Explicație pas cu pas:
[tex]A(- \sqrt{6} , 2), B( \sqrt{6} ,3). C(3 \sqrt{6} , 2) , D( \sqrt{6},1)[/tex]
[tex]AB = \sqrt{ {( - \sqrt{6} - \sqrt{6} )}^{2} + {(2 - 3)}^{2} } = \sqrt{24 + 1} = \sqrt{25} = 5[/tex]
[tex]BC = \sqrt{ {( \sqrt{6} - 3 \sqrt{6} )}^{2} + {(3 - 2)}^{2} } = \sqrt{24 + 1} = \sqrt{25} = 5[/tex]
[tex]CD = \sqrt{ {(3 \sqrt{6} - \sqrt{6} )}^{2} + {(2 - 1)}^{2} } = \sqrt{24 + 1} = \sqrt{25} = 5[/tex]
[tex]AD = \sqrt{ {( - \sqrt{6} - \sqrt{6})}^{2} + {(2 - 1)}^{2}} = \sqrt{24 + 1} = \sqrt{25} = 5[/tex]
=> AB = BC = CD = AD = 5cm
=> ABCD este romb
perimetrul = 4×5 = 20
diagonalele:
[tex]AC = \sqrt{ {( - \sqrt{6} - 3 \sqrt{6} )}^{2} + {(2 - 2)}^{2} } = \sqrt{ {4}^{2} \times 6 + 0 } = 4 \sqrt{6} [/tex]
[tex]BD = \sqrt{ {( \sqrt{6} - \sqrt{6} )}^{2} + {(3 - 1)}^{2} } = \sqrt{4} = 2[/tex]
[tex]Aria= \frac{AC \times BD}{2} = \frac{4 \sqrt{6} \times 2}{2} = 4\sqrt{6}[/tex]
