Question

Знайдіть периметр трикутника ABC

A (1; 0; 0) ,B (3; 0; 2 ),

C(1; 1; 0 ).

Answer 1

Ответ:

Р =  4 + 2√2 ≈ 6,83.

Объяснение:

A (1; 0; 0) ,B (3; 0; 2 ), C(1; 1; 0 ).

Периметр АВС - ?

$AB_x = x_B - x_A = 3-1 =2;~~~~~~AB_y = y_B - y_A = 0-0 =0;$

$AB_z = z_B - z_A = 2-0 =2;\\\\ AB = \sqrt{AB_x^2+AB_y^2+AB_z^2} =\sqrt{2^2+0^2+2^2} =\sqrt{8} =2\sqrt{2} .$

$BC_x = x_C - x_B = 1 -3 =-2;~~~~~~BC_y = y_C - y_B = 1-0 =1;$

$BC_z = z_C - z_B = 0-2 =-2;\\\\ BC = \sqrt{BC_x^2+BC_y^2+BC_z^2} =\sqrt{(-2)^2+1^2+(-2)^2} =\sqrt{9} =3.$

$AC_x = x_C - x_A = 1-1 =0;~~~~~~AC_y = y_B - y_A = 1-0 =1;$

$AC_z = z_C - z_A = 0-0 =0;\\\\ AC = \sqrt{AC_x^2+AC_y^2+AC_z^2} =\sqrt{0^2+1^2+0^2} =\sqrt{1} =1.$

Периметр АВС:

Р = АВ + ВС + АС = 2√2 + 3 + 1 = 4 + 2√2 ≈ 6,83.