Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по геометрии 8 класс Атанасян ФГОС Задание 1050
Задание 1050
\[\boxed{\mathbf{1050.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунки\ по\ условию\ задачи:\]
\[\mathbf{а)}\]
\(\mathbf{\ }\)
\[\mathbf{б)\ }\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[\left| \overrightarrow{a} \right| = 5;\]
\[\left| \overrightarrow{b} \right| = 8;\]
\[\widehat{\overrightarrow{a}\overrightarrow{b}} = 60{^\circ}.\]
\[\mathbf{Найти:}\]
\[\textbf{а)}\left| \overrightarrow{a} + \overrightarrow{b} \right| - ?;\]
\[\textbf{б)}\left| \overrightarrow{a} - \overrightarrow{b} \right| - ?\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ \mathrm{\Delta}ADK\ и\ \mathrm{\Delta}ACK -\]
\[прямоугольные:\]
\[\angle KAD = 90{^\circ} - 60{^\circ} = 30{^\circ};\]
\[KD = \frac{1}{2}AD = \frac{1}{2} \bullet 5 = 2,5.\]
\[2)\ KC = KD + DC = 2,5 + 8 =\]
\[= 10,5\ \left( AB = DC = \overrightarrow{b} \right).\]
\[3)\ \left. \ \frac{AK = \sqrt{AD^{2} - KD^{2}}}{AK = \sqrt{AC^{2} - CK^{2}}} \right| \Longrightarrow\]
\[\Longrightarrow AD^{2} - KD^{2} = AC^{2} - CK^{2}.\]
\[25 - 6,25 = AC^{2} - 110,25\]
\[AC^{2} = 110,25 + 25 - 6,25 = 129\]
\[AC = \sqrt{129}.\]
\[\left| \overrightarrow{a} + \overrightarrow{b} \right| = \sqrt{129}.\]
\[\textbf{б)}\ 1)\ \mathrm{\Delta}AKB\ и\ \mathrm{\Delta}AKD -\]
\[прямоугольные:\]
\[\angle KAB = 90{^\circ} - 60{^\circ} = 30{^\circ};\]
\[KB = \frac{1}{2}AB = \frac{1}{2} \bullet 5 = 2,5.\]
\[2)\ DB = DK + KB = 8\]
\[DK = 8 - 2,5 = 5,5.\]
\[3)\ \left. \ \frac{AK = \sqrt{AB^{2} - KB^{2}}}{AK = \sqrt{AD^{2} - DK^{2}}} \right| \Longrightarrow\]
\[\Longrightarrow AB^{2} - KB^{2} = AD^{2} - DK^{2};\]
\[25 - 6,25 = AD^{2} - 30,25\]
\[AD^{2} = 25 - 6,25 + 30,25 = 49\]
\[AD = \sqrt{49} = 7\]
\[\left| \overrightarrow{a} - \overrightarrow{b} \right| = 7.\]
\[\mathbf{Ответ:}а)\ \sqrt{129};б)\ 7.\]
\[\boxed{\mathbf{1050.еуроки - ответы\ на\ пятёрку}}\]
\[(x - 3)^{2} + (y - 5)^{2} = 25\]
\[\textbf{а)}\ x = 3:\]
\[(3 - 3)^{2} + {(y - 5)}^{2} = 25;\]
\[{(y - 5)}^{2} = 25;\]
\[y_{1} = 10;\]
\[y_{2} = 0.\]
\[A(3;10)\ и\ B(3;0).\]
\[\textbf{б)}\ y = 5:\]
\[{(x - 3)}^{2} + {(5 - 5)}^{2} = 25;\]
\[(x - 3)^{2} = 25;\]
\[x_{1} = - 2;\]
\[x_{2} = 8.\]
\[C( - 2;5)\ и\ D(8;5).\]
