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Question Number 65366 by Tawa1 last updated on 29/Jul/19

Commented by Prithwish sen last updated on 29/Jul/19

$the length is$ $= 3 + 2 \times 2 + 2 \times \frac{2^{2}}{3} + 2 \times \frac{2^{3}}{3^{2}} + 2 \times \frac{2^{4}}{3^{3}} + . . . . . . .$ $= 3 + 4 [1 + (\frac{2}{3}) + {(\frac{2}{3})}^{2} + {(\frac{2}{3})}^{3} + . . . . . . . .]$ $= 3 + 4 [\frac{1}{1 - \frac{2}{3}}] = 3 + 4 \times 3 = 15$
Commented by Tawa1 last updated on 29/Jul/19

$God bless you sir$
	Answered by JDamian last updated on 29/Jul/19

	$L = \frac{2 + 3}{1 - \frac{2}{3}} = \frac{5}{\frac{1}{3}} = 15 f e e t$
	Commented by KanhAshish last updated on 29/Jul/19

	$how ? ?$
	Commented by JDamian last updated on 29/Jul/19

	$a_{n} \equiv h o r i z o n t a l s i d e$ $b_{n} \equiv v e r t i c a l s i d e = \frac{2}{3} a_{n}$ $a_{n + 1} = b_{n} l_{n} = a_{n} + b_{n} l_{n + 1} = \frac{2}{3} l_{n}$ $You can't use 'macro parameter character #' in math mode$ $You can't use 'macro parameter character #' in math mode$ $You can't use 'macro parameter character #' in math mode$ $this is a geometric progression$
	Commented by KanhAshish last updated on 29/Jul/19

	$amazing explanation . . god bless you .$
	Commented by Tawa1 last updated on 29/Jul/19

	$God bless you sir$
	Answered by mr W last updated on 29/Jul/19

	Commented by mr W last updated on 29/Jul/19

	$a = b + 3$ $\frac{a}{b} = \frac{b + 3}{b} = \frac{3}{2}$ $\Rightarrow 1 + \frac{3}{b} = \frac{3}{2}$ $\Rightarrow b = 6$ $\Rightarrow a = 6 + 3 = 9$ $l e n g t h o f r e d p a t h = a + b = 9 + 6 = 15$
	Commented by mr W last updated on 29/Jul/19

	$f o r a z i g - z a g p a t h t h e p a t h l e n g t h i s$ $t h e s u m o f h o r i z o n t a l a n d v e r t i c a l$ $p r o j e c t i o n (a + b) . t h e s t e p s m a y b e$ $i r r e g u l a r . f o r e x a m p l e t h e r e d p a t h,$ $t h e b l u e p a t h a n d t h e g r e e n p a t h a l l$ $h a v e t h e s a m e l e n g t h .$
	Commented by mr W last updated on 29/Jul/19

	Commented by Tawa1 last updated on 29/Jul/19

	$God bless you sir . i appreciate$
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