A room has dimensions 3 m × 4 m × 5 m. A fly starting at one corner ends up at the diametrically opposite corner. If the fly were to walks, what is the length of the shortest path it can take?


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Question Number 14821 by Tinkutara last updated on 04/Jun/17

$A room has dimensions 3 m \times 4 m \times 5 m .$ $A fly starting at one corner ends up at$ $the diametrically opposite corner . If$ $the fly were to walks, what is the length$ $of the shortest path it can take ?$
	Answered by ajfour last updated on 04/Jun/17

	$d_{m i n} = \sqrt{{(3 + 4)}^{2} + {(5)}^{2}} = \sqrt{74}$ $\approx 8.6 u n i t s$
	Commented by RasheedSoomro last updated on 04/Jun/17

	$Sir the fly is walking not flying .$ $You have determine correct corner,$ $but the path {shouldn}^{'} t be through air .$
	Commented by ajfour last updated on 04/Jun/17

	$t h a t s w h y t h i s d i a g r a m . t r y a n d$ $u n d e r s t a n d . .$
	Commented by RasheedSoomro last updated on 04/Jun/17

	$Sir I understood!$
	Commented by ajfour last updated on 04/Jun/17

	Commented by mrW1 last updated on 05/Jun/17

	$I f t h e d i m e n s i o n s a r e a, b, c w i t h a ⩽ b ⩽ c .$ $T h e r e a r e 3 p o s s i b l e w a y s :$ $\sqrt{{(a + b)}^{2} + c^{2}} = \sqrt{a^{2} + b^{2} + c^{2} + 2 a b} (1)$ $\sqrt{{(a + c)}^{2} + b^{2}} = \sqrt{a^{2} + b^{2} + c^{2} + 2 a c} (2)$ $\sqrt{{(b + c)}^{2} + a^{2}} = \sqrt{a^{2} + b^{2} + c^{2} + 2 b c} (3)$ $t h e s h o r t e s t w a y i s (1), s i n c e$ $2 a b ⩽ 2 a c ⩽ 2 b c .$ $i n c u r r e n t c a s e a = 3, b = 4, c = 5$ $\Rightarrow \sqrt{{(3 + 4)}^{2} + 5^{2}} = 8.6$
	Commented by Tinkutara last updated on 05/Jun/17

	$Thanks Sir!$
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