ГДЗ по геометрии 7 класс Атанасян Задание 1094

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Год:2020-2021-2022
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Задание 1094

\(\boxed{\mathbf{1094.ОК\ ГДЗ - домашка\ на}\ 5}\)

\[\textbf{а)}\ n = 4;R = 3\sqrt{2}\ см:\]

\[1)\ a_{4} = 2 \bullet R \bullet \sin\frac{180{^\circ}}{4} =\]

\[= 2 \bullet 3\sqrt{2} \bullet \sin{45{^\circ}} =\]

\[= 2 \bullet 3\sqrt{2} \bullet \frac{\sqrt{2}}{2} = 6\ см;\]

\[2)\ P = 4 \bullet a_{4} = 4 \bullet 6 = 24\ см;\]

\[3)\ r = R \bullet \cos\frac{180{^\circ}}{4} =\]

\[= 3\sqrt{2} \bullet \cos{45{^\circ}} = 3\sqrt{2} \bullet \frac{\sqrt{2}}{2} =\]

\[= 3\ см;\]

\[4)\ S = \frac{1}{2}P \bullet r = \frac{1}{2} \bullet 24 \bullet 3 =\]

\[= 36\ см^{2}.\]

\[\textbf{б)}\ n = 3;P = 24\ см:\]

\[1)\ P = 3 \bullet a_{3} \Longrightarrow a_{3} = \frac{P}{3} = \frac{24}{3} =\]

\[= 8\ см;\]

\[2)\ 8 = 2R \bullet \sin\frac{180{^\circ}}{3}\]

\[8 = 2R \bullet \sin{60{^\circ}}\]

\[8 = 2R \bullet \frac{\sqrt{3}}{2}\ \]

\[8 = R\sqrt{3}\]

\[R = \frac{8}{\sqrt{3}}\ см;\]

\[3)\ r = R \bullet \cos\frac{180{^\circ}}{3} =\]

\[= \frac{8}{\sqrt{3}} \bullet \cos{60{^\circ}} = \frac{8}{\sqrt{3}} \bullet \frac{1}{2} = \frac{4}{\sqrt{3}}\ см;\]

\[4)\ S = \frac{1}{2}P \bullet r = \frac{1}{2} \bullet 24 \bullet \frac{4}{\sqrt{3}} =\]

\[= \frac{48}{\sqrt{3}} = \frac{48\sqrt{3}}{3} = 16\sqrt{3}\ см^{2}.\]

\[\textbf{в)}\ n = 6;r = 9\ см:\]

\[1)\ r = R \bullet \cos\frac{180{^\circ}}{n}\]

\[9 = R \bullet \cos\frac{180{^\circ}}{6}\]

\[9 = R \bullet \cos{30{^\circ}}\]

\[9 = R\frac{\sqrt{3}}{2}\]

\[18 = R\sqrt{3}\]

\[R = \frac{18}{\sqrt{3}}\ см;\]

\[2)\ a_{6} = 2 \bullet \frac{18}{\sqrt{3}} \bullet \sin\frac{180{^\circ}}{6} =\]

\[= 2 \bullet \frac{18}{\sqrt{3}} \bullet \sin{30{^\circ}} = 2 \bullet \frac{18}{\sqrt{3}} \bullet \frac{1}{2} =\]

\[= \frac{18}{\sqrt{3}}\ см;\]

\[3)\ P = 6 \bullet \frac{18}{\sqrt{3}} = 6 \bullet \frac{18}{3} \bullet \sqrt{3} =\]

\[= 36\sqrt{3}\ см;\]

\[4)\ S = \frac{1}{2} \bullet P \bullet r = \frac{1}{2} \bullet 36\sqrt{3} \bullet 9 =\]

\[= 162\sqrt{3}\ см^{2}.\]

\[\textbf{г)}\ n = 8;r = 5\sqrt{3}\ см:\]

\[1)\ r = R \bullet \cos\frac{180{^\circ}}{8}\]

\[5\sqrt{3} = R \bullet \cos{22{^\circ}30^{'}}\]

\[R = \frac{5\sqrt{3}}{\cos{22{^\circ}30^{'}}}\ см;\]

\[2)\ a_{6} =\]

\[= 2 \bullet \frac{5\sqrt{3}}{\cos{22{^\circ}30^{'}}} \bullet \sin{22{^\circ}30^{'}} =\]

\[= 10\sqrt{3} \bullet tg\ 22{^\circ}30'\ см;\]

\[3)\ P = 8 \bullet 10\sqrt{3} \bullet tg\ 22{^\circ}30^{'} =\]

\[= 8 \bullet 10\sqrt{3} \bullet 0,4142 \approx\]

\[\approx 33,136\sqrt{3}\ см;\]

\[4)\ S = \frac{1}{2} \bullet P \bullet r =\]

\[= \frac{1}{2} \bullet 33,136\sqrt{3} \bullet 5\sqrt{3} =\]

\[= 248,52\ см^{2}.\]

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