Question Number 89097 by TawaTawa1 last updated on 15/Apr/20
Commented by mr W last updated on 15/Apr/20
$$\sqrt{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{10}+\mathrm{5}\right)^{\mathrm{2}} }=\mathrm{5}\sqrt{\mathrm{10}} \\ $$$${s}=\mathrm{5}\sqrt{\mathrm{10}}−\frac{\mathrm{5}}{\mathrm{5}\sqrt{\mathrm{10}}}\left(\mathrm{15}+\mathrm{5}\right)=\mathrm{3}\sqrt{\mathrm{10}} \\ $$$${area}\:={s}^{\mathrm{2}} =\mathrm{9}×\mathrm{10}=\mathrm{90} \\ $$
Answered by $@ty@m123 last updated on 15/Apr/20
$${Let}\:{sides}\:{of}\:{the}\:{right}\:{angled}\:\bigtriangleup\:{with} \\ $$$${hypotaneous}\:\mathrm{5}\:{be} \\ $$$${x}\:{and}\:{y}\:{where}\:{y}>{x}.\:{Then} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \:…\left(\mathrm{1}\right) \\ $$$${Let}\:{z}\:{be}\:{the}\:{side}\:{of}\:{coloured}\:{square}. \\ $$$${Then}, \\ $$$$\frac{{x}}{{y}}=\frac{{y}}{{y}+{z}}=\frac{\mathrm{5}}{\mathrm{15}} \\ $$$$\frac{{x}}{{y}}=\frac{{y}}{{y}+{z}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3}{y}={y}+{z}\:\Rightarrow\mathrm{2}{y}={z}\:…\left(\mathrm{2}\right) \\ $$$$\:\&\:{y}=\mathrm{3}{x}\:…..\left(\mathrm{3}\right) \\ $$$${From}\left(\mathrm{1}\right)\:\&\left(\mathrm{3}\right), \\ $$$$\left(\frac{{y}}{\mathrm{3}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{25} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{9}}+{y}^{\mathrm{2}} =\mathrm{25} \\ $$$$\mathrm{10}{y}^{\mathrm{2}} =\mathrm{25}×\mathrm{9} \\ $$$${y}^{\mathrm{2}} =\frac{\mathrm{45}}{\mathrm{2}}…\left(\mathrm{4}\right) \\ $$$$\therefore\:{From}\:\left(\mathrm{2}\right), \\ $$$${z}^{\mathrm{2}} =\mathrm{4}{y}^{\mathrm{2}} \\ $$$$\Rightarrow{Area}\:{of}\:{coloured}\:{square}\:=\mathrm{4}×\frac{\mathrm{45}}{\mathrm{2}} \\ $$$$=\mathrm{90}\:{sq}.{unit} \\ $$
Commented by john santu last updated on 15/Apr/20
$${something}\:{wrong}\:{sir} \\ $$$${it}\:{should}\:{be}\:\frac{{x}}{{y}}\:=\:\frac{\mathrm{5}}{\mathrm{15}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${y}\:=\:\mathrm{3}{x}\:\Rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{10}{x}^{\mathrm{2}} \:=\:\mathrm{25} \\ $$$${x}^{\mathrm{2}} =\:\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow{z}\:=\:\mathrm{6}{x}\: \\ $$$$\Rightarrow\:{z}^{\mathrm{2}} \:=\:\mathrm{36}\:{x}^{\mathrm{2}} \:=\:\mathrm{36}×\frac{\mathrm{5}}{\mathrm{2}}=\:\mathrm{90}\:{sq}\:{unit} \\ $$
Commented by $@ty@m123 last updated on 15/Apr/20
$${Thanks}\:{for}\:{correction}. \\ $$$${I}\:{am}\:{expert}\:{in}\:{committing}\:{silly}\:{mistakes}. \\ $$$$\left.:\left.\right)\left.\:\::\right)\:\::\right) \\ $$
Commented by john santu last updated on 15/Apr/20
$$\left.\vdots\left.\right)\:\vdots\right)\:{hahaha} \\ $$