Решебник по алгебре 9 класс Мерзляк Задание 819

\[\boxed{\mathbf{819\ (819).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[b_{8} = 16;\ \ q = \frac{3}{4}\]

\[b_{8} = b_{7} \cdot q,\ \ \]

\[b_{7} = \frac{b_{8}}{q} = \frac{16 \cdot 4}{3} = \frac{64}{3} = 21\frac{1}{3}.\]

\[\boxed{\mathbf{819.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[По\ т.\ Пифагора\ \ a^{2} + b^{2} = c^{2}\text{\ \ \ }\]

\[или\ \ \ a^{2} + b^{2} = 41^{2}\]

\[S = \frac{1}{2}ab = 180.\]

\[Составим\ систему:\]

\[\left\{ \begin{matrix} a^{2} + b^{2} = 41^{2}\text{\ \ \ } \\ \frac{1}{2}ab = 180\ \ | \cdot 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} a² + b² = 1681\ \\ 2ab = 720\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \ +\]

\[a^{2} + 2ab + b^{2} = 2401\]

\[(a + b)^{2} = 2401\]

\[a + b = 49\ \ \ \ или\ \ \ \ a + b = - 49\]

\[\left\{ \begin{matrix} a + b = 49 \\ ab = 360 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} a = 49 - b\ \ \ \ \ \ \ \ \ \\ (49 - b)b = 360 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} a = 49 - b \\ b^{2} - 49b + 360 = 0 \\ \end{matrix} \right.\ \]

\[b_{1} + b_{2} = 49\]

\[b_{1} \cdot b_{2} = 360\]

\[\left\{ \begin{matrix} b = 40 \\ a = 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }или\ \ \ \ \ \ \ \left\{ \begin{matrix} b = 9 \\ a = 40 \\ \end{matrix} \right.\ \]

\[Ответ:9\ см,\ 40\ см.\]