Question

Determine o perímetro do triângulo ABC, cujas coordenadas são: A (3, 7), B (– 8, 8) e C (2, 10)

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{d_{AB} = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}}$

$\mathsf{d_{AB} = \sqrt{(3 - (-8))^2 + (7 - 8)^2}}$

$\mathsf{d_{AB} = \sqrt{(3 + 8)^2 + (7 - 8)^2}}$

$\mathsf{d_{AB} = \sqrt{(11)^2 + (-1)^2}}$

$\mathsf{d_{AB} = \sqrt{121 + 1}}$

$\boxed{\boxed{\mathsf{d_{AB} = \sqrt{122}}}}$

$\mathsf{d_{AC} = \sqrt{(x_A - x_C)^2 + (y_A - y_C)^2}}$

$\mathsf{d_{AC} = \sqrt{(3 - 2)^2 + (7 - 10)^2}}$

$\mathsf{d_{AC} = \sqrt{(1)^2 + (-3)^2}}$

$\mathsf{d_{AC} = \sqrt{(1)^2 + (-3)^2}}$

$\mathsf{d_{AC} = \sqrt{1 + 9}}$

$\boxed{\boxed{\mathsf{d_{AC} = \sqrt{10}}}}$

$\mathsf{d_{BC} = \sqrt{(x_B - x_C)^2 + (y_B - y_C)^2}}$

$\mathsf{d_{BC} = \sqrt{(-8 - 2)^2 + (8 - 10)^2}}$

$\mathsf{d_{BC} = \sqrt{(-10)^2 + (-2)^2}}$

$\mathsf{d_{BC} = \sqrt{100 + 4}}$

$\mathsf{d_{BC} = \sqrt{104}}$

$\boxed{\boxed{\mathsf{d_{BC} = 2\sqrt{26}}}}$

$\boxed{\boxed{\mathsf{P = \sqrt{122} + \sqrt{10} + 2\sqrt{26}}}}$