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МатематикаГДЗ по геометрии 9 класс Атанасян ФГОС Задание 476
Задание 476
\[\boxed{\mathbf{476.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\text{ABCD} - ромб;\]
\[\textbf{а)}\ \text{AC} = 14\ см;\]
\[\text{BD} = 3,2\ дм;\]
\[\textbf{б)}\ \text{AC} = 2\ дм;\]
\[\text{BD} = 4,6\ дм.\]
\[\mathbf{Доказать:}\]
\[S_{\text{ABCD}} = \frac{\text{BD} \bullet \text{AC}}{2}.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABCD}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ По\ свойству\ диагоналей\ \]
\[ромба:\]
\[\text{AO} = \text{OC};\]
\[\text{BO} = \text{OD};\]
\[\text{BD}\bot\text{AC}.\]
\[Треугольники\ \text{AOD},\text{DOC},\text{AOB},\]
\[\text{BOC} - прямоугольные.\]
\[2)\ ⊿\text{AOD} = ⊿\text{DOC} = ⊿\text{AOB} =\]
\[= ⊿\text{BOC}\ (по\ двум\ катетам).\]
\[3)\ S_{\text{AOD}} = \frac{1}{2} \bullet \frac{\text{BD}}{2} \bullet \frac{\text{AC}}{2} = \frac{\text{BD} \bullet \text{AC}}{8}.\]
\[4)\ S_{\text{ABCD}} =\]
\[= S_{\text{AOD}} + S_{\text{DOC}} + S_{\text{AOB}} + S_{\text{BOC}} =\]
\[= 4 \bullet S_{\text{AOD}} = 4 \bullet \frac{\text{BD} \bullet \text{AC}}{8} =\]
\[= \frac{\text{BD} \bullet \text{AC}}{2}.\]
\[Что\ и\ требовалось\ доказать.\]
\[\textbf{а)}\ S_{\text{ABCD}} = \frac{14 \bullet 32}{2} = 224\ см^{2}.\]
\[\textbf{б)}\ S_{\text{ABCD}} = \frac{2 \bullet 4,6}{2} = 4,6\ дм^{2}.\]
\[\boxed{\mathbf{476.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунки\ по\ условию\ задачи:\]
\[\textbf{а)} - г)\]
\[\textbf{д)}\ \]
\[\mathbf{Дано:}\]
\[\text{ABCD} - параллелограмм;\]
\[\textbf{а)}\ \angle A = 84{^\circ}\]
\[\textbf{б)}\ \angle A - \angle B = 55{^\circ}\]
\[\textbf{в)}\ \angle A + \angle C = 142{^\circ}\]
\[\textbf{г)}\ \angle A = 2\angle B\]
\[\textbf{д)}\ \angle CAD = 16{^\circ}\]
\[\angle ACD = 37{^\circ}.\]
\[\mathbf{Найти:}\ \]
\[\angle A,\angle B,\angle C,\angle D.\]
\[\mathbf{Решение.}\]
\[ABCD - праллелограмм:\]
\[\angle A = \angle C;\ \ \ \angle B = \angle D\ \]
\[\angle A + \angle B = 180{^\circ}\ \]
\[(как\ односторонние).\]
\[\textbf{а)}\ 1)\ \angle A = \angle C = 84{^\circ}.\]
\[2)\ \angle B = 180{^\circ} - 84{^\circ} = 96{^\circ};\ \ \]
\[\angle B = \angle D = 96{^\circ}.\]
\[\textbf{б)}\ 1)\ \angle A - \angle B = 55{^\circ}\]
\[\angle A = \angle B + 55{^\circ}.\]
\[2)\ \angle B + 55{^\circ} + \angle B = 180{^\circ}\]
\[2\angle B = 125\]
\[\angle B = 62{^\circ}30^{'}\]
\[\angle B = \angle D = 62{^\circ}30^{'}.\]
\[3)\ \angle A = 62{^\circ}30^{'} + 55{^\circ} = 117{^\circ}30^{'};\ \ \]
\[\ \angle A = \angle C = 117{^\circ}30^{'}.\]
\[\textbf{в)}\ 1)\ \angle A + \angle C = 142{^\circ}\]
\[2\angle A = 142{^\circ}\ \]
\[\angle A = 71{^\circ};\]
\[\angle A = \angle C = 71{^\circ}.\]
\[2)\ \angle B = 180{^\circ} - 71{^\circ} = 109{^\circ};\ \ \ \]
\[\angle B = \angle D = 109{^\circ}.\]
\[\textbf{г)}\ 1)\ 2\angle B + \angle B = 180{^\circ}\]
\[3\angle B = 180{^\circ}\]
\[\angle B = 60{^\circ}.\]
\[\angle B = \angle D = 60{^\circ}.\]
\[2)\ \angle A = 2\angle B = 2 \bullet 60{^\circ} = 120{^\circ};\ \]
\[\angle A = \angle C = 60{^\circ}.\]
\[\textbf{д)}\ 1)\ \angle ACD + \angle CDA + \angle DAC =\]
\[= 180{^\circ}\]
\[\angle CDA = 180 - (37 + 16) = 127\]
\[\angle D = \angle B = 127{^\circ}.\]
\[2)\ \angle A = 180{^\circ} - 127{^\circ} = 53{^\circ};\ \ \ \]
\[\angle A = \angle C = 53{^\circ}.\]
\[Ответ:а)\ \angle A = \angle C = 84{^\circ};\ \]
\[\angle B = \angle D = 96{^\circ};\]
\[\textbf{б)}\angle A = \angle C = 117{^\circ}30^{'};\ \]
\[\angle B = \angle D = 62{^\circ}30';\]
\[\textbf{в)}\angle A = \angle C = 71{^\circ};\]
\[\angle B = \angle D = 109{^\circ};\]
\[\textbf{г)}\ \angle A = \angle C = 120{^\circ};\]
\[\angle B = \angle D = 60{^\circ};\ \]
\[\ д)\angle A = \angle C = 53{^\circ};\]
\[\angle B = \angle D = 127{^\circ}.\]
