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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 ?
$${what}'{s}\:\:{the}\:\:{question}\:,\:{sir}\:? \\ $$
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. (wont edit any more!)
$${If}\:{each}\:{coloured}\:{region}\:{has}\:{unit} \\ $$$${area},\:{name}\:{the}\:{coloured}\:{regions} \\ $$$$\:{that}\:{have}\:{the}\:{largest}\:{perimeter},\: \\ $$$${and}\:{the}\:{smallest}. \\ $$$$\left({wont}\:{edit}\:{any}\:{more}!\right) \\ $$
Commented by ajfour last updated on 07/Apr/20
overall figure is a square.
$${overall}\:{figure}\:{is}\:{a}\:{square}. \\ $$
Commented by Tony Lin last updated on 07/Apr/20
Commented by Tony Lin last updated on 07/Apr/20
(1/2)×(√3)((√3)−((2(√3))/3))=(1/2) =half of the green area the hypotenuse of brown area is (√(((√3))^2 +(((2(√3))/3))^2 ))=((√(39))/3) the blue one and half of the green one share the same height ⇒ratio of the bottom side is 2 ⇒bottom side of the blue one is ((2(√(39)))/9) the green one is ((√(39))/9) blue area=1=(1/2)×(√3)×((2(√(39)))/9)sinθ ⇒sinθ=(3/( (√(13))))→cosθ=(2/( (√(13)))) l^2 =((√3))^2 +(((2(√(39)))/9))^2 −2×(√3)×((2(√(39)))/9)×(2/( (√(13)))) =((61)/(27)) ⇒l=((√(183))/9) brown perimeter =(√3)+((2(√3))/3)+((√(39))/3)=((5(√3)+(√(39)))/3)≈4.9684 green perimeter =(√3)+((√3)−((2(√3))/3))+((√(39))/9)+((√(183))/9) =((12(√3)+(√(39))+(√(183)))/9)≈4.5064 blue perimeter =(√3)+((2(√(39)))/9)+((√(183))/9) =((9(√3)+2(√(39))+(√(183)))/9)≈4.6229
$$\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$={half}\:{of}\:{the}\:{green}\:{area} \\ $$$${the}\:{hypotenuse}\:{of}\:{brown}\:{area}\:{is} \\ $$$$\sqrt{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{39}}}{\mathrm{3}} \\ $$$${the}\:{blue}\:{one}\:{and}\:{half}\:{of}\:{the}\:{green}\:{one} \\ $$$${share}\:{the}\:{same}\:{height} \\ $$$$\Rightarrow{ratio}\:{of}\:{the}\:{bottom}\:{side}\:{is}\:\mathrm{2} \\ $$$$\:\Rightarrow{bottom}\:{side}\:{of}\:{the}\:{blue}\:{one}\:{is}\:\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{the}\:{green}\:{one}\:{is}\:\frac{\sqrt{\mathrm{39}}}{\mathrm{9}} \\ $$$${blue}\:{area}=\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{3}}×\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}{sin}\theta \\ $$$$\Rightarrow{sin}\theta=\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}\rightarrow{cos}\theta=\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}} \\ $$$${l}^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}\right)^{\mathrm{2}} −\mathrm{2}×\sqrt{\mathrm{3}}×\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}} \\ $$$$=\frac{\mathrm{61}}{\mathrm{27}} \\ $$$$\Rightarrow{l}=\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$${brown}\:{perimeter} \\ $$$$=\sqrt{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\sqrt{\mathrm{39}}}{\mathrm{3}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}+\sqrt{\mathrm{39}}}{\mathrm{3}}\approx\mathrm{4}.\mathrm{9684} \\ $$$${green}\:{perimeter} \\ $$$$=\sqrt{\mathrm{3}}+\left(\sqrt{\mathrm{3}}−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)+\frac{\sqrt{\mathrm{39}}}{\mathrm{9}}+\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$$=\frac{\mathrm{12}\sqrt{\mathrm{3}}+\sqrt{\mathrm{39}}+\sqrt{\mathrm{183}}}{\mathrm{9}}\approx\mathrm{4}.\mathrm{5064} \\ $$$${blue}\:{perimeter} \\ $$$$=\sqrt{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}+\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$$=\frac{\mathrm{9}\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{39}}+\sqrt{\mathrm{183}}}{\mathrm{9}}\approx\mathrm{4}.\mathrm{6229} \\ $$
Commented by ajfour last updated on 07/Apr/20
Thanks Sir, perfect!
$${Thanks}\:{Sir},\:{perfect}! \\ $$
Commented by ajfour last updated on 07/Apr/20
Commented by ajfour last updated on 07/Apr/20
square side s=(√3) BE=(2/( (√3))) =GM=p AE=(√(3+(4/3))) = ((√(39))/3) AM=(p/s)(AE) =(2/3)(((√(39))/3))=((2(√(39)))/9) (q/p)=((BE)/s)⇒ q=(2/3)×(2/( (√3))) = ((4(√3))/9) MD=(√(p^2 +(s−q)^2 )) =(√((4/3)+((√3)−((4(√3))/9))^2 )) =((√(183))/9) CE=(√3)−(2/( (√3))) = ((√3)/3) ME=AE−AM=((√(39))/9) perimeter(brown)=(√3)+(2/( (√3)))+((√(39))/3) ≈ 4.9684 (blue)=(√3)+((2(√(39)))/9)+((√(183))/9) ≈ 4.6229 (green)=(√3)+((√3)/3)+((√(39))/9)+((√(183))/9) ≈ 4.5064 so perimeters of (brown) > (blue) > (green)
$${square}\:{side}\:{s}=\sqrt{\mathrm{3}} \\ $$$${BE}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:={GM}={p} \\ $$$${AE}=\sqrt{\mathrm{3}+\frac{\mathrm{4}}{\mathrm{3}}}\:=\:\frac{\sqrt{\mathrm{39}}}{\mathrm{3}} \\ $$$${AM}=\frac{{p}}{{s}}\left({AE}\right)\:=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\sqrt{\mathrm{39}}}{\mathrm{3}}\right)=\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}} \\ $$$$\frac{{q}}{{p}}=\frac{{BE}}{{s}}\Rightarrow\:\:{q}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$$${MD}=\sqrt{{p}^{\mathrm{2}} +\left({s}−{q}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{4}}{\mathrm{3}}+\left(\sqrt{\mathrm{3}}−\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$${CE}=\sqrt{\mathrm{3}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${ME}={AE}−{AM}=\frac{\sqrt{\mathrm{39}}}{\mathrm{9}} \\ $$$${perimeter}\left({brown}\right)=\sqrt{\mathrm{3}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}+\frac{\sqrt{\mathrm{39}}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\approx\:\mathrm{4}.\mathrm{9684} \\ $$$$\left({blue}\right)=\sqrt{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{9}}+\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\approx\:\mathrm{4}.\mathrm{6229} \\ $$$$\left({green}\right)=\sqrt{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\sqrt{\mathrm{39}}}{\mathrm{9}}+\frac{\sqrt{\mathrm{183}}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\approx\:\mathrm{4}.\mathrm{5064} \\ $$$${so}\:{perimeters}\:{of} \\ $$$$\:\:\left({brown}\right)\:>\:\left({blue}\right)\:>\:\left({green}\right)\:\:\: \\ $$
Commented by $@ty@m123 last updated on 07/Apr/20
ar(ABCD)=3 sq. unit ∴ each side=(√3) unit ...(i) ar(AEB)=1 sq. unit (1/2)×AB×EB=1 (1/2)×(√3)×EB=1 EB=(2/( (√3))) ....(ii) (1/2)×AD×GM=1 GM=(2/( (√3))) ...(iii) (Pl. ignore this comment)
$${ar}\left({ABCD}\right)=\mathrm{3}\:{sq}.\:{unit} \\ $$$$\therefore\:{each}\:{side}=\sqrt{\mathrm{3}}\:{unit}\:\:\:…\left({i}\right) \\ $$$${ar}\left({AEB}\right)=\mathrm{1}\:{sq}.\:{unit} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×{AB}×{EB}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{3}}×{EB}=\mathrm{1} \\ $$$${EB}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\:….\left({ii}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×{AD}×{GM}=\mathrm{1} \\ $$$${GM}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\:…\left({iii}\right) \\ $$$$\left({Pl}.\:{ignore}\:{this}\:{comment}\right) \\ $$

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