Respuesta :
The Quadrilateral is JKLM,
let [tex]M_{JK}, M_{KL}, M_{LM}, M_{JM}, [/tex] be the midpoints of JK, KL, LM and JM respectively.
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let [tex]M_{JK}, M_{KL}, M_{LM}, M_{JM}, [/tex] be the midpoints of JK, KL, LM and JM respectively.
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Given any 2 point P(m,n) and Q(k,l),
the coordinates of the midpoint of the line
segment PQ are given by the formula:
[tex]M_{PQ}=( \frac{m+k}{2} ,
\frac{n+l}{2})[/tex],
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thus the coordinates of points [tex]M_{JK}, M_{KL}, M_{LM}, M_{JM}, [/tex]
are as follows:
[tex]M_{JK}= (\frac{-3+1}{2}, \frac{1+3}{2})=(-1,2), \\\\M_{KL}= (\frac{1+5}{2}, \frac{3-1}{2}=(3,1), \\\\M_{LM}= (\frac{5-1}{2}, \frac{-1-3}{2})=(2, -2),\\\\ M_{JM}= (\frac{-3-1}{2}, \frac{1-3}{2})=(-2,-1)[/tex]
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The distance between any 2 points P(a,b) and
Q(c,d) in the coordinate plane, is given by the formula:
[tex]|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}
}[/tex]
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thus the distances connecting the opposite entrances can be calculated as follows:
[tex]|M_{JK},M_{LM}|= \sqrt{ (-1-2)^{2} + (2-(-2))^{2} }= \sqrt{9+16}=5 [/tex]
[tex]|M_{KL}M_{JM}|= \sqrt{ (3-(-2))^{2} + (1-(-1))^{2}}= \sqrt{25+4}= \sqrt{29}=5.39 [/tex]
Thus the total distance of the paths joining the opposite entrances is
5+5.39 units = 50 m + 53.9 m = 104 m (rounded to the nearest meter)
Answer: 104 m
