\[\boxed{\mathbf{1035.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[Окружность\ (O;r);\]
\[AB,CD - хорды;\]
\[AB \cap CD = E;\]
\[AB = 13\ см;\]
\[CE = 9\ см;\]
\[ED = 4\ см;\]
\[BD = 4\sqrt{3}\ см.\]
\[\mathbf{Найти:}\]
\[\angle BED - ?\]
\[\mathbf{Решение.}\]
\[1)\ Пусть\ AE = x;\ EB = 13 - x.\]
\[2)\ AE \bullet EB = DE \bullet EC\ \]
\[(по\ свойству\ пересекающихся\ хорд):\]
\[x(13 - x) = 9 \bullet 4\]
\[- x^{2} + 13x = 36\]
\[x^{2} - 13x + 36 = 0.\]
\[3)\ EB = 4 \Longrightarrow \mathrm{\Delta}DEB -\]
\[равнобедренный.\]
\[По\ теоремме\ косинусов:\]
\[\left( 4\sqrt{3} \right)^{2} = 32 - 32\cos{\angle DEB}\]
\[32\cos{\angle DEB} = - 16\]
\[\cos{\angle DEB} = - 0,5\]
\[\angle DEB = 120{^\circ}.\]
\[3)\ EB = 9.\]
\[По\ теореме\ косинусов:\]
\[\left( 4\sqrt{3} \right)^{2} =\]
\[= 4^{2} + 9^{2} - 72\cos{\angle DEB}\]
\[72\cos{\angle DEB} = 49\]
\[\cos{\angle DEB} \approx 0,6806\]
\[\angle DEB \approx 47{^\circ}7^{'}.\]
\[Ответ:120{^\circ}\ или\ 47{^\circ}7'.\]
\[\boxed{\mathbf{1035.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - четырехугольник;\]
\[\textbf{а)}\ A( - 3; - 1);\]
\[B(1; - 1);\]
\[C(1; - 3);\]
\[D( - 3; - 3).\]
\[\textbf{б)}\ A(4;1);\]
\[B(3;5);\]
\[C( - 1;4);\]
\[D(0;0).\]
\[\mathbf{Доказать:}\]
\[ABCD - прямоугольник.\]
\[Найти:\]
\[S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ AB =\]
\[= \sqrt{(1 + 3)^{2} + ( - 1 + 1)^{2}} =\]
\[= \sqrt{16} = 4\]
\[CD =\]
\[= \sqrt{( - 3 - 1)^{2} + ( - 1 + 1)^{2}} =\]
\[= \sqrt{16} = 4\]
\[BC = \sqrt{(1 - 1)^{2} + ( - 3 + 1)^{2}} =\]
\[= \sqrt{4} = 2\]
\[AD =\]
\[= \sqrt{( - 3 + 3)^{2} + ( - 3 + 1)^{2}} =\]
\[= \sqrt{4} = 2\]
\[AB = CD;\ \ BC = AD.\]
\[3)\ AC =\]
\[= \sqrt{(1 + 3)^{2} + ( - 3 + 1)^{2}} =\]
\[= \sqrt{16 + 4} = 2\sqrt{5};\]
\[BD =\]
\[= \sqrt{( - 3 - 1)^{2} + ( - 3 + 1)^{2}} =\]
\[= \sqrt{16 + 4} = 2\sqrt{5}.\]
\[4)\ AC = BD;\ \ ABCD -\]
\[параллелограмм:\]
\[5)\ S_{\text{ABCD}} = AB \bullet AD = 4 \bullet 2 = 8.\]
\[\textbf{б)}\ 1)\ AB =\]
\[= \sqrt{(3 - 4)^{2} + (5 - 1)^{2}} =\]
\[= \sqrt{1 + 16} = \sqrt{17}\]
\[CD = \sqrt{(0 + 1)^{2} + (0 - 4)^{2}} =\]
\[= \sqrt{1 + 16} = \sqrt{17}\]
\[BC = \sqrt{( - 1 - 3)^{2} + (4 - 5)^{2}} =\]
\[= \sqrt{16 + 1} = \sqrt{17}\]
\[AD = \sqrt{(0 - 4)^{2} + (0 - 1)^{2}} =\]
\[= \sqrt{16 + 1} = \sqrt{17}\]
\[AB = CD;\ \ BC = AD.\]
\[3)\ AC =\]
\[= \sqrt{( - 1 - 4)^{2} + (4 - 1)^{2}} =\]
\[= \sqrt{25 + 9} = \sqrt{34}\]
\[BD = \sqrt{(0 - 3)^{2} + (0 - 5)^{2}} =\]
\[= \sqrt{9 + 25} = \sqrt{34}\]
\[4)\ AC = BD;\ \ ABCD -\]
\[параллелограмм:\]
\[5)\ S_{\text{ABCD}} = AB \bullet AD =\]
\[= \sqrt{17} \bullet \sqrt{17} = 17.\]
\[\mathbf{Ответ:}\mathbf{\ }а)\ 8;\ \ б)\ 17.\]