Решебник по геометрии 8 класс Атанасян ФГОС Задание 1087

 

\[\boxed{\mathbf{1087.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[1)\ R = \frac{\sqrt{a^{2} + a^{2}}}{2} = \frac{a}{\sqrt{2}} = \frac{6}{\sqrt{2}} =\]

\[= 3\sqrt{2};\]

\[r = \frac{a}{2} = \frac{6}{2} = 3;\]

\[P = a \bullet 4 = 6 \bullet 4 = 24;\]

\[S = a^{2} = 6^{2} = 36.\]

\[2)\ a = r \bullet 2 = 4;\]

\[R = \frac{\sqrt{a^{2} + a^{2}}}{2} = \frac{a}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2};\]

\[P = a \bullet 4 = 4 \bullet 4 = 16;\]

\[S = a^{2} = 4^{2} = 16.\]

\[3)\ a = R \bullet \sqrt{2} = 4\sqrt{2};\]

\[r = \frac{a}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2};\]

\[P = a \bullet 4 = 4\sqrt{2} \bullet 4 = 16\sqrt{2};\]

\[S = a^{2} = \left( 4\sqrt{2} \right)^{2} = 32.\]

\[4)\ a = \frac{P}{4} = 7;\]

\[R = \frac{\sqrt{a^{2} + a^{2}}}{2} = \frac{a}{\sqrt{2}} = \frac{7}{\sqrt{2}} =\]

\[= \frac{7\sqrt{2}}{2};\]

\[r = \frac{a}{2} = \frac{7}{2} = 3,5;\]

\[S = a^{2} = 7^{2} = 49.\]

\[5)\ a = \sqrt{S} = 4;\]

\[R = \frac{\sqrt{a^{2} + a^{2}}}{2} = \frac{a}{\sqrt{2}} = \frac{4}{\sqrt{2}} =\]

\[= 2\sqrt{2};\]

\[r = \frac{a}{2} = \frac{4}{2} = 2;\]

\(P = a \bullet 4 = 4 \bullet 4 = 16.\)

\[\boxed{\mathbf{1087.еуроки - ответы\ на\ пятёрку}}\]

\[\textbf{а)}\ (x - 1)^{2} + (y + 2)^{2} = 25.\]

\[Окружность\ с\ центром\ в\ точке\ \]

\[(1; - 2)\ и\ R = 5.\]

\[\textbf{б)}\ x^{2} + (y + 7)^{2} = 1.\]

\[Окружность\ с\ центром\ в\ точке\]

\[\ (0; - 7)\ и\ R = 1.\]

\[\textbf{в)}\ x^{2} + y^{2} + 8x - 4y + 40 = 0\]

\[(x + 4)^{2} + (y - 2)^{2} = - 20.\]

\[Не\ окружность,\ так\ как\ R^{2} < 0.\]

\[\textbf{г)}\ x^{2} + y^{2} - 2x + 4y - 20 = 0;\]

\[(x - 1)^{2} + (y + 2)^{2} = 25.\]

\[Окружность\ с\ центром\ в\ точке\]

\[\ (1; - 2)\ и\ R = 5.\]

\[\textbf{д)}\ x^{2} + y^{2} - 4x - 2y + 1 = 0;\]

\[(x - 2)^{2} + (y - 1)^{2} = 4.\]

\[Окружность\ с\ центром\ в\ точке\ \]

\[(2;1)\ и\ R = 2.\]