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Question Number 87939 by ajfour last updated on 07/Apr/20

Commented by naka3546 last updated on 07/Apr/20

${w h a t}^{'} s t h e q u e s t i o n, s i r ?$
Commented by ajfour last updated on 07/Apr/20

$I f e a c h c o l o u r e d r e g i o n h a s u n i t$ $a r e a, n a m e t h e c o l o u r e d r e g i o n s$ $t h a t h a v e t h e l a r g e s t p e r i m e t e r,$ $a n d t h e s m a l l e s t .$ $(w o n t e d i t a n y m o r e!)$
Commented by ajfour last updated on 07/Apr/20

$o v e r a l l f i g u r e i s a s q u a r e .$
Commented by Tony Lin last updated on 07/Apr/20

Commented by Tony Lin last updated on 07/Apr/20

$\frac{1}{2} \times \sqrt{3} (\sqrt{3} - \frac{2 \sqrt{3}}{3}) = \frac{1}{2}$ $= h a l f o f t h e g r e e n a r e a$ $t h e h y p o t e n u s e o f b r o w n a r e a i s$ $\sqrt{{(\sqrt{3})}^{2} + {(\frac{2 \sqrt{3}}{3})}^{2}} = \frac{\sqrt{39}}{3}$ $t h e b l u e o n e a n d h a l f o f t h e g r e e n o n e$ $s h a r e t h e s a m e h e i g h t$ $\Rightarrow r a t i o o f t h e b o t t o m s i d e i s 2$ $\Rightarrow b o t t o m s i d e o f t h e b l u e o n e i s \frac{2 \sqrt{39}}{9}$ $t h e g r e e n o n e i s \frac{\sqrt{39}}{9}$ $b l u e a r e a = 1 = \frac{1}{2} \times \sqrt{3} \times \frac{2 \sqrt{39}}{9} s i n θ$ $\Rightarrow s i n θ = \frac{3}{\sqrt{13}} \to c o s θ = \frac{2}{\sqrt{13}}$ $l^{2} = {(\sqrt{3})}^{2} + {(\frac{2 \sqrt{39}}{9})}^{2} - 2 \times \sqrt{3} \times \frac{2 \sqrt{39}}{9} \times \frac{2}{\sqrt{13}}$ $= \frac{61}{27}$ $\Rightarrow l = \frac{\sqrt{183}}{9}$ $b r o w n p e r i m e t e r$ $= \sqrt{3} + \frac{2 \sqrt{3}}{3} + \frac{\sqrt{39}}{3} = \frac{5 \sqrt{3} + \sqrt{39}}{3} \approx 4.9684$ $g r e e n p e r i m e t e r$ $= \sqrt{3} + (\sqrt{3} - \frac{2 \sqrt{3}}{3}) + \frac{\sqrt{39}}{9} + \frac{\sqrt{183}}{9}$ $= \frac{12 \sqrt{3} + \sqrt{39} + \sqrt{183}}{9} \approx 4.5064$ $b l u e p e r i m e t e r$ $= \sqrt{3} + \frac{2 \sqrt{39}}{9} + \frac{\sqrt{183}}{9}$ $= \frac{9 \sqrt{3} + 2 \sqrt{39} + \sqrt{183}}{9} \approx 4.6229$
Commented by ajfour last updated on 07/Apr/20

$T h a n k s S i r, p e r f e c t!$
Commented by ajfour last updated on 07/Apr/20

Commented by ajfour last updated on 07/Apr/20

$s q u a r e s i d e s = \sqrt{3}$ $B E = \frac{2}{\sqrt{3}} = G M = p$ $A E = \sqrt{3 + \frac{4}{3}} = \frac{\sqrt{39}}{3}$ $A M = \frac{p}{s} (A E) = \frac{2}{3} (\frac{\sqrt{39}}{3}) = \frac{2 \sqrt{39}}{9}$ $\frac{q}{p} = \frac{B E}{s} \Rightarrow q = \frac{2}{3} \times \frac{2}{\sqrt{3}} = \frac{4 \sqrt{3}}{9}$ $M D = \sqrt{p^{2} + {(s - q)}^{2}}$ $= \sqrt{\frac{4}{3} + {(\sqrt{3} - \frac{4 \sqrt{3}}{9})}^{2}}$ $= \frac{\sqrt{183}}{9}$ $C E = \sqrt{3} - \frac{2}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ $M E = A E - A M = \frac{\sqrt{39}}{9}$ $p e r i m e t e r (b r o w n) = \sqrt{3} + \frac{2}{\sqrt{3}} + \frac{\sqrt{39}}{3}$ $\approx 4.9684$ $(b l u e) = \sqrt{3} + \frac{2 \sqrt{39}}{9} + \frac{\sqrt{183}}{9}$ $\approx 4.6229$ $(g r e e n) = \sqrt{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{39}}{9} + \frac{\sqrt{183}}{9}$ $\approx 4.5064$ $s o p e r i m e t e r s o f$ $(b r o w n) > (b l u e) > (g r e e n)$
Commented by $@ty@m123 last updated on 07/Apr/20

$a r (A B C D) = 3 s q . u n i t$ $∴ e a c h s i d e = \sqrt{3} u n i t . . . (i)$ $a r (A E B) = 1 s q . u n i t$ $\frac{1}{2} \times A B \times E B = 1$ $\frac{1}{2} \times \sqrt{3} \times E B = 1$ $E B = \frac{2}{\sqrt{3}} . . . . (i i)$ $\frac{1}{2} \times A D \times G M = 1$ $G M = \frac{2}{\sqrt{3}} . . . (i i i)$ $(P l . i g n o r e t h i s c o m m e n t)$
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