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Question Number 53570 by ajfour last updated on 23/Jan/19

Commented by ajfour last updated on 23/Jan/19

$R e g u l a r p e n t a g o n s i d e a . F i n d t h e$ $c e n t r a l u n c o l o u r e d a r e a .$
Commented by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19

$n o c o m m e n t s r e c e i v e d . . h e n c e d e l e t e d$
Commented by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19

Commented by ajfour last updated on 23/Jan/19

$a s s o o n a s i s a w i h a d c o m m e n t e d,$ $t h a n k s s i r, i r e m e m b e r y o u r a n s w e r$ $e v e n .$ $A_{w h i t e} = \frac{a^{2}}{4} \tan^{2} 27 ° (5 \cot 36 ° - \frac{3 π}{2})$ $i t m a t c h e d w i t h w h a t i c o u l d o b t a i n .$
Commented by ajfour last updated on 23/Jan/19

$T a n m a y S i r, w h y y o u d e l e t e d y o u r$ $p o s t e d s o l u t i o n ?$
Commented by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19

$t a n α = \frac{r}{\frac{a}{2}} r = \frac{a}{2} t a n α . . . . (1)$ $5 θ = 2 π θ = \frac{2 π}{5} . . . . . (2)$ $2 β + θ = π β = \frac{π - \frac{2 π}{5}}{2} = \frac{3 π}{10} . . . . (3)$ $n o w a r e a o f s m a l l p e n t a g o n . . .$ $5 \times \frac{1}{2} \times 2 r \times h$ $t a n \frac{θ}{2} = \frac{r}{h} s o h = r c o t \frac{θ}{2}$ $s o a r e a o f s m a l l p e n t a g o n = 5 \times \frac{1}{2} \times 2 r \times r c o t (\frac{θ}{2})$ $a r e a o f f i v e s e c t o r o f c i r c l e = 5 \times \frac{π r^{2}}{2 π} \times 2 β$ $= 5 r^{2} β$ $s o a r e a o f w h i t e c e n t r e s t a r$ $= a r e a o f s m a l l p e n t a g o n - f i v e s e c t o r a r e a$ $= 5 r^{2} c o t (\frac{θ}{2}) - 5 r^{2} β$ $= 5 r^{2} [c o t (\frac{θ}{2}) - β]$ $= 5 \times {(\frac{a}{2} t a n α)}^{2} [c o t (\frac{θ}{2}) - β]$ $n o w θ = \frac{2 π}{5} β = \frac{3 π}{10}$ $t o f i n d α . . .$ $u s i n g f o r m u l a . . .$ $n u m b e r o f s i d e = \frac{360}{e x t e r n a l a n g k e}$ $e x t e r n a l a n g k e = \frac{360}{5} = 72^{o}$ $i n t e r n a l + e x t e r n a l a n g l d = 180^{o}$ $i n y e r n s l a n g l d = 180^{o} - 72 = 108^{o}$ $4 α = 108 = s o α = 27^{o}$ $h d n c e r e q u i r e d a n s w e r i s$ $= ★ ★ 5 \times {(\frac{a}{2} t a n α)}^{2} [c o t (\frac{θ}{2}) - β] ★ ★$ $= 5 \times {(\frac{a}{2} t a n 27^{o})}^{2} [c o t (\frac{2 π}{5 \times 2}) - \frac{3 π}{10}]$ $= \frac{a^{2}}{4} {t a n}^{2} 27^{o} \times [5 c o t 36^{o} - \frac{3 π}{2}]$ $n o w p l s c h e c k . . .$
Commented by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19

$t h a n k y o u . .$
	Answered by mr W last updated on 23/Jan/19

	Commented by mr W last updated on 23/Jan/19

	$α = \frac{2 π}{10} = \frac{π}{5}$ $(r + \frac{r}{\sin α}) \tan α = \frac{a}{2}$ $r = \frac{a \cos α}{2 (1 + \sin α)}$ $A = 10 [\frac{r^{2}}{2 \tan α} - \frac{r^{2}}{2} (\frac{π}{2} - α)]$ $A = 5 r^{2} [α + \frac{1}{\tan α} - \frac{π}{2}]$ $A = \frac{5 a^{2} \cos^{2} α}{4 {(1 + \sin α)}^{2}} [\frac{π}{5} + \frac{1}{\tan α} - \frac{π}{2}]$ $\cos α = \frac{\sqrt{5} + 1}{4}$ $\sin α = \frac{\sqrt{2 (5 - \sqrt{5})}}{4}$ $\tan α = \sqrt{5 - 2 \sqrt{5}}$ $A = \frac{a^{2} \cos^{2} α}{8 {(1 + \sin α)}^{2}} [\frac{10}{\tan α} - 3 π]$ $A = \frac{a^{2} {(\sqrt{5} + 1)}^{2}}{8 {(4 + \sqrt{2 (5 - \sqrt{5}})}^{2}} [\frac{10 \sqrt{5 + 2 \sqrt{5}}}{\sqrt{5}} - 3 π]$ $\Rightarrow A = \frac{a^{2} (3 + \sqrt{5}) [2 \sqrt{5 (5 + 2 \sqrt{5}}) - 3 π]}{8 [13 - \sqrt{5} + 4 \sqrt{2 (5 - \sqrt{5})}]} \approx 0.14081 a^{2}$
	Commented by ajfour last updated on 23/Jan/19

	$T h a n k s, b e a u t i f u l w a y s i r .$
	Commented by mr W last updated on 23/Jan/19

	$t h a n k s s i r!$
	Commented by Otchere Abdullai last updated on 24/Jan/19

	$m y p r o f$
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