\[\boxed{\mathbf{1023.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - прямоугольник;\]
\[AC = 10\ см;\]
\[\angle AOB = 30{^\circ}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ ABCD - прямоугольник:\]
\[BD = AC = 10\ см\ (по\ свойству).\]
\[2)\ BO = OD = AO = OC =\]
\[= 5\ см\ (по\ свойству).\]
\[3)\angle COD = \angle AOB =\]
\[= 30{^\circ}\ (как\ вертикальные);\]
\[\angle BOC = \angle AOD = 180{^\circ} - 30{^\circ} =\]
\[= 150{^\circ}\ (как\ смежные);\]
\[\sin{30{^\circ}} = \sin{150{^\circ}} = \frac{1}{2}.\]
\[4)\ S_{\text{AOB}} =\]
\[= \frac{1}{2} \bullet AO \bullet BO \bullet \sin{\angle AOB} =\]
\[= \frac{1}{2} \bullet 5 \bullet 5 \bullet \frac{1}{2} = \frac{25}{4}\ см^{2};\]
\[S_{\text{COD}} = \frac{1}{2} \bullet CO \bullet OD \bullet \sin{\angle COD} =\]
\[= \frac{1}{2} \bullet 5 \bullet 5 \bullet \frac{1}{2} = \frac{25}{4}\ см^{2};\]
\[S_{\text{BOC}} = \frac{1}{2} \bullet BO \bullet OC \bullet \sin{\angle BOC} =\]
\[= \frac{1}{2} \bullet 5 \bullet 5 \bullet \frac{1}{2} = \frac{25}{4}\ см^{2};\]
\[S_{\text{AOD}} = \frac{1}{2} \bullet AO \bullet OD \bullet \sin{\angle AOD} =\]
\[= \frac{1}{2} \bullet 5 \bullet 5 \bullet \frac{1}{2} = \frac{25}{4}\ см^{2}.\]
\[5)\ S_{\text{ABCD}} = 4 \bullet \frac{25}{4} = 25\ см^{2}.\]
\[Ответ:25\ см^{2}.\]
\[\boxed{\mathbf{1023.еуроки - ответы\ на\ пятёрку}}\]
\[AM = MB \Longrightarrow\]
\[\Longrightarrow M - середина\ \text{AB.}\]
\[A\] | \[(2; - 3)\] | \[( - 10; - 11)\] | \[(0;1)\] | \[(0;0)\] | \[(c,d)\] | \[(3;5)\] | \[(3t + 5;7)\] | \[(1;3)\] |
---|---|---|---|---|---|---|---|---|
\[B\] | \[( - 3;1)\] | \[(4;7)\] | \[(6; - 11)\] | \[( - 3;7)\] | \[(2a - c;2b - d)\] | \[(3;8)\] | \[(t + 7; - 7)\] | \[( - 1; - 3)\] |
\[M\] | \[( - 0,5; - 1)\] | \[( - 3; - 2)\] | \[(3; - 5)\] | \[( - 1,5;3,5)\] | \[(a;b)\] | \[(3;6,5)\] | \[(2t + 6;0)\] | \[(0;0)\] |