Question

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Answer 1

La calcule am scris putin direct, pt ca nu am avut loc

Answer 2

Explicație pas cu pas:

A(2;-2), B(-4;4), C(-4;0)

$p = \frac{AB + AC + BC}{2} =$

$AB = \sqrt{ {(2 - ( - 4))}^{2} + {( - 2 - 4)}^{2} } = \sqrt{ {6}^{2} + {6}^{2} } \\ = 6 \sqrt{2}$

$AC = \sqrt{ {(2 - ( - 4))}^{2} + {( - 2 - 0)}^{2} } \\ = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}$

$BC = \sqrt{ {( - 4 - ( - 4))}^{2} + {(4 - 0)}^{2} } \\ = \sqrt{ {4}^{2} } = 4$

semiperimetrul:

$p = \frac{6 \sqrt{2} + 2\sqrt{10} + 4}{2} \\ = 3 \sqrt{2} + \sqrt{10} + 2$

$Aria = \sqrt{(3 \sqrt{2} + \sqrt{10} + 2)(3 \sqrt{2} + \sqrt{10} + 2 - 6 \sqrt{2} )(3 \sqrt{2} + \sqrt{10} + 2 - 2 \sqrt{10} )(3 \sqrt{2} + \sqrt{10} + 2 - 4)}$

$= \sqrt{(\sqrt{10} + 2 + 3 \sqrt{2})(\sqrt{10} + 2 - 3 \sqrt{2})(3 \sqrt{2} - \sqrt{10} + 2)(3 \sqrt{2} + \sqrt{10} - 2)}$

$= \sqrt{4( \sqrt{10} + 1) \times 4( \sqrt{10} + 1)}$

$= \sqrt{16(10 - 1)} = \sqrt{16 \times 9} = \sqrt{{12}^{2}} = 12$

unde:

$(\sqrt{10} + 2 + 3 \sqrt{2})(\sqrt{10} + 2 - 3 \sqrt{2}) = {(\sqrt{10} + 2)}^{2} - {(3 \sqrt{2})}^{2} = 10 + 4 + 4 \sqrt{10} - 18 = 4 \sqrt{10} - 4 = 4(\sqrt{10} - 1)$

și

$(3 \sqrt{2} - \sqrt{10} + 2)(3 \sqrt{2} + \sqrt{10} - 2) = {(3 \sqrt{2} )}^{2} - {(\sqrt{10} - 2)}^{2} = 18 - 10 - 4 + 4 \sqrt{10} = 4 + 4 \sqrt{10} = 4( \sqrt{10} + 1)$