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-

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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