Решебник по геометрии 7 класс Атанасян ФГОС Задание 1124

 

\[\boxed{\mathbf{1124.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[r_{1} = 1;r_{2} = 2;\]

\[r_{3} = 3;r_{4} = 4.\]

\[\mathbf{Найти:}\]

\[S_{1};S_{2кольца};\]

\[S_{3к};S_{4к}.\]

\[\mathbf{Решение.}\]

\[1)\ S_{1} = \pi \bullet \left( r_{1} \right)^{2} = \pi \bullet 1^{2} = \pi;\]

\[2)\ S_{2} = \pi \bullet \left( r_{2} \right)^{2} = \pi \bullet 2^{2} = 4\pi;\]

\[3)\ S_{3} = \pi \bullet \left( r_{3} \right)^{2} = \pi \bullet 3^{2} = 9\pi;\]

\[4)\ S_{4} = \pi \bullet \left( r_{4} \right)^{2} = \pi \bullet 4^{2} = 16\pi;\]

\[5)\ S_{2к} = S_{2} - S_{1} = 4\pi - \pi = 3\pi;\]

\[6)\ S_{3к} = S_{3} - S_{2} = 9\pi - 4\pi =\]

\[= 5\pi;\]

\[7)\ S_{4к} = S_{4} - S_{3} = 16\pi - 9\pi =\]

\[= 7\pi.\]

\[Ответ:S_{1} = \pi;\ S_{2к} = 3\pi;\ \]

\[S_{3к} = 5\pi;\ S_{4к} = 7\pi.\]

\[\boxed{\mathbf{1124.еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[Окружность\ (O;r);\]

\[AB,CD - хорды;\]

\[AB \cap CD = E;\]

\[AB = 13\ см;\]

\[CE = 9\ см;\]

\[ED = 4\ см;\]

\[BD = 4\sqrt{3}\ см.\]

\[\mathbf{Найти:}\]

\[\angle BED - ?\]

\[\mathbf{Решение.}\]

\[1)\ Пусть\ AE = x;\ EB = 13 - x.\]

\[x(13 - x) = 9 \bullet 4\]

\[- x^{2} + 13x = 36\]

\[x^{2} - 13x + 36 = 0.\]

\[3)\ EB = 4 \Longrightarrow \mathrm{\Delta}DEB -\]

\[равнобедренный.\]

\[По\ теоремме\ косинусов:\]

\[\left( 4\sqrt{3} \right)^{2} = 32 - 32\cos{\angle DEB}\]

\[32\cos{\angle DEB} = - 16\]

\[\cos{\angle DEB} = - 0,5\]

\[\angle DEB = 120{^\circ}.\]

\[3)\ EB = 9.\]

\[По\ теореме\ косинусов:\]

\[\left( 4\sqrt{3} \right)^{2} =\]

\[= 4^{2} + 9^{2} - 72\cos{\angle DEB}\]

\[72\cos{\angle DEB} = 49\]

\[\cos{\angle DEB} \approx 0,6806\]

\[\angle DEB \approx 47{^\circ}7^{'}.\]

\[Ответ:120{^\circ}\ или\ 47{^\circ}7'.\]