\[\boxed{\mathbf{636.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\text{p\ }и\ m - касательные;\]
\[AB - хорда;\]
\[AB = r.\]
\[\mathbf{Найти:}\]
\[\angle ACB - ?\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}AOB - равносторонний:\]
\[OA = OB\ \]
\[\left( так\ как\ \text{OA\ }и\ OB - радиусы \right);\ \]
\[AB = r.\]
\[Отсюда:\]
\[\ \angle A = 60{^\circ};\]
\[\angle B = 60{^\circ}.\]
\[2)\ Рассмотрим\ \mathrm{\Delta}ACB:\]
\[\angle A = 90{^\circ} - 60{^\circ} = 30{^\circ};\ \]
\[\angle B = 90{^\circ} - 60{^\circ} = 30{^\circ}.\]
\[Значит:\]
\[\angle ACB = 180{^\circ} - (30{^\circ} + 30{^\circ}) =\]
\[= 120{^\circ}.\]
\[Ответ:\ \angle ACB = 120{^\circ}.\]
\[\boxed{\mathbf{636.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - прямоугольник;\]
\[AB = 6\ см;\]
\[BC = 8\ см;\]
\[a\bot BD;C \in a;\]
\[a \cap AD = M;\]
\[BD \cap a = K.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABKM}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABD}} = S_{\text{BCD}}\ \]
\[(так\ как\ BD - диагональ).\]
\[2)\ BD^{2} = AB^{2} + AD^{2} =\]
\[= 36 + 64 = 100\]
\[BD = 10\ см.\]
\[3)\ S_{\text{ABD}} = S_{\text{BCD}} = \frac{1}{2}BA \bullet AD =\]
\[= \frac{1}{2} \bullet 6 \bullet 8 = 24\ см^{2}.\]
\[4)\ S_{\text{BCD}} = \frac{1}{2} \bullet BD \bullet CK\]
\[24 = \frac{1}{2} \bullet 10 \bullet CK\]
\[24 = 5CK\]
\[CK = 4,8\ см.\]
\[5)\ \mathrm{\Delta}CKD - прямоугольный:\]
\[KD^{2} = CD^{2} - CK^{2} =\]
\[= 36 - 23,04 = 12,96\]
\[KD = 3,6\ см.\]
\[6)\ S_{\text{CDM}} = \frac{1}{2}CD \bullet DM =\]
\[= \frac{1}{2} \bullet 6 \bullet DM = 3DM;\]
\[S_{\text{CDM}} = \frac{1}{2} \bullet CM \bullet KD =\]
\[= \frac{1}{2} \bullet 3,6 \bullet CM = 1,8\ CM.\]
\[7)\ CM = \sqrt{CD^{2} + MD^{2}} =\]
\[= \sqrt{36 + MD^{2}}.\]
\[8)\ 3MD = 1,8\sqrt{36 + MD^{2}}\]
\[MD = 0,6\sqrt{36 + MD^{2}}\]
\[MD^{2} = 0,36\left( 36 + MD^{2} \right)\]
\[MD^{2} = 12,96 + 0,36MD^{2}\]
\[0,64MD^{2} = 12,96\]
\[MD^{2} = 20,25\]
\[MD = 4,5\ см.\]
\[9)\ KM^{2} = MD^{2} - KD^{2} =\]
\[= 20,25 - 12,96 = 7,29\]
\[KM = 2,7\ см.\]
\[10)\ S_{\text{DKM}} = \frac{1}{2}KD \bullet KM =\]
\[= \frac{1}{2} \bullet 3,6 \bullet 2,7 = 4,86\ см^{2}.\]
\[11)\ S_{\text{ABKM}} = S_{\text{ABD}} - S_{\text{DKM}} =\]
\[= 24 - 4,86 = 19,14\ см^{2}.\]
\[\mathbf{Ответ:}19,14\ см^{2}.\]