Question

qual e o perimetro do triangulo ABC, sendo A( 1,0) B ( 3,4) C ( -2,4)

Answer 1

Resposta:

P = 10 + 2√5

Explicação passo-a-passo:

P = AB + BC + AC

$AB^{2} = (x_B - x_A)^{2} + (y_B - y_A)^{2}\\\\AB^{2} = (3 - 1)^{2} + (4 - 0)^{2}\\\\AB^{2} = (2)^{2} + (4)^{2}\\\\AB^{2} = 4 + 16\\\\AB^{2} = 20 = 2^{2} 5\\\\AB = 2\sqrt{5}$

BC ⇒ podemos usar a mesma formula de cima, mas observe que

yB = yC = 4, portanto a parcela (yB - yC)² = 0

logo BC² = (xB - xC)²

BC = |xB - xC|

BC = 3 - (-2)

BC = 5

$AC^{2} = (x_C - x_A)^{2} + (y_C - y_A)^{2}\\\\AC^{2} = (-2 - 1)^{2} + (4 - 0)^{2}\\\\AC^{2} = (3)^{2} + (4)^{2}\\\\AC^{2} = 9 + 16\\\\AC^{2} = 25 \\\\AC = 5$

P = AB + BC + AC

P = 2√5 + 5 + 5

P = 10 + 2√5