Explicație pas cu pas:
A(2;-2), B(-4;4), C(-4;0)
[tex]p = \frac{AB + AC + BC}{2} = [/tex]
[tex]AB = \sqrt{ {(2 - ( - 4))}^{2} + {( - 2 - 4)}^{2} } = \sqrt{ {6}^{2} + {6}^{2} } \\ = 6 \sqrt{2} [/tex]
[tex]AC = \sqrt{ {(2 - ( - 4))}^{2} + {( - 2 - 0)}^{2} } \\ = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10} [/tex]
[tex]BC = \sqrt{ {( - 4 - ( - 4))}^{2} + {(4 - 0)}^{2} } \\ = \sqrt{ {4}^{2} } = 4 [/tex]
semiperimetrul:
[tex]p = \frac{6 \sqrt{2} + 2\sqrt{10} + 4}{2} \\ = 3 \sqrt{2} + \sqrt{10} + 2[/tex]
[tex]Aria = \sqrt{(3 \sqrt{2} + \sqrt{10} + 2)(3 \sqrt{2} + \sqrt{10} + 2 - 6 \sqrt{2} )(3 \sqrt{2} + \sqrt{10} + 2 - 2 \sqrt{10} )(3 \sqrt{2} + \sqrt{10} + 2 - 4)} \\[/tex]
[tex]= \sqrt{(\sqrt{10} + 2 + 3 \sqrt{2})(\sqrt{10} + 2 - 3 \sqrt{2})(3 \sqrt{2} - \sqrt{10} + 2)(3 \sqrt{2} + \sqrt{10} - 2)} \\[/tex]
[tex]= \sqrt{4( \sqrt{10} + 1) \times 4( \sqrt{10} + 1)} \\ [/tex]
[tex]= \sqrt{16(10 - 1)} = \sqrt{16 \times 9} = \sqrt{{12}^{2}} = 12 \\ [/tex]
unde:
[tex](\sqrt{10} + 2 + 3 \sqrt{2})(\sqrt{10} + 2 - 3 \sqrt{2}) = {(\sqrt{10} + 2)}^{2} - {(3 \sqrt{2})}^{2} = 10 + 4 + 4 \sqrt{10} - 18 = 4 \sqrt{10} - 4 = 4(\sqrt{10} - 1)[/tex]
și
[tex](3 \sqrt{2} - \sqrt{10} + 2)(3 \sqrt{2} + \sqrt{10} - 2) = {(3 \sqrt{2} )}^{2} - {(\sqrt{10} - 2)}^{2} = 18 - 10 - 4 + 4 \sqrt{10} = 4 + 4 \sqrt{10} = 4( \sqrt{10} + 1)[/tex]
