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МатематикаГДЗ по геометрии 7 класс Атанасян ФГОС Задание 1141
Задание 1141
\[\boxed{\mathbf{1141.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[AB = 6\ см;\]
\[\text{I\ }окр:\ AB = a_{4};\]
\[\text{II\ \ }окр:AB = a_{6};\]
\[l_{1} = \cup AA_{1}B;\]
\[l_{2} = \cup AB_{1}\text{B.}\]
\[\mathbf{Найти:}\]
\[\mathbf{(}l_{1} + l_{2}) - ?\]
\[\mathbf{Решение.}\]
\[1)\ AB = a_{6} = 6\ см.\]
\[2)\ a_{6} = 2R \bullet \sin\frac{180{^\circ}}{6} =\]
\[= 2R \bullet \sin{30{^\circ}} = 2 \bullet \frac{1}{2}R = R =\]
\[= 6\ см.\]
\[3)\ l_{1} = \frac{\text{πR}}{180{^\circ}} \bullet \alpha =\]
\[= \frac{6\pi}{180{^\circ}} \bullet \left( \frac{360{^\circ} \bullet 5}{6} \right) =\]
\[= \frac{6\pi \bullet 300{^\circ}}{180{^\circ}} = 10\pi\ см.\]
\[4)\ AB = a_{4} = 6\ см.\]
\[5)\ a_{4} = 2R \bullet \sin\frac{180{^\circ}}{4} =\]
\[= 2R \bullet \sin{45{^\circ}} = 2R\frac{\sqrt{2}}{2} = R\sqrt{2}\]
\[6 = R\sqrt{2}\]
\[R = \frac{6}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}\ см.\]
\[6)\ l_{2} = \frac{\text{πR}}{180{^\circ}} \bullet \alpha =\]
\[= \frac{3\sqrt{2}\pi}{180{^\circ}} \bullet \left( \frac{360{^\circ} \bullet 3}{4} \right) =\]
\[= \frac{3\sqrt{2}\pi \bullet 270{^\circ}}{180{^\circ}} = \frac{9\sqrt{2}\pi}{2}\ см.\]
\[7)\ l_{1} + l_{2} = 10\pi + \frac{9\sqrt{2}\pi}{2} =\]
\[= \pi\left( 10 + \frac{9\sqrt{2}}{2} \right) =\]
\[= \frac{\pi}{2}\left( 20 + 9\sqrt{2} \right)\ см.\]
\[Ответ:\ \frac{\pi}{2}\left( 20 + 9\sqrt{2} \right)\ см.\]
\[\boxed{\mathbf{1141.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\left| \overrightarrow{a} \right| = 5;\left| \overrightarrow{b} \right| = 2;\]
\[\left| \overrightarrow{c} \right| = 4;\overrightarrow{a}\bot\overrightarrow{b};\]
\[\overrightarrow{p} = \overrightarrow{a} - \overrightarrow{b} - \overrightarrow{c};\]
\[\overrightarrow{q} = \overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c}.\]
\[\mathbf{Найти:}\]
\[\overrightarrow{p} \bullet \overrightarrow{q} - ?\]
\[\mathbf{Решение.}\]
\[1)\ \overrightarrow{a}\bot\overrightarrow{b} \Longrightarrow \overrightarrow{a} \bullet \overrightarrow{b} = 0.\]
\[2)\ \overrightarrow{p} \bullet \overrightarrow{q} = \left( \overrightarrow{a} - \overrightarrow{b} - \overrightarrow{c} \right)\left( \overrightarrow{a} - \overrightarrow{b} + \overrightarrow{c} \right)\]
\[\overrightarrow{p} \bullet \overrightarrow{q} = \overrightarrow{a^{2}} + \overrightarrow{b^{2}} - \overrightarrow{c^{2}}.\]
\[3)\ \overrightarrow{p} \bullet \overrightarrow{q} = 5^{2} + 2^{2} - 4^{2} =\]
\[= 25 + 4 - 16 = 13.\]
\[\mathbf{Ответ:}13.\]
