Question Number 87939 by ajfour last updated on 07/Apr/20
Commented by naka3546 last updated on 07/Apr/20
$${what}'{s}\:\:{the}\:\:{question}\:,\:{sir}\:? \\ $$
Commented by ajfour last updated on 07/Apr/20
$${If}\:{each}\:{coloured}\:{region}\:{has}\:{unit} \\ $$$${area},\:{name}\:{the}\:{coloured}\:{regions} \\ $$$$\:{that}\:{have}\:{the}\:{largest}\:{perimeter},\: \\ $$$${and}\:{the}\:{smallest}. \\ $$$$\left({wont}\:{edit}\:{any}\:{more}!\right) \\ $$
Commented by ajfour last updated on 07/Apr/20
$${overall}\:{figure}\:{is}\:{a}\:{square}. \\ $$
Commented by Tony Lin last updated on 07/Apr/20
Commented by Tony Lin last updated on 07/Apr/20
$$\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$={half}\:{of}\:{the}\:{green}\:{area} \\ $$$${the}\:{hypotenuse}\:{of}\:{brown}\:{area}\:{is} \\ $$$$\sqrt{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{39}}}{\mathrm{3}} \\ $$$${the}\:{blue}\:{one}\:{and}\:{half}\:{of}\:{the}\:{green}\:{one} \\ $$$${share}\:{the}\:{same}\:{height} \\ $$$$\Rightarrow{ratio}\:{of}\:{the}\:{bottom}\:{side}\:{is}\:\mathrm{2} \\ $$$$\:\Rightarrow{bottom}\:{side}\:{of}\:{the}\:{blue}\:{one}\:{is}\:\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{the}\:{green}\:{one}\:{is}\:\frac{\sqrt{\mathrm{39}}}{\mathrm{9}} \\ $$$${blue}\:{area}=\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{3}}×\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}{sin}\theta \\ $$$$\Rightarrow{sin}\theta=\frac{\mathrm{3}}{\sqrt{\mathrm{13}}}\rightarrow{cos}\theta=\frac{\mathrm{2}}{\sqrt{\mathrm{13}}} \\ $$$${l}^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}\right)^{\mathrm{2}} −\mathrm{2}×\sqrt{\mathrm{3}}×\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}×\frac{\mathrm{2}}{\sqrt{\mathrm{13}}} \\ $$$$=\frac{\mathrm{61}}{\mathrm{27}} \\ $$$$\Rightarrow{l}=\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$${brown}\:{perimeter} \\ $$$$=\sqrt{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\sqrt{\mathrm{39}}}{\mathrm{3}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}+\sqrt{\mathrm{39}}}{\mathrm{3}}\approx\mathrm{4}.\mathrm{9684} \\ $$$${green}\:{perimeter} \\ $$$$=\sqrt{\mathrm{3}}+\left(\sqrt{\mathrm{3}}−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)+\frac{\sqrt{\mathrm{39}}}{\mathrm{9}}+\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$$=\frac{\mathrm{12}\sqrt{\mathrm{3}}+\sqrt{\mathrm{39}}+\sqrt{\mathrm{183}}}{\mathrm{9}}\approx\mathrm{4}.\mathrm{5064} \\ $$$${blue}\:{perimeter} \\ $$$$=\sqrt{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}+\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$$=\frac{\mathrm{9}\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{39}}+\sqrt{\mathrm{183}}}{\mathrm{9}}\approx\mathrm{4}.\mathrm{6229} \\ $$
Commented by ajfour last updated on 07/Apr/20
$${Thanks}\:{Sir},\:{perfect}! \\ $$
Commented by ajfour last updated on 07/Apr/20
Commented by ajfour last updated on 07/Apr/20
$${square}\:{side}\:{s}=\sqrt{\mathrm{3}} \\ $$$${BE}=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:={GM}={p} \\ $$$${AE}=\sqrt{\mathrm{3}+\frac{\mathrm{4}}{\mathrm{3}}}\:=\:\frac{\sqrt{\mathrm{39}}}{\mathrm{3}} \\ $$$${AM}=\frac{{p}}{{s}}\left({AE}\right)\:=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\sqrt{\mathrm{39}}}{\mathrm{3}}\right)=\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}} \\ $$$$\frac{{q}}{{p}}=\frac{{BE}}{{s}}\Rightarrow\:\:{q}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$$${MD}=\sqrt{{p}^{\mathrm{2}} +\left({s}−{q}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{4}}{\mathrm{3}}+\left(\sqrt{\mathrm{3}}−\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$${CE}=\sqrt{\mathrm{3}}−\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${ME}={AE}−{AM}=\frac{\sqrt{\mathrm{39}}}{\mathrm{9}} \\ $$$${perimeter}\left({brown}\right)=\sqrt{\mathrm{3}}+\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}+\frac{\sqrt{\mathrm{39}}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\approx\:\mathrm{4}.\mathrm{9684} \\ $$$$\left({blue}\right)=\sqrt{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}+\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\approx\:\mathrm{4}.\mathrm{6229} \\ $$$$\left({green}\right)=\sqrt{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\sqrt{\mathrm{39}}}{\mathrm{9}}+\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\approx\:\mathrm{4}.\mathrm{5064} \\ $$$${so}\:{perimeters}\:{of} \\ $$$$\:\:\left({brown}\right)\:>\:\left({blue}\right)\:>\:\left({green}\right)\:\:\: \\ $$
Commented by $@ty@m123 last updated on 07/Apr/20
$${ar}\left({ABCD}\right)=\mathrm{3}\:{sq}.\:{unit} \\ $$$$\therefore\:{each}\:{side}=\sqrt{\mathrm{3}}\:{unit}\:\:\:...\left({i}\right) \\ $$$${ar}\left({AEB}\right)=\mathrm{1}\:{sq}.\:{unit} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×{AB}×{EB}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{3}}×{EB}=\mathrm{1} \\ $$$${EB}=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\:....\left({ii}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×{AD}×{GM}=\mathrm{1} \\ $$$${GM}=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\:...\left({iii}\right) \\ $$$$\left({Pl}.\:{ignore}\:{this}\:{comment}\right) \\ $$