\[\boxed{\mathbf{1396.еуроки - ответы\ на\ пятёрку}}\]
\[Дано:\]
\[ABCD - трапеция,\ \]
\[AD||BC;\ \]
\[AD < BC;\ \]
\[AD = 3;\ \]
\[AB\bot BC;\ \]
\[CD = 6;\]
\[E \in CD;\ \]
\[EC = ED;\ \]
\[\angle CBE = a.\]
\[Найти:\]
\[S - ?\]
\[Решение.\]
\[1)\ Опустим\ перпендикуляры\ \]
\[DH\bot BC,\ EK\bot BC,\ \left\{ H,K \right\} \in BC.\]
\[\left. \ \begin{matrix} \angle DHC = \angle EKC = 90{^\circ} \\ \angle C - общий\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right\} \Longrightarrow\]
\[\Longrightarrow \mathrm{\Delta}DHC\sim\mathrm{\Delta}EKC - по\ двум\ \]
\[углам.\]
\[Коэффициент\ подобия:\]
\[k = \frac{\text{CD}}{\text{CE}} = 2.\]
\[\ EK - средняя\ линия\ \mathrm{\Delta}DHC.\]
\[Обозначим:\ \]
\[HK = KC = a.\]
\[2)\ В\ \mathrm{\Delta}CEK:\]
\[EK = \sqrt{EC^{2} - KC^{2}} =\]
\[= \sqrt{3^{2} - a^{2}} = \sqrt{9 - a^{2}}.\]
\[3)\ Получаем:\]
\[\sqrt{9 - a^{2}} = (3 + a) \cdot tga\]
\[9 - a^{2} = (3 + a)^{2}tg^{2}a\]
\[9 - a^{2} = \left( 9 + 6a + a^{2} \right)tg^{2}a\]
\[a = \frac{- 6tg^{2}a \pm 6}{2\left( tg^{2}a + 1 \right)}\ (a > 0)\]
\[a = 3\frac{1 - tg^{2}a}{1 + tg^{2}a} =\]
\[= 3\frac{\frac{\cos^{2}{a - sin^{2}a}}{\cos^{2}a}}{\frac{\cos^{2}{a + \sin^{2}a}}{\cos^{2}a}} = 3cos2a.\]
\[EK = \sqrt{9 - a^{2}} =\]
\[= \sqrt{9 - 9\cos^{2}{2a}} = 3sin2a.\]
\[h = DH = 2EK = 6sin2a.\]
\[S = S_{\text{ADHB}} + S_{\text{DCN}} =\]
\[= h \cdot AD + \frac{1}{2}h \cdot HC =\]
\[= h\left( AD + \frac{\text{HC}}{2} \right) =\]
\[= 6sin2a(3 + 3cos2a) =\]
\[= 18sin2a( + cos2a).\]
\[Ответ:18 \cdot sin2a(1 + cos2a).\]