Решебник по геометрии 8 класс Мерзляк Задание 484

\[Схематический\ рисунок.\]

\[Дано:\]

\[MNKP - прямоугольник;\]

\[AD - высота;\]

\[BC = 72\ см;\]

\[AD = 24\ см;\]

\[MP\ :MN = 9\ :5.\]

\[Найти:\]

\[MN;\ MP.\]

\[Решение.\]

\[1)\ MNKP - прямоугольник:\]

\[MP \parallel NK;\]

\[\angle M = 90{^\circ};\]

\[NK = MP = 1,8MN.\]

\[2)\ Для\ BC\ и\ NK\ и\ секущей\ AB:\]

\[\angle ABC = \angle ANK.\]

\[3)\ \mathrm{\Delta}ANK\sim\mathrm{\Delta}ABC - первый\ \]

\[признак:\]

\[\angle NAK = \angle BAC;\]

\[\angle ANK = \angle ABC.\]

\[Отсюда:\]

\[\frac{\text{NK}}{\text{BC}} = \frac{\text{AN}}{\text{AB}}\]

\[AB = \frac{BC \bullet AN}{\text{NK}}\]

\[AB = \frac{72 \bullet AN}{1,8 \bullet MN} = \frac{40 \bullet AN}{\text{MN}}.\]

\[4)\ \mathrm{\Delta}ABD\sim\mathrm{\Delta}NBM - первый\ \]

\[признак:\]

\[\angle ABD = \angle NBM;\]

\[\angle ADB = \angle NMB.\]

\[Отсюда:\]

\[\frac{\text{NM}}{\text{AD}} = \frac{\text{NB}}{\text{AB}}\]

\[AB = \frac{AD \bullet BN}{\text{MN}}\]

\[AB = \frac{24 \bullet (AB - AN)}{\text{MN}}.\]

\[5)\ \frac{40 \bullet AN}{\text{MN}} = \frac{24 \bullet (AB - AN)}{\text{MN}}\]

\[5AN = 3AB - 3AN\]

\[8AN = 3AB\]

\[\frac{\text{AN}}{\text{AB}} = \frac{3}{8}.\]

\[MN = \frac{40 \bullet AN}{\text{AB}} = \frac{40 \bullet 3}{8} = 15\ см.\]

\[MP = 1,8 \bullet 15 = 27\ см.\]

\[Ответ:\ \ 15\ см;\ 27\ см.\]