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МатематикаГДЗ по геометрии 8 класс Атанасян Задание 1105
Задание 1105
\[\boxed{\mathbf{1105}\mathbf{.}\mathbf{ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Дано:\]
\[окружность\ (O;r) - вписанная.\]
\[Найти:\]
\[C - вписанной\ окружности.\]
\[Решение.\]
\[\textbf{а)}\ ABCD - квадрат;AB = a:\]
\[1)\ HH_{1} = AB = d = 2r\]
\[a = 2r\]
\[r = \frac{a}{2}.\]
\[2)\ C = 2\pi r = 2\pi\frac{a}{2} = \pi a.\]
\[\textbf{б)}\ \mathrm{\Delta}ABC - равнобедренный\ и\ \]
\[прямоугольный;\]
\[AC = CB;\ \angle C = 90{^\circ};\ AB = c:\]
\[1)\ ON = OM = OD = r.\]
\[2)\ NODC - прямоугольник \Longrightarrow\]
\[\Longrightarrow \ CD = NO = DO = CN:\]
\[NODC - квадрат.\]
\[3)\ AB^{2} = AC^{2} + BC^{2}\]
\[AB^{2} = 2AC^{2}\]
\[AC^{2} = \frac{AB^{2}}{2}\]
\[AC = \sqrt{\frac{c^{2}}{2}} = \frac{c}{\sqrt{2}} = \frac{c\sqrt{2}}{2}.\]
\[4)\ CN = AC - AN:\]
\[5)\ CD = CB - BD:\]
\[6)\ \left. \ + \frac{r = \frac{c\sqrt{2}}{2} - AM}{r = \frac{c\sqrt{2}}{2} - MB} \right| \Longrightarrow 2r =\]
\[= \frac{2c\sqrt{2}}{2} - (AM + MB).\]
\[2r = \frac{2c\sqrt{2}}{2} - AB\]
\[2r = \frac{2\sqrt{2}c}{2} - c\]
\[2r = \frac{2\sqrt{2}c - 2c}{2} = \sqrt{2c} - c\]
\[r = \frac{\sqrt{2}c - c}{2} = \frac{c\left( \sqrt{2} - 1 \right)}{2}.\]
\[7)\ C = 2\pi r = 2\pi\ \frac{c\left( \sqrt{2} - 1 \right)}{2} =\]
\[= \pi c\left( \sqrt{2} - 1 \right).\]
\[\textbf{в)}\ \mathrm{\Delta}ABC - \ прямоугольный;\]
\[AB = c;\ \angle A = \alpha;\ \angle C = 90{^\circ}:\]
\[1)\ ON = OM = OD = r.\]
\[2)\ NODC - прямоугольник \Longrightarrow\]
\[\Longrightarrow CD = NO = DO = CN:\]
\[NODC - квадра \Longrightarrow\]
\[\Longrightarrow CN = ON = r.\]
\[3)\sin{\angle A} = \frac{\text{BC}}{\text{AB}}\]
\[BC = AB \bullet \sin{\angle A} = c \bullet \sin\alpha.\]
\[4)\cos{\angle A} = \frac{\text{AC}}{\text{AB}}\]
\[AC = AB \bullet \sin{\angle A} = c \bullet \cos\alpha.\]
\[5)\ CN = AC - AN:\]
\[6)\ CD = CB - BD:\]
\[7)\ \left. \ + \frac{r = c \bullet \cos\alpha - AM}{r = c \bullet \sin\alpha - MB} \right| \Longrightarrow\]
\[\Longrightarrow 2r =\]
\[= c\left( \cos\alpha + \sin\alpha \right) - (AM + MB);\]
\[2r = c\left( \cos\alpha + \sin\alpha \right) - c\]
\[r = \frac{c\left( \cos\alpha + \sin\alpha - 1 \right)}{2}.\]
\[8)\ C = 2\pi r =\]
\[= 2\pi\ \frac{c\left( \cos\alpha + \sin{\alpha - 1} \right)}{2} =\]
\[= \pi c\left( \cos\alpha + \sin\alpha - 1 \right).\]
\[\textbf{г)}\ \mathrm{\Delta}ABC - \ равнобедренный;\]
\[AB = BC;\ \]
\[\angle A = \angle C = \alpha;\ \]
\[BH = h:\]
\[1)\ OH = r.\]
\[2)\ Рассмотрим\ \mathrm{\Delta}ABH:\]
\[tg\ \angle A = \frac{\text{BH}}{\text{AH}}\]
\[AH = \frac{\text{BH}}{tg\ \angle A} = \frac{h}{\text{tg\ α}}.\]
\[3)\ Рассмотрим\ \mathrm{\Delta}AOH:\]
\[\angle OAH = \frac{\alpha}{2}\ \]
\[(так\ как\ AO - биссектрисса);\]
\[tg\ \angle OAH = \frac{\text{OH}}{\text{AH}}\]
\[OH = AH \bullet \ tg\ \angle OAH =\]
\[= \frac{h}{\text{tg\ α}} \bullet tg\frac{\alpha}{2}.\]
\[4)\ C = 2\pi r = \frac{2\pi \bullet h\ tg\frac{\alpha}{2}}{\text{tg\ α}}.\]
