Find the length of diagonal HJ. Round to the nearest hundredth.

Answer:The length of the diagonal HJ is 10.82 unitsStep-by-step explanation:* Lets revise the rule of the distance between two points- [tex]d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex], where  [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] are the two points* Lets use this rule to find the length of the diagonal HJ∵ The coordinates of point H are (-4 , 3)∵ The coordinates of point J are (5 , -3)∴ [tex]x_{1}=-4[/tex] and [tex]x_{2}=5[/tex]∴ [tex]y_{1}=3[/tex] and [tex]y_{2}=-3[/tex]- Lets find the length of the diagonal HJ by using the rule above∴ HJ = [tex]\sqrt{(5-(-4))^{2}+(-3-3)^{2}}=\sqrt{(5+4)^{2}+(-6)^{2}}[/tex]∴ HJ = [tex]\sqrt{(9)^{2}+36}=\sqrt{81+36}=\sqrt{117}=10.81665[/tex]∴ HJ = 10.82* The length of the diagonal HJ is 10.82 units