\[\boxed{\mathbf{491.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дан}\mathbf{о:\ }\]
\[\mathrm{\Delta}\text{ABC};\ \]
\[\angle C = 90{^\circ};\]
\[\text{CH}\bot\text{AB};\]
\[\text{AC} = a;\]
\[\text{CB} = b;\]
\[Найти:\]
\[\text{CH} - ?\]
\[Решение.\]
\[\textbf{а)}\ a = 5;b = 12:\]
\[1)\ \text{AB} = c = \sqrt{144 + 25} =\]
\[= \sqrt{169} = 13;\]
\[2)\ S_{\text{ABC}} = \frac{1}{2}\text{ab} = \frac{1}{2} \bullet 5 \bullet 12 = 30;\]
\[S_{\text{ABC}} = \frac{1}{2} \bullet c \bullet \text{CH} = 30;\]
\[\text{CH} = 30\ :\left( \frac{1}{2} \bullet 13 \right) = \frac{30 \bullet 2}{13} =\]
\[= 4\frac{8}{13}.\]
\[\textbf{б)}\ a = 12;b = 16:\]
\[1)\ \text{AB} = c = \sqrt{144 + 256} =\]
\[= \sqrt{400} = 20;\]
\[2)\ S_{\text{ABC}} = \frac{1}{2}\text{ab} = \frac{1}{2} \bullet 12 \bullet 16 = 96;\]
\[S_{\text{ABC}} = \frac{1}{2} \bullet c \bullet \text{CH} = 96;\]
\[\text{CH} = \frac{192}{20} = \frac{96}{10} = 9,6.\]
\[Ответ:а)\ 4\frac{8}{13};\ \ б)\ 9,6.\]