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МатематикаРешебник по геометрии 9 класс Атанасян ФГОС Задание 1105
Задание 1105
\[\boxed{\mathbf{1105}\mathbf{.}\mathbf{ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Дано:\]
\[окружность\ (O;r) - вписанная.\]
\[Найти:\]
\[C - вписанной\ окружности.\]
\[Решение.\]
\[\textbf{а)}\ ABCD - квадрат;AB = a:\]
\[1)\ HH_{1} = AB = d = 2r\]
\[a = 2r\]
\[r = \frac{a}{2}.\]
\[2)\ C = 2\pi r = 2\pi\frac{a}{2} = \pi a.\]
\[\textbf{б)}\ \mathrm{\Delta}ABC - равнобедренный\ и\ \]
\[прямоугольный;\]
\[AC = CB;\ \angle C = 90{^\circ};\ AB = c:\]
\[1)\ ON = OM = OD = r.\]
\[2)\ NODC - прямоугольник \Longrightarrow\]
\[\Longrightarrow \ CD = NO = DO = CN:\]
\[NODC - квадрат.\]
\[3)\ AB^{2} = AC^{2} + BC^{2}\]
\[AB^{2} = 2AC^{2}\]
\[AC^{2} = \frac{AB^{2}}{2}\]
\[AC = \sqrt{\frac{c^{2}}{2}} = \frac{c}{\sqrt{2}} = \frac{c\sqrt{2}}{2}.\]
\[4)\ CN = AC - AN:\]
\[5)\ CD = CB - BD:\]
\[6)\ \left. \ + \frac{r = \frac{c\sqrt{2}}{2} - AM}{r = \frac{c\sqrt{2}}{2} - MB} \right| \Longrightarrow 2r =\]
\[= \frac{2c\sqrt{2}}{2} - (AM + MB).\]
\[2r = \frac{2c\sqrt{2}}{2} - AB\]
\[2r = \frac{2\sqrt{2}c}{2} - c\]
\[2r = \frac{2\sqrt{2}c - 2c}{2} = \sqrt{2c} - c\]
\[r = \frac{\sqrt{2}c - c}{2} = \frac{c\left( \sqrt{2} - 1 \right)}{2}.\]
\[7)\ C = 2\pi r = 2\pi\ \frac{c\left( \sqrt{2} - 1 \right)}{2} =\]
\[= \pi c\left( \sqrt{2} - 1 \right).\]
\[\textbf{в)}\ \mathrm{\Delta}ABC - \ прямоугольный;\]
\[AB = c;\ \angle A = \alpha;\ \angle C = 90{^\circ}:\]
\[1)\ ON = OM = OD = r.\]
\[2)\ NODC - прямоугольник \Longrightarrow\]
\[\Longrightarrow CD = NO = DO = CN:\]
\[NODC - квадра \Longrightarrow\]
\[\Longrightarrow CN = ON = r.\]
\[3)\sin{\angle A} = \frac{\text{BC}}{\text{AB}}\]
\[BC = AB \bullet \sin{\angle A} = c \bullet \sin\alpha.\]
\[4)\cos{\angle A} = \frac{\text{AC}}{\text{AB}}\]
\[AC = AB \bullet \sin{\angle A} = c \bullet \cos\alpha.\]
\[5)\ CN = AC - AN:\]
\[6)\ CD = CB - BD:\]
\[7)\ \left. \ + \frac{r = c \bullet \cos\alpha - AM}{r = c \bullet \sin\alpha - MB} \right| \Longrightarrow\]
\[\Longrightarrow 2r =\]
\[= c\left( \cos\alpha + \sin\alpha \right) - (AM + MB);\]
\[2r = c\left( \cos\alpha + \sin\alpha \right) - c\]
\[r = \frac{c\left( \cos\alpha + \sin\alpha - 1 \right)}{2}.\]
\[8)\ C = 2\pi r =\]
\[= 2\pi\ \frac{c\left( \cos\alpha + \sin{\alpha - 1} \right)}{2} =\]
\[= \pi c\left( \cos\alpha + \sin\alpha - 1 \right).\]
\[\textbf{г)}\ \mathrm{\Delta}ABC - \ равнобедренный;\]
\[AB = BC;\ \]
\[\angle A = \angle C = \alpha;\ \]
\[BH = h:\]
\[1)\ OH = r.\]
\[2)\ Рассмотрим\ \mathrm{\Delta}ABH:\]
\[tg\ \angle A = \frac{\text{BH}}{\text{AH}}\]
\[AH = \frac{\text{BH}}{tg\ \angle A} = \frac{h}{\text{tg\ α}}.\]
\[3)\ Рассмотрим\ \mathrm{\Delta}AOH:\]
\[\angle OAH = \frac{\alpha}{2}\ \]
\[(так\ как\ AO - биссектрисса);\]
\[tg\ \angle OAH = \frac{\text{OH}}{\text{AH}}\]
\[OH = AH \bullet \ tg\ \angle OAH =\]
\[= \frac{h}{\text{tg\ α}} \bullet tg\frac{\alpha}{2}.\]
\[4)\ C = 2\pi r = \frac{2\pi \bullet h\ tg\frac{\alpha}{2}}{\text{tg\ α}}.\]
\[\boxed{\mathbf{1105.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\textbf{а)}\ OA = 3;\ \alpha = 45{^\circ};\]
\[\textbf{б)}\ OA = 1,5;\ \alpha = 90{^\circ};\]
\[\textbf{в)}\ OA = 5;\ \alpha = 150{^\circ};\]
\[\textbf{г)}\ OA = 1;\ \alpha = 180{^\circ};\]
\[\textbf{д)}\ OA = 2;\ \alpha = 30{^\circ}.\]
\[\mathbf{Найти:}\]
\[A(x;y) - ?\]
\[\mathbf{Решение.}\]
\[1)\ Вычисление\ координат\ \]
\[точки\ A:\]
\[x = OA \bullet \cos\alpha;\ \ \ \]
\[y = OA \bullet \sin\alpha.\]
\[\textbf{а)}\sin{45{^\circ}} = \frac{\sqrt{2}}{2};\]
\[\cos{45{^\circ}} = \frac{\sqrt{2}}{2}.\]
\[\textbf{б)}\sin{90{^\circ}} = 1;\cos{90{^\circ}} = 0.\]
\[\left\{ \begin{matrix} x = 1,5 \bullet 0 = 0\ \ \ \\ y = 1,5 \bullet 1 = 1,5 \\ \end{matrix} \right.\ \Longrightarrow A(0;1,5).\]
\[\textbf{в)}\sin{150{^\circ}} = \frac{1}{2};\cos{150{^\circ}} =\]
\[= - \frac{\sqrt{3}}{2}.\]
\[\textbf{г)}\sin{180{^\circ}} = 0;\cos{180{^\circ}} = - 1.\]
\[\left\{ \begin{matrix} x = 1 \bullet ( - 1) = - 1 \\ y = 1 \bullet 0 = 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow A( - 1;0).\]
\[\textbf{д)}\sin{30{^\circ}} = \frac{1}{2};\cos{30{^\circ}} = \frac{\sqrt{3}}{2}.\]
\[\left\{ \begin{matrix} x = 2 \bullet \frac{\sqrt{3}}{2} = \sqrt{3} \\ y = 2 \bullet \frac{1}{2} = 1\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow A\left( \sqrt{3};1 \right).\]
\[Ответ:а)\ A\left( \frac{3\sqrt{2}}{2};\frac{3\sqrt{2}}{2} \right);\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ A(0;1,5);\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ в)\ A\left( - \frac{5\sqrt{3}}{2};\frac{5}{2} \right);\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ г)\ A( - 1;0);\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ д)\ A(\sqrt{3};1).\]
