Question

Determine o perimetro do triangulo cujos vertices sao A (1,-2) B (-3,1) e C (1,4).

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{A(1,-2)\iff B(-3,1)\iff C(1,4)}$

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

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

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

$\mathsf{d_{AB} = \sqrt{(4)^2 + (-3)^2}}$

$\mathsf{d_{AB} = \sqrt{16 + 9}}$

$\mathsf{d_{AB} = \sqrt{25}}$

$\boxed{\boxed{\mathsf{d_{AB} = 5}}}$

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

$\mathsf{d_{AC} = \sqrt{(1 - 1)^2 + (-2 - 4)^2}}$

$\mathsf{d_{AC} = \sqrt{(0)^2 + (-6)^2}}$

$\mathsf{d_{AC} = \sqrt{0 + 36}}$

$\mathsf{d_{AC} = \sqrt{36}}$

$\boxed{\boxed{\mathsf{d_{AC} = 6}}}$

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

$\mathsf{d_{BC} = \sqrt{(-3 - 1)^2 + (1 - 4)^2}}$

$\mathsf{d_{BC} = \sqrt{(-4)^2 + (-3)^2}}$

$\mathsf{d_{BC} = \sqrt{16 + 9}}$

$\mathsf{d_{BC} = \sqrt{25}}$

$\boxed{\boxed{\mathsf{d_{BC} = 5}}}$

$\mathsf{P = d_{AB} + d_{AC} + d_{BC}}$

$\mathsf{P = 5 + 6 + 5}$

$\boxed{\boxed{\mathsf{P = 16}}}$