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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 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?
$$\mathrm{A}\:\mathrm{park}\:\mathrm{has}\:\mathrm{the}\:\mathrm{shape}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{hexagon} \\ $$$$\mathrm{of}\:\mathrm{sides}\:\mathrm{2km}\:\mathrm{each}.\:\mathrm{A}\:\mathrm{boy}\:\mathrm{walks}\:\mathrm{a}\:\mathrm{distance} \\ $$$$\mathrm{of}\:\mathrm{5km}\:\mathrm{along}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{park}.\: \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{direct}\:\mathrm{distance}\:\mathrm{between}\: \\ $$$$\mathrm{the}\:\mathrm{start}\:\mathrm{point}\:\mathrm{and}\:\mathrm{the}\:\mathrm{end}\:\mathrm{point}? \\ $$
Commented by malwaan last updated on 21/Oct/18
distance=(√(13 )) km ≈3.61 km angle≈46.1°
$$\mathrm{distance}=\sqrt{\mathrm{13}\:}\:\mathrm{km}\:\approx\mathrm{3}.\mathrm{61}\:\mathrm{km} \\ $$$$\mathrm{angle}\approx\mathrm{46}.\mathrm{1}° \\ $$
Answered by ajfour last updated on 21/Oct/18
let side of hexagon , 2km=a d = (√((2a−(a/2)cos 60°)^2 +((a/2)sin 60°)^2 )) = ( (√((2−(1/4))^2 +(((√3)/4))^2 )) )a d = (((√(52))/4))a = (√(13)) km .
$${let}\:{side}\:{of}\:{hexagon}\:,\:\mathrm{2}{km}={a} \\ $$$$\:{d}\:=\:\sqrt{\left(\mathrm{2}{a}−\frac{{a}}{\mathrm{2}}\mathrm{cos}\:\mathrm{60}°\right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{2}}\mathrm{sin}\:\mathrm{60}°\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:=\:\left(\:\sqrt{\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)^{\mathrm{2}} }\:\right){a} \\ $$$$\:\:\:\:{d}\:=\:\left(\frac{\sqrt{\mathrm{52}}}{\mathrm{4}}\right){a}\:=\:\sqrt{\mathrm{13}}\:{km}\:. \\ $$
Answered by MrW3 last updated on 22/Oct/18
case 1 let′s say the boy starts his walk a km before a vertex of hexagon. then he ends the walk 1−a km after an other vertex. with 0≤a≤1 we have d=(√(12+(1−2a)^2 )) d_(max) =(√(13)) ≈3.606 km by a=0 or 1 km d_(min) =(√(12)) ≈3.464 km by a=0.5 km case 2 let′s say the boy starts his walk a km after a vertex of hexagon. then he ends the walk 1+a km after an other vertex. with 0≤a≤1 we have d^2 =(4−a)^2 +(3+a)^2 −2(4−a)(3+a)cos 60° d^2 =3(a^2 −a)+13 ⇒d=(√(3a(a−1)+13)) d_(min) =3.5 km by a=0.5 km d_(max) =(√(13)) km by a=0 or 1 km summarised we have d_(max) =(√(13)) km d_(min) =(√(12)) km
$${case}\:\mathrm{1} \\ $$$${let}'{s}\:{say}\:{the}\:{boy}\:{starts}\:{his}\:{walk}\:{a}\:{km} \\ $$$${before}\:{a}\:{vertex}\:{of}\:{hexagon}.\:{then}\:{he}\:{ends} \\ $$$${the}\:{walk}\:\mathrm{1}−{a}\:{km}\:{after}\:{an}\:{other}\:{vertex}. \\ $$$${with}\:\mathrm{0}\leqslant{a}\leqslant\mathrm{1}\:{we}\:{have} \\ $$$${d}=\sqrt{\mathrm{12}+\left(\mathrm{1}−\mathrm{2}{a}\right)^{\mathrm{2}} } \\ $$$${d}_{{max}} =\sqrt{\mathrm{13}}\:\approx\mathrm{3}.\mathrm{606}\:{km}\:{by}\:{a}=\mathrm{0}\:{or}\:\mathrm{1}\:{km} \\ $$$${d}_{{min}} =\sqrt{\mathrm{12}}\:\approx\mathrm{3}.\mathrm{464}\:{km}\:{by}\:{a}=\mathrm{0}.\mathrm{5}\:{km} \\ $$$$ \\ $$$${case}\:\mathrm{2} \\ $$$${let}'{s}\:{say}\:{the}\:{boy}\:{starts}\:{his}\:{walk}\:{a}\:{km} \\ $$$${after}\:{a}\:{vertex}\:{of}\:{hexagon}.\:{then}\:{he}\:{ends} \\ $$$${the}\:{walk}\:\mathrm{1}+{a}\:{km}\:{after}\:{an}\:{other}\:{vertex}. \\ $$$${with}\:\mathrm{0}\leqslant{a}\leqslant\mathrm{1}\:{we}\:{have} \\ $$$${d}^{\mathrm{2}} =\left(\mathrm{4}−{a}\right)^{\mathrm{2}} +\left(\mathrm{3}+{a}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{4}−{a}\right)\left(\mathrm{3}+{a}\right)\mathrm{cos}\:\mathrm{60}° \\ $$$${d}^{\mathrm{2}} =\mathrm{3}\left({a}^{\mathrm{2}} −{a}\right)+\mathrm{13} \\ $$$$\Rightarrow{d}=\sqrt{\mathrm{3}{a}\left({a}−\mathrm{1}\right)+\mathrm{13}} \\ $$$${d}_{{min}} =\mathrm{3}.\mathrm{5}\:{km}\:{by}\:{a}=\mathrm{0}.\mathrm{5}\:{km} \\ $$$${d}_{{max}} =\sqrt{\mathrm{13}}\:{km}\:{by}\:{a}=\mathrm{0}\:{or}\:\mathrm{1}\:{km} \\ $$$$ \\ $$$${summarised}\:{we}\:{have} \\ $$$${d}_{{max}} =\sqrt{\mathrm{13}}\:{km} \\ $$$${d}_{{min}} =\sqrt{\mathrm{12}}\:\:{km} \\ $$
Commented by MrW3 last updated on 21/Oct/18

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