Решебник по геометрии 9 класс Мерзляк Задание 58

\[Схематический\ рисунок.\]

\[Дано:\]

\[ABCD - параллелограмм;\]

\[BD = 8\ см;\]

\[AC = 14\ см;\]

\[AB = AD + 2\ см.\]

\[Найти:\]

\[AB;\ AD.\]

\[1)\ AO = \frac{1}{2}AC = 7\ см;\ \ \ \]

\[DO = BO = \frac{1}{2}BD = 4\ см.\]

\[2)\ В\ \mathrm{\Delta}AOD:\]

\[= 49 + 16 - 2 \bullet 7 \bullet 4 \bullet \cos{\angle AOD} =\]

\[= 65 - 56 \bullet \cos{\angle AOD}\]

\[\cos{\angle AOD} = \frac{65 - AD^{2}}{56};\]

\[\angle BOA = 180{^\circ} - \cos{\angle AOD}\]

\[\cos{\angle BOA} = - \cos{\angle AOD} =\]

\[= \frac{AD^{2} - 65}{56}.\]

\[3)\ В\ \mathrm{\Delta}AOB:\]

\[= 49 + 16 - 2 \bullet 7 \bullet 4 \bullet \frac{AD^{2} - 65}{56}\]

\[(AD + 2)^{2} = 65 - AD^{2} + 65\]

\[AD^{2} + 4AD + 4 = 130 - AD^{2}\]

\[2AD^{2} + 4AD - 126 = 0\]

\[AD^{2} + 2AD - 63 = 0\]

\[D = 2^{2} + 4 \bullet 63 = 4 + 252 = 256\]

\[AD_{1} = \frac{- 2 - 16}{2} = - 9;\]

\[AD_{2} = \frac{- 2 + 16}{2} = 7\ см.\]

\[AB_{1} = - 9 + 2 = - 7;\]

\[AB_{2} = 7 + 2 = 9\ см.\]

\[Ответ:\ \ 7\ см;\ 9\ см.\]