Question

2 - São dados A(3,-1), B(1,1) e C(5,5).
a) Calcule o perimetro do triângulo ABC.
b) Mostre que ABC é um triângulo retângulo.​

Answer 1

Resposta:

$dAB=\sqrt{(x_{b}-x_{a})^{2}+(y_{b}-y_{a})^{2}}\\dAB=\sqrt{(1-3)^{2}+(1+1)^{2}}\\dAB=\sqrt{(-2)^{2}+(2)^{2}}\\dAB=\sqrt{8}}=\sqrt{2^{2}*2}}\\dAB=2\sqrt{2}\\\\dBC=\sqrt{(x_{c}-x_{b})^{2}+(y_{c}-y_{b})^{2}}\\dBC=\sqrt{(5-1)^{2}+(5-1)^{2}}\\dBC=\sqrt{(4)^{2}+(4)^{2}}\\dBC=\sqrt{16+16}}\\dBC=\sqrt{32}=\sqrt{2^{2}*2^{2}*2}\\dBC=4\sqrt{2}$$dAC=\sqrt{(x_{c}-x_{a})^{2}+(y_{c}-y_{a})^{2}}\\dAC=\sqrt{(5-3)^{2}+(5+1)^{2}}\\dAC=\sqrt{(2)^{2}+(6)^{2}}\\dAC=\sqrt{4+36}}\\dAC=\sqrt{40}}=\sqrt{2^{2*10}}}\\dAC=2\sqrt{10}$

$perimetro;\\2\sqrt{2}+4\sqrt{2}+2\sqrt{10}\\6\sqrt{2}+2\sqrt{10}\\\\a^{2} =b^{2}+c^{2} \\(2\sqrt{10})^{2}=(4\sqrt{2})^{2}+(2\sqrt{2})^{2}\\40=32+8\\40=40$

Explicação passo-a-passo: