The perimeter of the quadrilateral is approximately 50.47 units
Given the vertices of the quadrilateral: H(– 4,6), I(4,9), J(2,3), K(7, – 4).
Let us calculate the distances of HI , IJ , JK and KH
[tex]{\displaystyle HI={\sqrt {(4-(-4))^{2}+(9-(6))^{2}}}.}[/tex]
or, HI = 17 units
[tex]{\displaystyle IJ={\sqrt {(2-(4))^{2}+(3-(9))^{2}}}.}[/tex]
or, IJ = 10
[tex]{\displaystyle JK={\sqrt {(7-(2))^{2}+(-4-(3))^{2}}}.}[/tex]
or, JK = √74 = 8.602... ≈ 8.60
[tex]{\displaystyle KH={\sqrt {(7-(-4))^{2}+(-4-(6))^{2}}}.}[/tex]
or, KH = √221 = 14.866... ≈ 14.87
Now perimeter = HI + IJ + JK + KH
or, perimeter = 8.60 + 14.87 + 10 + 17 = 50.47 units.
Therefore the perimeter of the quadrilateral is 50.47 units.
To learn more about perimeter visit:
https://brainly.com/question/20521578
#SPJ9