\[\boxed{\mathbf{1241.ОК\ ГДЗ - домашка\ н}а\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[PABCD - пирамида;\]
\[ABCD - параллелограмм;\]
\[AB = 5\ м;\]
\[AD = 4\ м;\]
\[BD = 3\ м;\]
\[H = AC \cap BD;\]
\[PH - высота;\]
\[PH = 2\ м.\]
\[\mathbf{Найти:}\]
\[S_{поверхности} - ?\]
\[\mathbf{Решение.}\]
\[1)\ \mathrm{\Delta}ADB - египетский\ \]
\[треугольник:\]
\[\angle ADB = 90{^\circ}.\]
\[2)\ S_{\text{ABCD}} = 2S_{\mathrm{\Delta}ADB} =\]
\[= 2 \cdot \frac{1}{2}AD \cdot BD = 4 \cdot 3 = 12\ м^{2}.\]
\[3)\ AD\bot DBP \Longrightarrow AD\bot PD:\]
\[\angle ADP = 90{^\circ}.\]
\[4)\ DP = \sqrt{DH^{2} + PH^{2}} =\]
\[= \sqrt{\left( \frac{3}{2} \right)^{2} + 2^{2}} = \sqrt{6,25} = 2,5\ м.\]
\[5)\ AP = \sqrt{AD^{2} + DP^{2}} =\]
\[= \sqrt{4^{2} + 2,^{2}} = \sqrt{22,25} = \sqrt{\frac{89}{4}} =\]
\[= \frac{\sqrt{89}}{2}\ м.\]
\[6)\ \mathrm{\Delta}APH = \mathrm{\Delta}CPH - по\ первому\ \]
\[признаку:\]
\[AH = HC;\]
\[PH - общая\ сторона;\]
\[\angle PHA = \angle PHC.\]
\[Отсюда:\ \]
\[AP = CP.\]
\[7)\ \mathrm{\Delta}ABPH = \mathrm{\Delta}DPH - по\ \]
\[первому\ признаку:\ \]
\[BH = HD;\]
\[PH - общая\ сторона;\]
\[\angle PHB = \angle PHD\]
\[Отсюда:\ \]
\[BP = DP.\]
\[8)\ Получаем:\]
\[\mathrm{\Delta}APD = \mathrm{\Delta}CPB - по\ трем\ \]
\[сторонам;\]
\[\mathrm{\Delta}APB = \mathrm{\Delta}CPD - по\ трем\ \]
\[сторонам.\]
\[9)\ S_{\mathrm{\Delta}APD} = \frac{1}{2}AD \cdot DP =\]
\[= \frac{1}{2} \cdot 4 \cdot 2,5 = 5\ \left( м^{2} \right).\]
\[10)\ В\ \mathrm{\Delta}APB:\ \]
\[cos\angle PAB = \frac{AP^{2} + AB^{2} - BP^{2}}{2AP \cdot AB} =\]
\[= \frac{\frac{89}{4} + 25 - \frac{25}{4}}{2 \cdot \frac{\sqrt{89}}{2} \cdot 5} = \frac{41}{5\sqrt{89}}\ .\]
\[11)\ sin\angle PAB =\]
\[= \sqrt{1 - \cos^{2}{\angle PAB}} =\]
\[= \sqrt{1 - \frac{41^{2}}{5^{2} \cdot 89}} = \frac{\sqrt{544}}{5\sqrt{89}} =\]
\[= \frac{4\sqrt{34}}{5\sqrt{89}}.\]
\[12)\ S_{\mathrm{\Delta}APB} =\]
\[= \frac{1}{2}AP \cdot AB \cdot sin\angle PAB =\]
\[= \frac{1}{2} \cdot \frac{\sqrt{89}}{2} \cdot 5 \cdot \frac{4\sqrt{34}}{5\sqrt{89}} = \sqrt{34}\ \left( м^{2} \right).\]
\[S_{пов} =\]
\[= S_{\text{ABCD}} + 2S_{\mathrm{\Delta}APD} + 2S_{\mathrm{\Delta}APB} =\]
\[= 12 + 2 \cdot 5 + 2 \cdot \sqrt{34} =\]
\[= 2 \cdot \left( 11 + \sqrt{34} \right)\ м^{2} \approx 33\frac{2}{3}\ м^{2}.\]
\[Ответ:\ \ S_{поверхности} \approx 33\frac{2}{3}\ м^{2}.\]