ГДЗ по геометрии 8 класс Атанасян Задание 1088

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Год:2020-2021-2022
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Задание 1088

\[\boxed{\mathbf{1088.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[N\] \[R\] \[r\] \[a_{3}\] \[P\] \[S\]
\[1\] \[3\] \[1,5\] \[3\sqrt{3}\] \[9\sqrt{3}\] \[\frac{27\sqrt{3}}{4}\]
\[2\] \[\frac{2}{3}\sqrt{10\sqrt{3}}\] \[\frac{1}{3}\sqrt{10\sqrt{3}}\] \[2\sqrt{\frac{10\sqrt{3}}{3}}\] \[6\sqrt{\frac{10\sqrt{3}}{3}}\] \[10\]
\[3\] \[4\] \[2\] \[4\sqrt{3}\] \[12\sqrt{3}\] \[12\sqrt{3}\]
\[4\] \[\frac{5\sqrt{3}}{3}\] \[\frac{5\sqrt{3}}{6}\] \[5\] \[15\] \[\frac{25\sqrt{3}}{4}\]
\[5\] \[\frac{2\sqrt{3}}{3}\] \[\frac{\sqrt{3}}{3}\] \[2\] \[6\] \[\sqrt{}3\]

\[1)\ a = \frac{3R}{\sqrt{3}} = \frac{9}{\sqrt{3}} = 3\sqrt{3};\]

\[r = \frac{a\sqrt{3}}{6} = \frac{3\sqrt{3} \bullet \sqrt{3}}{6} = 1,5;\]

\[P = 3a = 3 \bullet 3\sqrt{3} = 9\sqrt{3};\]

\[S = \frac{a^{2}\sqrt{3}}{4} = \frac{\left( 3\sqrt{3} \right)^{2}\sqrt{3}}{4} = \frac{27\sqrt{3}}{4}.\]

\[2)\ a = \sqrt{\frac{4S}{\sqrt{3}}} = \sqrt{\frac{4 \bullet 10}{\sqrt{3}}} =\]

\[= 2\sqrt{\frac{10\sqrt{3}}{3}};\]

\[R = \frac{a\sqrt{3}}{3} = \frac{2\sqrt{\frac{10\sqrt{3}}{3}} \bullet \sqrt{3}}{3} =\]

\[= \frac{2}{3}\sqrt{10\sqrt{3}};\]

\[r = \frac{a\sqrt{3}}{6} = \frac{2\sqrt{\frac{10\sqrt{3}}{3}} \bullet \sqrt{3}}{6} =\]

\[= \frac{1}{3}\sqrt{10\sqrt{3}};\]

\[P = 3a = 3 \bullet 2\sqrt{\frac{10\sqrt{3}}{3}} = 6\sqrt{\frac{10\sqrt{3}}{3}}.\]

\[3)\ a = \frac{6r}{\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3};\]

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

\[P = 3a = 3 \bullet 4\sqrt{3} = 12\sqrt{3};\]

\[S = \frac{a^{2}\sqrt{3}}{4} = \frac{\left( 4\sqrt{3} \right)^{2}\sqrt{3}}{4} = 12\sqrt{3}.\]

\[4)\ R = \frac{a\sqrt{3}}{3} = \frac{5\sqrt{3}}{3};\]

\[r = \frac{a\sqrt{3}}{6} = \frac{5\sqrt{3}}{6};\]

\[P = 3a = 3 \bullet 5 = 15;\]

\[S = \frac{a^{2}\sqrt{3}}{4} = \frac{(5)^{2}\sqrt{3}}{4} = \frac{25\sqrt{3}}{4};\]

\[5)\ a = \frac{P}{3} = \frac{6}{3} = 2;\]

\[R = \frac{a\sqrt{3}}{3} = \frac{2\sqrt{3}}{3};\]

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

\[S = \frac{a^{2}\sqrt{3}}{4} = \frac{(2)^{2}\sqrt{3}}{4} = \sqrt{3}.\]

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