Решебник по геометрии 10 класс Атанасян ФГОС 668

\[\boxed{\mathbf{668.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Найдем\ длины\ сторон\ каждого\ \]

\[треугольника\ по\ формуле:\]

\[\textbf{а)}\ A(9;3; - 5);\ \ B(2;10; - 5);\ \ \]

\[C(2;3;2).\]

\[\left| \text{AB} \right| = \left| \text{BC} \right| = \left| \text{AC} \right| -\]

\[равносторонний\ треугольник.\]

\[\textbf{б)}\ A(3;7; - 4);\ \ B(5; - 3;2);\ \ \]

\[C(1;3; - 10).\]

\[\left| \text{AB} \right| = \left| \overrightarrow{\text{AB}} \right| =\]

\[= \sqrt{(5 - 3)^{2} + ( - 3 - 7)^{2} + (2 + 4)^{2}} =\]

\[= \sqrt{140};\]

\[\left| \text{AB} \right|^{2} + \left| \text{AC} \right|^{2} = 140 + 56 =\]

\[= 196 = \left| \text{AC} \right|^{2}.\]

\[\mathrm{\Delta}ABC - прямоугольный;\ \]

\[\angle A = 90{^\circ}.\]

\[\textbf{в)}\ A(5; - 5; - 1);\ \ B(5; - 3; - 1);\ \ \]

\[C(4; - 3;0).\]

\[\left| \text{AB} \right|^{2} + \left| \text{BC} \right|^{2} = 4 + 2 = 6 =\]

\[= \left| \text{AC} \right|^{2}.\]

\[⊿ABC - прямоугольный;\ \ \]

\[\angle B = 90{^\circ}.\]

\[\textbf{г)}\ A( - 5;2;0);\ \ B( - 4;3;0);\ \ \]

\[C( - 5;2; - 2).\]

\[\left| \text{AB} \right|^{2} + \left| \text{AC} \right|^{2} = 2 + 4 = 6 =\]

\[= \left| \text{BC} \right|^{2}.\]

\[⊿ABC - прямоугольный;\ \]

\[\angle A = 90{^\circ}.\]