ГДЗ по геометрии 11 класс. Атанасян ФГОС 676

Авторы:
Год:2023
Тип:учебник

676

\[\boxed{\mathbf{676.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[O(0;0;0);\ \ A(4;0;0);\ \ B(0;6;0);\ \ \]

\[C(0;0; - 2).\]

\[\textbf{а)}\ \mathrm{\Delta}ABC - лежит\ в\ плоскости\ \]

\[Oxy;\]

\[D(x;y);\]

\[AD = BD = OD.\]

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

\[= \sqrt{(x - 0)^{2} + (y - 0)^{2}} =\]

\[= \sqrt{x^{2} + y^{2}};\]

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

\[= \sqrt{(x - 4)^{2} + (y - 0)^{2}} =\]

\[= \sqrt{x^{2} - 8x + 16 + y^{2}};\]

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

\[= \sqrt{(x - 0)^{2} + (y - 6)^{2}} =\]

\[= \sqrt{x^{2} + y^{2} - 12y + 36}.\]

\[x^{2} + y^{2} = x^{2} - 8x + y^{2} + 16 =\]

\[= x^{2} + y^{2} - 12y + 36\]

\[0 = - 8x + 16 = - 12y + 36\]

\[x = 2;\ \ y = 3.\]

\[R = \left| \text{OD} \right| = \sqrt{2^{2} + 3^{2}} = \sqrt{13}.\]

\[D(2;3;0) - центр\ окружности,\ \]

\[описанной\ около\ \mathrm{\Delta}ABC.\]

\[\textbf{б)}\ D(x;y;z) - равноудалена\ от\ \]

\[вершин\ тетраэдра:\]

\[OD = AD = BD = CD.\]

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

\[= \sqrt{(x - 0)^{2} + (y - 0)^{2} + (z - 0)^{2}} =\]

\[= \sqrt{x^{2} + y^{2} + z^{2}};\]

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

\[= \sqrt{(x - 4)^{2} + (y - 0)^{2} + (z - 0)^{2}} =\]

\[= \sqrt{x^{2} - 8x + 16 + y^{2} + z^{2}};\]

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

\[= \sqrt{(x - 0)^{2} + (y - 6)^{2} + (z - 0)^{2}} =\]

\[= \sqrt{x^{2} + y^{2} - 12y + 36 + z^{2}};\]

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

\[= \sqrt{(x - 0)^{2} + (y - 0)^{2} + (z + 2)^{2}} =\]

\[= \sqrt{x^{2} + y^{2} + z^{2} + 4z + 4}.\]

\[x^{2} + y^{2} + z^{2} =\]

\[= x^{2} - 8x + 16 + y^{2} + z^{2} =\]

\[= x^{2} + y^{2} - 12y + 36 + z^{2} =\]

\[= x^{2} + y^{2} + z^{2} + 4z + 4;\]

\[0 = - 8x + 16 = - 12y + 36 =\]

\[= 4z + 4\]

\[x = 2;\ \ y = 3;\ \ z = - 1.\]

\[D(2;3; - 1) - равноудалена\ от\ \]

\[вершин\ тетраэдра\ \text{OABC.}\]


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