Question Number 189808 by normans last updated on 22/Mar/23
Answered by mr W last updated on 22/Mar/23
Commented by mr W last updated on 22/Mar/23
$$\mathrm{7}^{\mathrm{2}} +\left(\mathrm{7}+{x}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{13}}{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{13}{a}^{\mathrm{2}} −\mathrm{49}}−\mathrm{7} \\ $$$$ \\ $$$$\sqrt{\left(\mathrm{2}{a}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }+\frac{\mathrm{3}{a}×{x}}{\mathrm{2}{a}}=\mathrm{7}+{x} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} −\mathrm{28}{x}+\mathrm{196}−\mathrm{16}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{14}−\mathrm{4}\sqrt{\mathrm{5}{a}^{\mathrm{2}} −\mathrm{49}}}{\mathrm{5}} \\ $$$$\frac{\mathrm{14}−\mathrm{4}\sqrt{\mathrm{5}{a}^{\mathrm{2}} −\mathrm{49}}}{\mathrm{5}}=\sqrt{\mathrm{13}{a}^{\mathrm{2}} −\mathrm{49}}−\mathrm{7} \\ $$$$\mathrm{49}−\mathrm{4}\sqrt{\mathrm{5}{a}^{\mathrm{2}} −\mathrm{49}}=\mathrm{5}\sqrt{\mathrm{13}{a}^{\mathrm{2}} −\mathrm{49}} \\ $$$$\mathrm{58}−\mathrm{5}{a}^{\mathrm{2}} =\mathrm{8}\sqrt{\mathrm{5}{a}^{\mathrm{2}} −\mathrm{49}} \\ $$$${a}^{\mathrm{4}} −\mathrm{36}{a}^{\mathrm{2}} +\mathrm{260}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\mathrm{10} \\ $$$${blue}\:{area}\:=\mathrm{6}{a}^{\mathrm{2}} =\mathrm{60}\:\checkmark \\ $$
Answered by HeferH last updated on 22/Mar/23
Commented by HeferH last updated on 22/Mar/23
$$\mathrm{7}\:+\:\frac{\mathrm{2k}}{\mathrm{3}}\:=\:\mathrm{3k} \\ $$$$\:\mathrm{k}\:=\:\mathrm{3} \\ $$$$\:\mathrm{k}^{\mathrm{2}} +\mathrm{9k}^{\mathrm{2}} =\:\mathrm{9a}^{\mathrm{2}} \\ $$$$\:\mathrm{a}^{\mathrm{2}} \:=\:\frac{\mathrm{10k}^{\mathrm{2}} }{\mathrm{9}}\:=\:\mathrm{10} \\ $$$$\:\mathrm{2a}\:×\:\mathrm{3a}\:=\mathrm{6a}^{\mathrm{2}} \:=\:\mathrm{60u}^{\mathrm{2}} \\ $$
Commented by manxsol last updated on 22/Mar/23
$$ \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{liked}\:\mathrm{your}\:\mathrm{solution}.\: \\ $$$$\mathrm{proportio}{n}:\:\left(\mathrm{AB90AB}\right):\:\mathrm{tg}\:. \\ $$