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Question Number 51658 by ajfour last updated on 29/Dec/18

Commented by ajfour last updated on 29/Dec/18

$F i n d i n t e r m s o f a a n d b t h e s i d e$ $l e n g t h s o f t h e e q u a l s i d e d p e n t a g o n .$
Commented by ajfour last updated on 29/Dec/18

$e s p e c i a l l y M j S S i r . .$
Commented by afachri last updated on 29/Dec/18

$all pentagon sides have the same S lenght$ $sir ? ? ?$
Commented by ajfour last updated on 29/Dec/18

$y e s .$
Commented by afachri last updated on 29/Dec/18

$i have an answer Sir . Wait, i^{'} ll post it$ $and please correct me if i am mistaken .$
Commented by Kunal12588 last updated on 29/Dec/18

$s i r i s i t a e q u i l a t e r a l p e n t a g o n w i t h n o n e q u a l$ $a n g l e s ? o r a r e g u l a r p e n t a g o n (i t w i l l b e e a s i e r) ?$
Commented by ajfour last updated on 29/Dec/18

$e q u a l s i d e d o n l y, g i v e n i n q u e s t i o n .$
	Answered by afachri last updated on 29/Dec/18

	Commented by mr W last updated on 29/Dec/18

	${i t}^{'} s n o t c o r r e c t s i r .$ $y o u a s s u m e a = 2 x + s a n d b = x + y . b u t$ $t h i s i s n o t t r u e . t h e t r i a n g l e o n p o i n t$ $C i s n o t t h e s a m e a s t h e t r i a n g l e o n$ $p o i n t B . o t h e r w i s e D C = a = 2 y!$
	Commented by afachri last updated on 29/Dec/18

	$thanks for your correction Mr W .$ $so that i thought D C = a = 2 y .$ $i^{'} ll fix it Mr W . really appreciate your$ $correction Sir .$
	Answered by afachri last updated on 29/Dec/18

	$side a 2 x + s = a$ $x = \frac{a - s}{2}$ $side b x + y = b$ $y = b - x$ $y = b - \frac{a - s}{2}$ $based on phytagoras theorm : s^{2} = x^{2} + y^{2}$ $s^{2} = {(\frac{a - s}{2})}^{2} + {(b - \frac{a - s}{2})}^{2}$ $s^{2} = \frac{{(a - s)}^{2}}{4} + b^{2} - b (a - s) + \frac{{(a - s)}^{2}}{4}$ $s^{2} = b^{2} - a b + b s + \frac{{(a - s)}^{2}}{2} then multiple by 2$ $2 s^{2} = 2 b^{2} - 2 a b + 2 b s + (a^{2} - 2 a s + s^{2})$ $2 s^{2} = s^{2} + 2 (b - a) s + (a^{2} + 2 b^{2} - 2 a b)$ $s^{2} - 2 (b - a) s = a^{2} + 2 b^{2} - 2 a b$ $s^{2} - 2 (b - a) s + {(b - a)}^{2} = a^{2} + 2 b^{2} - 2 a b + {(b - a)}^{2}$ ${[s - (b - a)]}^{2} = 2 a^{2} - 4 a b + 3 b^{2}$ ${[s - (b - a)]}^{2} = 2 {(a - b)}^{2} + b^{2}$ $s - (b - a) = \sqrt{2 {(a - b)}^{2} + b^{2}}$ $s = (b - a) + \sqrt{2 {(a - b)}^{2} + b^{2}}$
	Answered by mr W last updated on 29/Dec/18

	$\sqrt{s^{2} - \frac{a^{2}}{4}} + \sqrt{s^{2} - {(\frac{a - s}{2})}^{2}} = b$ $w i t h μ = \frac{b}{a} a n d λ = \frac{s}{a}$ $\Rightarrow \sqrt{λ^{2} - \frac{1}{4}} + \sqrt{λ^{2} - {(\frac{1 - λ}{2})}^{2}} = μ$ $\Rightarrow λ = f (μ) (o n l y n u m e r i c a l l y, I t h i n k)$ $e x a m p l e s :$ $b = 0.5 a \Rightarrow s \approx 0.5038 a$ $b = a \Rightarrow s \approx 0.6351 a$ $b = 2 a \Rightarrow s \approx 1.0627 a$
	Commented by mr W last updated on 29/Dec/18

	Commented by ajfour last updated on 29/Dec/18

	$V e r y w e l l S i r, w h a t s t h i s a p p y o u$ $g r a p h o n, t h a n k y o u .$
	Commented by mr W last updated on 29/Dec/18

	Commented by mr W last updated on 29/Dec/18

	${i t}^{'} s a f r e e a p p w h i c h y o u c a n d o w n l o a d$ $f r o m g o o g l e p l a y . a v e r y v e r y g o o d a p p!$
	Commented by afachri last updated on 30/Dec/18

	$what about an app used for draw in geometry ? ?$ $any suggestion, Sir ?$
	Commented by ajfour last updated on 30/Dec/18

	$L e k h D i a g r a m$
	Commented by afachri last updated on 31/Dec/18

	$thNks Mr Ajfour$
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