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МатематикаГДЗ по геометрии 8 класс Атанасян ФГОС Задание 950
Задание 950
\[\boxed{\mathbf{950.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[MNPQ - четырехугольник;\]
\[\textbf{а)}\ M(1;1);\]
\[N(6;1);\]
\[P(7;4);\]
\[Q(2;4).\]
\[\textbf{б)}\ M( - 5;1);\]
\[N( - 4;4);\]
\[P( - 1;5);\]
\[Q( - 2;2).\]
\[\mathbf{Доказать:}\]
\[MNPQ - параллелограмм.\]
\[Найти:\]
\[MN\ и\ \text{PQ.}\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ MQ =\]
\[= \sqrt{(2 - 1)^{2} + (4 - 1)^{2}} =\]
\[= \sqrt{1 + 9} = \sqrt{10}\]
\[NP = \sqrt{(7 - 6)^{2} + (4 - 1)^{2}} =\]
\[= \sqrt{1 + 9} = \sqrt{10}\]
\[MN = \sqrt{(6 - 1)^{2} + (1 - 1)^{2}} =\]
\[= \sqrt{25 + 0} = 5\]
\[PQ = \sqrt{(2 - 7)^{2} + (4 - 4)^{2}} =\]
\[= \sqrt{25 + 0} = 5\]
\[3)\ NQ = \sqrt{(2 - 6)^{2} + (4 - 1)^{2}} =\]
\[= \sqrt{16 + 9} = 5;\]
\[MP = \sqrt{(7 - 1)^{2} + (4 - 1)^{2}} =\]
\[= \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}.\]
\[\textbf{б)}\ 1)\ MQ =\]
\[= \sqrt{( - 2 + 5)^{2} + (2 - 1)^{2}} =\]
\[= \sqrt{9 + 1} = \sqrt{10}\]
\[NP = \sqrt{( - 1 + 4)^{2} + (5 - 4)^{2}} =\]
\[= \sqrt{9 + 1} = \sqrt{10}\]
\[MN = \sqrt{( - 4 + 5)^{2} + (4 - 1)^{2}} =\]
\[= \sqrt{1 + 9} = \sqrt{10}\]
\[PQ = \sqrt{( - 2 + 1)^{2} + (2 - 5)^{2}} =\]
\[= \sqrt{1 + 9} = \sqrt{10}\]
\[3)\ NQ =\]
\[= \sqrt{( - 2 + 4)^{2} + (2 - 4)^{2}} =\]
\[= \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2};\]
\[MP = \sqrt{( - 1 + 5)^{2} + (5 - 1)^{2}} =\]
\[= \sqrt{16 + 16} = 4\sqrt{2}.\]
\[\mathbf{Ответ:}\mathbf{\ }а)\ NQ = 5;MP = 3\sqrt{2};\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ NQ = 2\sqrt{2};MP = 4\sqrt{2}.\]
\[\boxed{\mathbf{950.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ \mathbf{задачи:}\]
\[Дано:\ \]
\[\mathrm{\Delta}ABC - равносторонний;\]
\[AB = BC = AC = a.\]
\[Найти:\]
\[\textbf{а)}\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} \right|;\ \]
\[\textbf{б)}\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{AC}} \right|;\ \]
\[\textbf{в)}\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{CB}} \right|;\ \]
\[\textbf{г)}\left| \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} \right|;\ \]
\[\textbf{д)}\left| \overrightarrow{\text{AB}} - \overrightarrow{\text{AC}} \right|\text{.\ }\]
\[Решение.\]
\[\textbf{а)}\ \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} = \overrightarrow{\text{AC}} =\]
\[= \text{a\ }(по\ правилу\ треугольника).\]
\[\left| \overrightarrow{\text{AD}} \right| = 2AO;\]
\[AO = \sqrt{a^{2} - \left( \frac{a}{2} \right)^{2}} = \sqrt{\frac{3a^{2}}{4}} =\]
\[= \frac{a\sqrt{3}}{2}\]
\[\left| \overrightarrow{\text{AD}} \right| = 2 \bullet \frac{a\sqrt{3}}{2} = a\sqrt{3}.\]
\[\textbf{в)}\ \overrightarrow{\text{AB}} + \overrightarrow{\text{CB}} = \overrightarrow{\text{CD}} + \overrightarrow{\text{DE}} =\]
\[= \overrightarrow{\text{CE}}\ (DE \parallel BC;DE = BC):\]
\[CDEB -\]
\[ромб\ (по\ построению) \Longrightarrow\]
\[\Longrightarrow CE = AD.\]
\[Следовательно:\]
\[\left| \overrightarrow{\text{CE}} \right| = a\sqrt{3}.\]
\[\textbf{г)}\ \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} = \overrightarrow{\text{CA}} =\]
\[= \text{a\ }(по\ правилу\ треугольника).\]
\[\textbf{д)}\ \overrightarrow{\text{AB}} - \overrightarrow{\text{AC}} = \overrightarrow{\text{CB}} =\]
\[= \text{a\ }(по\ правилу\ треугольника).\]
\[Ответ:а)\ a;б)\ a\sqrt{3};в)\ a\sqrt{3};\]
\[\textbf{г)}\ a;д)\ \text{a.}\]
