Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по геометрии 7 класс Атанасян ФГОС Задание 678
Задание 678
\[\boxed{\mathbf{678.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AA_{1},BB_{1} - биссектриса;\]
\[AA_{1} \cap BB_{1} = M;\]
\[\textbf{а)}\ \angle AMB = 136{^\circ};\]
\[\textbf{б)}\ \angle AMB = 111{^\circ}.\]
\[\mathbf{Найти:}\]
\[\angle ACM - ?\]
\[\angle BCM - ?\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ M - точка\ пересечения\ \]
\[биссектрис\ AA_{1}\ и\ BB_{1}:\ \]
\[CM - биссектриса \Longrightarrow\]
\[\Longrightarrow \angle BCM = \angle MCA.\]
\[2)\ Рассмотрим\ \mathrm{\Delta}\text{ABM.\ }\]
\[По\ теореме\ о\ сумме\ углов\ в\ \]
\[треугольнике:\]
\[\angle BAM + \angle ABM =\]
\[= 180{^\circ} - 136{^\circ} = 44{^\circ}.\]
\[3)\ \angle C = 180{^\circ} - (\angle A + \angle B);\]
\[\angle A = \angle BAM + \angle MAC\ \]
\[\left( так\ как\ AA_{1} - биссектриса \right);\]
\[\angle A = 2\angle BAM.\]
\[\angle B = \angle ABM + \angle MBC\ \]
\[\left( так\ как\ BB_{1} - биссектриса \right);\]
\[\angle B = 2\angle ABM.\]
\[= 180{^\circ} - (44{^\circ} + 44{^\circ}) =\]
\[= 180{^\circ} - 88{^\circ} = 92{^\circ}.\]
\[5)\ \angle ACM = \angle BCM = \frac{92{^\circ}}{2} = 46{^\circ}.\]
\[\textbf{б)}\ 1)\ M - точка\ пересечения\ \]
\[биссектрис\ AA_{1}\ и\ BB_{1}:\ \]
\[CM - биссектриса \Longrightarrow\]
\[\Longrightarrow \ \angle BCM = \angle MCA.\]
\[2)\ Рассмотрим\ \mathrm{\Delta}\text{ABM.\ }По\ \]
\[теореме\ о\ сумме\ углов\ в\ \]
\[треугольнике:\]
\[\angle BAM + \angle ABM =\]
\[= 180{^\circ} - 111{^\circ} = 69{^\circ}.\]
\[3)\ \angle C = 180{^\circ} - (\angle A + \angle B);\]
\[\angle A = \angle BAM + \angle MAC\ \]
\[\left( так\ как\ AA_{1} - биссектриса \right);\]
\[\angle A = 2\angle BAM.\]
\[\angle B = \angle ABM + \angle MBC\ \]
\[\left( так\ как\ BB_{1} - биссектриса \right);\]
\[\angle B = 2\angle ABM.\]
\[= 180{^\circ} - (69{^\circ} + 69{^\circ}) =\]
\[= 180{^\circ} - 138{^\circ} = 42{^\circ}.\]
\[5)\ \angle ACM = \angle BCM = \frac{42{^\circ}}{2} = 21{^\circ}.\]
\[Отсюда:а)\ \angle ACM = \angle BCM =\]
\[= 46{^\circ};\]
\[\textbf{б)}\ \angle ACM = \angle BCM = 21{^\circ}.\ \]
\[\boxed{\mathbf{678.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\ \]
\[\mathrm{\Delta}ABC - прямоугольный;\ \]
\[\angle C = 90{^\circ};\]
\[CH\bot AB.\]
\[Решение:\]
\[\textbf{а)}\ b_{c} = 25;a_{c} = 16:\]
\[1)\ h = \sqrt{a_{c} \bullet b_{c}} = \sqrt{25 \bullet 16} =\]
\[= 5 \bullet 4 = 20;\]
\[2)\ c = b_{c} + a_{c} = 25 + 16 = 41;\]
\[3)\ b = \sqrt{c \bullet b_{c}} = \sqrt{41 \bullet 25} =\]
\[= 5\sqrt{41};\]
\[4)\ a = \sqrt{c \bullet a_{c}} = \sqrt{41 \bullet 16} =\]
\[= 4\sqrt{41}.\]
\[Ответ:h = 20;a = 4\sqrt{41};\]
\[b = 5\sqrt{41}.\]
\[\textbf{б)}\ b_{c} = 36;a_{c} = 64:\]
\[1)\ h = \sqrt{a_{c} \bullet b_{c}} = \sqrt{64 \bullet 36} =\]
\[= 6 \bullet 8 = 48;\]
\[2)\ c = b_{c} + a_{c} = 36 + 64 = 100;\]
\[3)\ a = \sqrt{c \bullet a_{c}} = \sqrt{64 \bullet 100} =\]
\[= 8 \bullet 10 = 80;\]
\[4)\ b = \sqrt{c \bullet b_{c}} = \sqrt{36 \bullet 100} =\]
\[= 6 \bullet 10 = 60.\]
\[Ответ:h = 48;a = 80;b = 60.\]
\[\textbf{в)}\ b = 12;b_{c} = 6:\]
\[1)\ b^{2} = b_{c} \bullet c\]
\[c = \frac{b^{2}}{b_{c}} = \frac{144}{6} = 24;\]
\[2)\ c = b_{c} + a_{c}\]
\[a_{c} = c - b_{c} = 24 - 6 = 18;\]
\[3)\ a = \sqrt{a_{c} \bullet c} = \sqrt{18 \bullet 24} =\]
\[= \sqrt{6 \bullet 3 \bullet 6 \bullet 4} = 6 \bullet 2\sqrt{3} = 12\sqrt{3}.\]
\[Ответ:a = 12\sqrt{3};a_{c} = 18;\]
\[c = 24.\]
\[\textbf{г)}\ a = 8;a_{c} = 4:\]
\[1)\ a = \sqrt{a_{c} \bullet c} = > c = \frac{a^{2}}{a_{c}} =\]
\[= \frac{64}{4} = 16;\]
\[2)\ c = b_{c} + a_{c}\]
\[b_{c} = c - a_{c} = 16 - 4 = 12;\]
\[3)\ b = \sqrt{b_{c} \bullet c} = \sqrt{12 \bullet 16} =\]
\[= 2 \bullet 4\sqrt{3} = 8\sqrt{3}.\]
\[Ответ:b = 8\sqrt{3};b_{c} = 12;c = 16.\]
\[\textbf{д)}\ a = 6;c = 9:\]
\[1)\ b = \sqrt{c^{2} - a^{2}} = \sqrt{81 - 36} =\]
\[= \sqrt{45} = 3\sqrt{5};\]
\[2)\ a = \sqrt{a_{c} \bullet c}\]
\[a_{c} = \frac{a^{2}}{c} = \frac{36}{9} = 4;\]
\[3)\ b = \sqrt{b_{c} \bullet c}\]
\[b_{c} = \frac{b^{2}}{c} = \frac{45}{9} = 5;\]
\[4)\ h = \sqrt{a_{c} \bullet b_{c}} = \sqrt{4 \bullet 5} = 2\sqrt{5}.\]
\[Ответ:h = 2\sqrt{5};b = 3\sqrt{5};\]
\[a_{c} = 4;b_{c} = 5.\ \]
