Ответ:
[tex]AB=\sqrt{(3-2)^2+(-1-2)^2}=\sqrt{1^2+(-3)^2} =\sqrt{1+9} =\sqrt{10}[/tex]
[tex]BC=\sqrt{(-3-3)^2+(-3+1)^2}=\sqrt{(-6)^2+(-2)^2}=\sqrt{36+4} =\sqrt{40} =2\sqrt{10}[/tex]
[tex]CD=\sqrt{(-4+3)^2+(0+3)^2}=\sqrt{(-1)^2+3^2} =\sqrt{1+9} =\sqrt{10}[/tex]
[tex]AD=\sqrt{(-4-2)^2+(0-2)^2}=\sqrt{(-6)^2+(-2)^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}[/tex]
AB=CD, BC=AD отже АВСD прямокутник
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Ответ:
[tex]AB=\sqrt{(3-2)^2+(-1-2)^2}=\sqrt{1^2+(-3)^2} =\sqrt{1+9} =\sqrt{10}[/tex]
[tex]BC=\sqrt{(-3-3)^2+(-3+1)^2}=\sqrt{(-6)^2+(-2)^2}=\sqrt{36+4} =\sqrt{40} =2\sqrt{10}[/tex]
[tex]CD=\sqrt{(-4+3)^2+(0+3)^2}=\sqrt{(-1)^2+3^2} =\sqrt{1+9} =\sqrt{10}[/tex]
[tex]AD=\sqrt{(-4-2)^2+(0-2)^2}=\sqrt{(-6)^2+(-2)^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}[/tex]
AB=CD, BC=AD отже АВСD прямокутник