Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по геометрии 9 класс Атанасян ФГОС Задание 1104
Задание 1104
\[\boxed{\mathbf{1104.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\textbf{а)}\ \mathrm{\Delta}ABC - правильный;AB = a:\]
\[a_{3} = 2R \bullet \sin{60{^\circ}} = 2R\frac{\sqrt{3}}{2} =\]
\[= \sqrt{3}R = > R = \frac{a_{3}}{\sqrt{3}} = \frac{a}{\sqrt{3}};\]
\[C = 2\pi R = 2\pi\frac{a}{\sqrt{3}} = \frac{2\sqrt{3}\text{πa}}{3}.\]
\[\textbf{б)}\ \mathrm{\Delta}ABC - прямоугольный;\]
\[\angle C = 90{^\circ};AC = a;CB = b:\]
\[AB = d \Longrightarrow AB = \sqrt{a^{2} + b^{2}};\]
\[AO = OB = \frac{\text{AB}}{2} = \frac{\sqrt{a^{2} + b^{2}}}{2} = R;\]
\[C = 2\pi R = \frac{2\pi\sqrt{a^{2} + b^{2}}}{2} =\]
\[= \pi\sqrt{a^{2} + b^{2}}.\]
\[\textbf{в)}\ \mathrm{\Delta}ABC - равнобедренный;\]
\[AC = a;AB = BC = b:\]
\[1)\ OB = R;BM = MC;OM\bot BC.\]
\[2)\ \mathrm{\Delta}BHC\sim\mathrm{\Delta}OBM - по\ двум\ \]
\[углам:\]
\[\angle HBC - общий;\ \]
\[\angle OMB = \angle BHC = 90{^\circ}.\]
\[Получаем:\]
\[\frac{\text{OB}}{\text{BC}} = \frac{\text{BM}}{\text{BH}}\]
\[OB = \frac{BC \bullet BM}{\text{BH}}.\]
\[3)\ \mathrm{\Delta}HBC - прямоугольный:\]
\[BH = \sqrt{BC^{2} - HC^{2}} =\]
\[= \sqrt{b^{2} - \left( \frac{a}{2} \right)^{2}} = \sqrt{b^{2} - \frac{a^{2}}{4}} =\]
\[= \sqrt{\frac{4b^{2} - a^{2}}{4}} = \frac{1}{2}\sqrt{4b^{2} - a^{2}}.\]
\[4)\ OB =\]
\[= (b \bullet b \bullet 2)/(2\sqrt{4b^{2} - a^{2}} =\]
\[= \frac{b^{2}}{\sqrt{4b^{2} - a^{2}}} = R.\]
\[5)\ C = 2\pi R = \frac{2\pi b^{2}}{\sqrt{4b^{2} - a^{2}}}.\]
\[\textbf{г)}\ ABCD - прямоугольник;\]
\[AB = a;\left( \widehat{\text{BD\ AC}} \right) = \alpha:\]
\[1)\ AC = BD = d.\]
\[2)\ BO = OC = AO = OD = R\ \]
\[(по\ свойству\ прямоугольника).\]
\[3)\ \mathrm{\Delta}AOB - равнобедренный:\]
\[OB = OA;\]
\[OH - высота,\ медиана\ и\ \]
\[биссектрисса\ (по\ свойству).\]
\[4)\ Рассмотрим\ \mathrm{\Delta}BOH:\]
\[\angle BOH = \frac{\alpha}{2}\ \]
\[(так\ как\ OH - биссектрисса);\]
\[BH = \frac{a}{2}\ \]
\[(так\ как\ OH - медиана);\]
\[\sin\frac{\alpha}{2} = \frac{\text{BH}}{\text{BO}}\]
\[BO = \frac{\text{BH}}{\sin\frac{\alpha}{2}} = \frac{a}{2\sin\frac{\alpha}{2}} = R.\]
\[5)\ C = 2\pi R = 2\pi\frac{a}{2\sin\frac{\alpha}{2}} = \frac{\text{πa}}{\sin\frac{\alpha}{2}}.\]
\[\textbf{д)}\ A_{1}A_{2}A_{3}A_{4}A_{5}A_{6} -\]
\[правильный\ шестиугольник;\]
\[S = 24\sqrt{3}\ см^{2}:\]
\[1)\ S = \frac{1}{2}\Pr\]
\[r = R \bullet \cos\frac{180{^\circ}}{6} = R \bullet \cos{30{^\circ}} =\]
\[= R\frac{\sqrt{3}}{2};\]
\[a_{6} = 2R \bullet \sin\frac{180{^\circ}}{6} =\]
\[= 2R \bullet \sin{30{^\circ}} = 2R\frac{1}{2} = R.\]
\[P = 6 \bullet R;\]
\[S = \frac{1}{2} \bullet 6R \bullet R\frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}R^{2}.\]
\[2)\ 24\sqrt{3} = \frac{3\sqrt{3}}{2}R^{2};\]
\[R^{2} = \frac{24\sqrt{3} \bullet 2}{3\sqrt{3}} = 16 \Longrightarrow R = 4\ см.\]
\[3)\ C = 2\pi R = 2 \bullet \pi \bullet 4 = 8\pi\ см.\]
\[\boxed{\mathbf{1104.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\sin{\angle A} = \frac{2}{3}.\]
\[Начертим\ единичную\ \]
\[окружность:\]
\[2)\ sin^{2}\angle A + cos^{2\ }\angle A = 1;\]
\[\left( \frac{2}{3} \right)^{2} + cos^{2}\angle A = 1\]
\[\text{co}s^{2}\angle A = 1 - \frac{4}{9} = \frac{5}{9}\]
\[\cos{\angle A} = \frac{\sqrt{5}}{3} \approx 0,75.\]
\[x_{M} = 0,75;\ \ y_{M} = \frac{2}{3}.\]
\[\textbf{б)}\cos{\angle A} = \frac{3}{4}.\]
\[Начертим\ единичную\ \]
\[окружность.\]
\[\text{si}n^{2}\angle A + cos^{2\ }\angle A = 1\]
\[\sin^{2}\angle A + \left( \frac{3}{4} \right)^{2} = 1\]
\[\sin^{2}{\angle A} = 1 - \frac{9}{16} = \frac{7}{16}\]
\[\sin{\angle A} = \frac{\sqrt{7}}{4} \approx 0,66 \approx \frac{2}{3}.\]
\[x_{M} = \frac{2}{3};y_{M} = \frac{3}{4}.\]
\[\textbf{в)}\cos{\angle A} = - \frac{2}{5};\]
\[Начертим\ единичную\ \]
\[окружность.\]
\[\text{si}n^{2}\angle A + cos^{2\ }\angle A = 1;\]
\[\sin^{2}\angle A + \left( - \frac{2}{5} \right)^{2} = 1\]
\[\sin^{2}{\angle A} = 1 - \frac{4}{25} = \frac{21}{25}\]
\[\sin{\angle A} = \frac{\sqrt{21}}{5} \approx 0,91.\]
\[x_{M} = - \frac{2}{5};y_{M} = 0,91.\]
