last updated on 21/Feb/23
Answered by mr W last updated on 21/Feb/23
Commented by mr W last updated on 22/Feb/23
![(2a−2)^2 +[((√5)−1)a]^2 =(2a)^2 +2^2 ⇒a=(4/(3−(√5)))=3+(√5) area of square =(2a)^2 =4(3+(√5))^2 =56+24(√5)≈109.666](https://www.tinkutara.com/question/Q187802.png)
$$\left(\mathrm{2}{a}−\mathrm{2}\right)^{\mathrm{2}} +\left[\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){a}\right]^{\mathrm{2}} =\left(\mathrm{2}{a}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{4}}{\mathrm{3}−\sqrt{\mathrm{5}}}=\mathrm{3}+\sqrt{\mathrm{5}} \\ $$$${area}\:{of}\:{square}\:=\left(\mathrm{2}{a}\right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$\:\:=\mathrm{56}+\mathrm{24}\sqrt{\mathrm{5}}\approx\mathrm{109}.\mathrm{666} \\ $$
Commented by cherokeesay last updated on 22/Feb/23
$${so}\:{nice}\:{thank}\:{you}\:{sir}. \\ $$
Answered by HeferH last updated on 22/Feb/23
![let 2x be the side of the square: (2x−2)^2 +(x(√5)−x)^2 =2^2 +(2x)^2 4x^2 +4−8x + 5x^2 +x^2 −2x^2 (√5) = 4 + 4x^2 6x^2 −8x−2x^2 (√5) = 0 6x −8 −2x(√5) = 0 3x−x(√5) = 4 x(3−(√5)) = 4 x = (4/(3−(√5))) ; 4x^2 =4[ ((4(3+(√5)))/(9−5))]^2 = 4(14+6(√5)) u^2 = 56+24(√5)u^2](https://www.tinkutara.com/question/Q187801.png)
$${let}\:\mathrm{2}{x}\:{be}\:{the}\:{side}\:{of}\:{the}\:{square}: \\ $$$$\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{2}} +\left({x}\sqrt{\mathrm{5}}−{x}\right)^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} +\left(\mathrm{2}{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}−\mathrm{8}{x}\:+\:\mathrm{5}{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} \sqrt{\mathrm{5}}\:=\:\mathrm{4}\:+\:\mathrm{4}{x}^{\mathrm{2}} \\ $$$$\:\mathrm{6}{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{2}{x}^{\mathrm{2}} \sqrt{\mathrm{5}}\:=\:\mathrm{0} \\ $$$$\:\mathrm{6}{x}\:−\mathrm{8}\:−\mathrm{2}{x}\sqrt{\mathrm{5}}\:=\:\mathrm{0} \\ $$$$\:\mathrm{3}{x}−{x}\sqrt{\mathrm{5}}\:=\:\mathrm{4} \\ $$$$\:{x}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)\:=\:\mathrm{4} \\ $$$$\:{x}\:=\:\frac{\mathrm{4}}{\mathrm{3}−\sqrt{\mathrm{5}}}\:;\:\mathrm{4}{x}^{\mathrm{2}} =\mathrm{4}\left[\:\frac{\mathrm{4}\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)}{\mathrm{9}−\mathrm{5}}\right]^{\mathrm{2}} \:=\:\mathrm{4}\left(\mathrm{14}+\mathrm{6}\sqrt{\mathrm{5}}\right)\:{u}^{\mathrm{2}} = \\ $$$$\:\mathrm{56}+\mathrm{24}\sqrt{\mathrm{5}}{u}^{\mathrm{2}} \\ $$
Commented by cherokeesay last updated on 22/Feb/23
$${perfect}\:! \\ $$$${thank}\:{you}\:{sir}. \\ $$