\[\boxed{\mathbf{876.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Дано:\]
\[MNPQ - квадрат;\]
\[AC\ :BD = m\ :n;\]
\[S_{\text{ABCD}} = S_{\text{MNPQ}}.\]
\[Построить:\]
\[ромб\ \text{ABCD.}\]
\[Построение.\]
\[1)\ S_{\text{ABCD}} = \frac{1}{2}AC \bullet BD\ и\ S_{\text{MNPQ}} =\]
\[= \frac{1}{2}MQ \bullet PN \Longrightarrow AC \bullet BD =\]
\[= MQ \bullet PN.\]
\[MQ \bullet PN = (m \bullet n) \bullet y \Longrightarrow\]
\[\Longrightarrow m\sqrt{y} = AC\ и\ n\sqrt{y} = BD.\]
\[2)\ Построим\ произведение\ \text{MQ\ }\]
\[и\ \text{PN\ }при\ единичном\ отрезке\ e:\]
\[FA_{1} = FA_{2} = PN;\]
\[FE_{1} = - единичный\ отрезок;\ \]
\[FE - искомое\ произведение;\]
\[FE_{1} = e.\]
\[3)\ Произведение\ m \bullet n:\]
\[FM_{1} = m;\]
\[FN_{1} = n;\]
\[FE_{2} = m \bullet n.\]
\[4)\ \frac{\text{FE}}{FE_{2}} = FY = y.\]
\[5)\ Найдем\ квадратный\ корень\ \]
\[из\ y:\ \]
\[Y_{1}A_{2} = FY;\]
\[A_{2}E_{3} = FE_{1};\]
\[A_{2}Y\bot Y_{1}O;\]
\[A_{2}Y = \sqrt{\text{FY}}.\]
\[6)\ Построим\ произведение\ \]
\[AN \bullet A_{2}\text{Y\ }и\ AM \bullet A_{2}Y:\]
\[7)\ Таким\ образом:\]
\[BD = A_{2}D;AC = A_{2}\text{C.}\]
\[8)\ Построим\ ромб\ с\ данными\ \]
\[диагоналями:\]