Question Number 189114 by normans last updated on 12/Mar/23
Answered by mr W last updated on 12/Mar/23
Commented by mr W last updated on 12/Mar/23
$${c}=\sqrt{{a}^{\mathrm{2}} +\left({b}−{a}\right)^{\mathrm{2}} } \\ $$$$\frac{\left({b}−{a}\right){a}}{{c}}+{c}+\frac{{ab}}{{c}}=\mathrm{13} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{13}{c}\:\:\:…\left({i}\right) \\ $$$$\frac{\left({b}−{a}\right){b}}{{c}}+\frac{{ab}}{{c}}=\mathrm{9} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\mathrm{9}{c}\:\:\:…\left({ii}\right) \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{4}{b}^{\mathrm{2}} }{\mathrm{9}}\:\Rightarrow{a}=\frac{\mathrm{2}{b}}{\mathrm{3}} \\ $$$${b}^{\mathrm{2}} =\mathrm{9}\sqrt{{a}^{\mathrm{2}} +\left({b}−{a}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{b}=\mathrm{3}\sqrt{\mathrm{5}}\:\Rightarrow{a}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\mathrm{45},\:{a}^{\mathrm{2}} =\mathrm{20} \\ $$
Answered by HeferH last updated on 12/Mar/23
Commented by HeferH last updated on 12/Mar/23
$$\ast\:\frac{{x}}{\mathrm{4}}\:=\:\frac{{b}}{{a}};\:{x}\:=\:\frac{\mathrm{4}{b}}{{a}} \\ $$$$\ast\:\frac{{b}}{{b}\:+\:{a}}\:=\:\frac{\mathrm{9}−\frac{\mathrm{4}{b}}{{a}}}{\mathrm{5}} \\ $$$$\:\mathrm{5}{b}\:=\:\mathrm{9}\left({b}+{a}\right)\:−\:\frac{\mathrm{4}{b}\left({b}+{a}\right)}{{a}} \\ $$$$\:\mathrm{5}{ab}\:=\:\mathrm{9}{ab}\:+\:\mathrm{9}{a}^{\mathrm{2}} \:−\mathrm{4}{b}^{\mathrm{2}} \:−\mathrm{4}{ab} \\ $$$$\:\mathrm{4}{b}^{\mathrm{2}} \:=\:\mathrm{9}{a}^{\mathrm{2}} \\ $$$$\:\frac{{b}}{{a}}\:=\:\sqrt{\frac{\mathrm{9}}{\mathrm{4}}}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\Rightarrow\:{x}\:=\:\mathrm{6} \\ $$$$\:\ast\:\mathrm{now}: \\ $$$$\:\left(\mathrm{5}−\mathrm{3}\right)^{\mathrm{2}} \:+\:\mathrm{4}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:=\:\mathrm{20}{u}^{\mathrm{2}} \\ $$$$\:\left(\mathrm{9}\:−\:\mathrm{6}\right)^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} \:=\:\mathrm{45}{u}^{\mathrm{2}} \\ $$