A park has the shape of a regular hexagon of sides 2km each. A boy walks a distance of 5km along the sides of the park. What is the direct distance between the start point and the end point?


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Question Number 46139 by pieroo last updated on 21/Oct/18

$A park has the shape of a regular hexagon$ $of sides 2 km each . A boy walks a distance$ $of 5 km along the sides of the park .$ $What is the direct distance between$ $the start point and the end point ?$
Commented by malwaan last updated on 21/Oct/18

$distance = \sqrt{13} km \approx 3.61 km$ $angle \approx 46.1 °$
	Answered by ajfour last updated on 21/Oct/18

	$l e t s i d e o f h e x a g o n, 2 k m = a$ $d = \sqrt{{(2 a - \frac{a}{2} \cos 60 °)}^{2} + {(\frac{a}{2} \sin 60 °)}^{2}}$ $= (\sqrt{{(2 - \frac{1}{4})}^{2} + {(\frac{\sqrt{3}}{4})}^{2}}) a$ $d = (\frac{\sqrt{52}}{4}) a = \sqrt{13} k m .$
	Answered by MrW3 last updated on 22/Oct/18

	$c a s e 1$ ${l e t}^{'} s s a y t h e b o y s t a r t s h i s w a l k a k m$ $b e f o r e a v e r t e x o f h e x a g o n . t h e n h e e n d s$ $t h e w a l k 1 - a k m a f t e r a n o t h e r v e r t e x .$ $w i t h 0 ⩽ a ⩽ 1 w e h a v e$ $d = \sqrt{12 + {(1 - 2 a)}^{2}}$ $d_{m a x} = \sqrt{13} \approx 3.606 k m b y a = 0 o r 1 k m$ $d_{m i n} = \sqrt{12} \approx 3.464 k m b y a = 0.5 k m$ $c a s e 2$ ${l e t}^{'} s s a y t h e b o y s t a r t s h i s w a l k a k m$ $a f t e r a v e r t e x o f h e x a g o n . t h e n h e e n d s$ $t h e w a l k 1 + a k m a f t e r a n o t h e r v e r t e x .$ $w i t h 0 ⩽ a ⩽ 1 w e h a v e$ $d^{2} = {(4 - a)}^{2} + {(3 + a)}^{2} - 2 (4 - a) (3 + a) \cos 60 °$ $d^{2} = 3 (a^{2} - a) + 13$ $\Rightarrow d = \sqrt{3 a (a - 1) + 13}$ $d_{m i n} = 3.5 k m b y a = 0.5 k m$ $d_{m a x} = \sqrt{13} k m b y a = 0 o r 1 k m$ $s u m m a r i s e d w e h a v e$ $d_{m a x} = \sqrt{13} k m$ $d_{m i n} = \sqrt{12} k m$
	Commented by MrW3 last updated on 21/Oct/18

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