Question

Uestion

Example
A city planner designs a park that is a quadrilateral with vertices
at J(-3, 1), K (1, 3),
L(5, -1), and M(-1, -3). There is
an entrance to the park at the
midpoint of each side of the
park. A straight path connects each entrance to the entrance on the opposite side.
Assuming each unit of the coordinate plane represents 10 meters,
what is the total length
of the paths to the nearest meter? Round your answer to the
nearest whole number.
The total length of the paths is approximately
meters.

Answer 1

Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JK(\stackrel{x_1}{-1}~,~\stackrel{y_1}{2})\qquad LM(\stackrel{x_2}{2}~,~\stackrel{y_2}{-2})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ JKLM=√((~~2 - (-1)~~)^2 + (~~-2 - 2~~)^2) \\\\\\ JKLM=√((2 +1)^2 + (-2 - 2)^2) \implies JKLM=√(( 3 )^2 + ( -4 )^2) \\\\\\ JKLM=√( 9 + 16 ) \implies JKLM=√( 25 )\implies \boxed{JKLM=5}`

now, let's check the other path, JM and KL

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JM(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad KL(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ JMKL=√((~~3 - (-2)~~)^2 + (~~1 - (-1)~~)^2) \\\\\\ JMKL=√((3 +2)^2 + (1 +1)^2) \implies JMKL=√(( 5 )^2 + ( 2 )^2) \\\\\\ JMKL=√( 25 + 4 ) \implies \boxed{JMKL=√( 29 )}`

so the red path will be
`5~~ + ~~√(29) ~~ \approx ~~ \blacksquare~~ 10 ~~\blacksquare`