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МатематикаГДЗ по геометрии 7 класс Атанасян ФГОС Задание 1088
Задание 1088
\[\boxed{\mathbf{1088.ОК\ ГДЗ - домашка\ на}\ 5}\]
| \[N\] | \[R\] | \[r\] | \[a_{3}\] | \[P\] | \[S\] |
|---|---|---|---|---|---|
| \[1\] | \[3\] | \[1,5\] | \[3\sqrt{3}\] | \[9\sqrt{3}\] | \[\frac{27\sqrt{3}}{4}\] |
| \[2\] | \[\frac{2}{3}\sqrt{10\sqrt{3}}\] | \[\frac{1}{3}\sqrt{10\sqrt{3}}\] | \[2\sqrt{\frac{10\sqrt{3}}{3}}\] | \[6\sqrt{\frac{10\sqrt{3}}{3}}\] | \[10\] |
| \[3\] | \[4\] | \[2\] | \[4\sqrt{3}\] | \[12\sqrt{3}\] | \[12\sqrt{3}\] |
| \[4\] | \[\frac{5\sqrt{3}}{3}\] | \[\frac{5\sqrt{3}}{6}\] | \[5\] | \[15\] | \[\frac{25\sqrt{3}}{4}\] |
| \[5\] | \[\frac{2\sqrt{3}}{3}\] | \[\frac{\sqrt{3}}{3}\] | \[2\] | \[6\] | \[\sqrt{}3\] |
\[1)\ a = \frac{3R}{\sqrt{3}} = \frac{9}{\sqrt{3}} = 3\sqrt{3};\]
\[r = \frac{a\sqrt{3}}{6} = \frac{3\sqrt{3} \bullet \sqrt{3}}{6} = 1,5;\]
\[P = 3a = 3 \bullet 3\sqrt{3} = 9\sqrt{3};\]
\[S = \frac{a^{2}\sqrt{3}}{4} = \frac{\left( 3\sqrt{3} \right)^{2}\sqrt{3}}{4} = \frac{27\sqrt{3}}{4}.\]
\[2)\ a = \sqrt{\frac{4S}{\sqrt{3}}} = \sqrt{\frac{4 \bullet 10}{\sqrt{3}}} =\]
\[= 2\sqrt{\frac{10\sqrt{3}}{3}};\]
\[R = \frac{a\sqrt{3}}{3} = \frac{2\sqrt{\frac{10\sqrt{3}}{3}} \bullet \sqrt{3}}{3} =\]
\[= \frac{2}{3}\sqrt{10\sqrt{3}};\]
\[r = \frac{a\sqrt{3}}{6} = \frac{2\sqrt{\frac{10\sqrt{3}}{3}} \bullet \sqrt{3}}{6} =\]
\[= \frac{1}{3}\sqrt{10\sqrt{3}};\]
\[P = 3a = 3 \bullet 2\sqrt{\frac{10\sqrt{3}}{3}} = 6\sqrt{\frac{10\sqrt{3}}{3}}.\]
\[3)\ a = \frac{6r}{\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3};\]
\[R = \frac{a\sqrt{3}}{3} = \frac{4\sqrt{3} \bullet \sqrt{3}}{3} = 4;\]
\[P = 3a = 3 \bullet 4\sqrt{3} = 12\sqrt{3};\]
\[S = \frac{a^{2}\sqrt{3}}{4} = \frac{\left( 4\sqrt{3} \right)^{2}\sqrt{3}}{4} = 12\sqrt{3}.\]
\[4)\ R = \frac{a\sqrt{3}}{3} = \frac{5\sqrt{3}}{3};\]
\[r = \frac{a\sqrt{3}}{6} = \frac{5\sqrt{3}}{6};\]
\[P = 3a = 3 \bullet 5 = 15;\]
\[S = \frac{a^{2}\sqrt{3}}{4} = \frac{(5)^{2}\sqrt{3}}{4} = \frac{25\sqrt{3}}{4};\]
\[5)\ a = \frac{P}{3} = \frac{6}{3} = 2;\]
\[R = \frac{a\sqrt{3}}{3} = \frac{2\sqrt{3}}{3};\]
\[r = \frac{a\sqrt{3}}{6} = \frac{2\sqrt{3}}{6};\]
\[S = \frac{a^{2}\sqrt{3}}{4} = \frac{(2)^{2}\sqrt{3}}{4} = \sqrt{3}.\]
\[\boxed{\mathbf{1088.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[A(3;0);\]
\[B( - 1;2);\]
\[A\ и\ B \in окружности;\]
\[O \in y = x + 2.\]
\[\mathbf{Найти:}\]
\[уравнение\ окружности.\]
\[\mathbf{Решение.}\]
\[1)\ R = AO =\]
\[= \sqrt{(3 - x)^{2} + (0 - y)^{2}} =\]
\[= \sqrt{(3 - x)^{2} + y^{2}};\]
\[R = BO =\]
\[= \sqrt{( - 1 - x)^{2} + (2 - y)^{2}}.\]
\[2)\ (3 - x)^{2} + y^{2} =\]
\[= ( - 1 - x)^{2} + (2 - y)^{2}\]
\[9 - 6x + x^{2} + y^{2} =\]
\[= 1 + 2x + x^{2} + 4 - 4y + y^{2}\]
\[9 - 6x = 2x - 4y + 5\]
\[\left. \ 8x - 4y - 4 = 0\ \ \ \ \ \ \right|\ :4\]
\[2x - y - 1 = 0.\]
\[3)\ \left\{ \begin{matrix} 2x - y - 1 = 0 \\ y = x + 2\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} 2x - x - 2 - 1 = 0 \\ y = x + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = 3 \\ y = 5 \\ \end{matrix} \Longrightarrow O(3;5). \right.\ \]
\[4)\ R = AO = \sqrt{(3 - 3)^{2} + 5^{2}} =\]
\[= \sqrt{25} = 5.\]
\[5)\ Уравнение\ окружности:\]
\[(x - 3)^{2} + (y - 5)^{2} = 25.\]
\[Ответ:(x - 3)^{2} + (y - 5)^{2} =\]
\[= 25.\]
