\[\boxed{\mathbf{1120.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[окружность\ \left( O;R_{1} \right);\]
\[окружность\ \left( O;R_{2} \right);\]
\[R_{1} = 1,5\ см;\]
\[R_{2} = 2,5\ см.\]
\[\mathbf{Найти:}\]
\[S_{кольца} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{1} = \pi\left( R_{1} \right)^{2};\ \ \ \]
\[S_{2} = \pi\left( R_{2} \right)^{2}.\]
\[2)\ S_{кольца} = S_{2} - S_{1} =\]
\[= \pi\left( R_{2} \right)^{2} - \pi\left( R_{1} \right)^{2} =\]
\[= \pi\left( \left( R_{2} \right)^{2} - \left( R_{1} \right)^{2} \right)\]
\[S_{кольца} = 3,14 \bullet \left( {2,5}^{2} - {1,5}^{2} \right) =\]
\[= 3,14 \bullet (6,25 - 2,25) =\]
\[= 3,14 \bullet 4 = 12,56\ см^{2}.\]
\[Ответ:S_{кольца} = 12,56\ см^{2}.\]
\[\boxed{\mathbf{1120.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{а})\ a = 5;b = c = 4:\]
\[= b^{2} + c^{2} - 2bc \bullet cos\angle A\]
\[25 = 16 + 16 - 2 \bullet 16 \bullet cos\angle A\]
\[32cos\angle A = 32 - 25\]
\[32cos\angle A = 7\]
\[cos\angle A = \frac{7}{32} \approx 0,2188;\]
\[\angle A = 77{^\circ}22^{'}.\]
\[Напртив\ большей\ стороны\ \]
\[лежит\ больший\ угол:\]
\[\ \mathrm{\Delta}ABC - остроугольный.\]
\[\mathbf{б})\ a = 17;b = 8;с = 15:\]
\[a^{2} = b^{2} + c^{2} - 2bc \bullet cos\angle A\]
\[289 = 64 + 225 - 240 \bullet cos\angle A\]
\[240cos\angle A = 0\]
\[cos\angle A = 0\]
\[\angle A = 90{^\circ}.\]
\[\mathrm{\Delta}ABC - прямоугольный.\]
\[\mathbf{в})\ a = 9;b = 5;c = 6:\]
\[a^{2} = b^{2} + c^{2} - 2bc \bullet cos\angle A\]
\[81 = 52 + 36 - 60 \bullet cos\angle A\]
\[60cos\angle A = - 20\]
\[cos\angle A = - \frac{1}{3} < 0\]
\[\angle A - тупой.\]
\[\mathrm{\Delta}ABC - тупоугольный.\]