ГДЗ по геометрии 8 класс Атанасян ФГОС Задание 617

\[\boxed{\mathbf{617}\mathbf{.}\mathbf{еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - трапеция;\]

\[BC = a;AD = b;\]

\[MN \parallel AD;\]

\[S_{\text{AMND}} = S_{\text{MBCN}}.\]

\[\mathbf{Найти:}\]

\[MN - ?\]

\[\mathbf{Решение.}\]

\[1)\ MN = x.\]

\[(b + x)BH = (a + b + 2x)\text{BF.}\]

\[4)\ S_{\text{ABCD}} = S_{\text{AMND}} + S_{\text{MBCN}}\]

\[aBH - xBH = aBF - bBF;\]

\[(a - x)BH = (a - b)\text{BF.}\]

\[5)\ \left\{ \begin{matrix} (b + x)BH = (a + b + 2x)\text{BF} \\ (a - x)BH = (a - b)\text{BF\ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} BH = \frac{a + b + 2x}{b + x} \bullet BF \\ BH = \frac{a - b}{a - x} \bullet BF\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[6)\ (a - x)(a + b + 2x) =\]

\[= (b + x)(a - b)\]

\[a^{2} + ab + 2ax - ax - bx - 2x^{2} =\]

\[= ba - b^{2} + ax - bx\]

\[a^{2} + 2ax - ax - 2x^{2} + b^{2} - ax =\]

\[= 0\]

\[a^{2} - 2x^{2} + b^{2} = 0\]

\[2x^{2} = a^{2} + b^{2}\]

\[x^{2} = \frac{a^{2} + b^{2}}{2}\]

\[x = \frac{\sqrt{a^{2} + b^{2}}}{2}.\]

\[Ответ:MN = \frac{\sqrt{a^{2} + b^{2}}}{2}.\]