\[\boxed{\mathbf{607.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[P_{\text{ABCD}} = 42\ см;\]
\[BH = 4\ см;\]
\[BF = 5\ см.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ P_{\text{ABCD}} = 2AB + 2AD =\]
\[= 2(AB + AD) = 42\ см.\]
\[AB + AD = 21\ см \Longrightarrow AB =\]
\[= 21 - AD.\]
\[2)\ S_{\text{ABCD}} = BH \bullet AD = BF \bullet CD =\]
\[= BF \bullet AB.\]
\[4 \bullet AD = 5 \bullet (21 - AD)\]
\[4AD = 105 - 5AD;\]
\[9AD = 105\]
\[AD = \frac{105}{9} = 11\frac{6}{9} = 11\frac{2}{3}\ см.\]
\[3)\ S_{\text{ABCD}} = 4 \bullet 11\frac{2}{3} = 44\frac{8}{3} =\]
\[= 46\frac{2}{3}\ см^{2}.\]
\(\mathbf{Ответ:}46\frac{2}{3}\ см^{2}\mathbf{.}\)