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Question Number 163522 by nurtani last updated on 07/Jan/22

Answered by ajfour last updated on 08/Jan/22

$$sinceAisnotfixed,let\mathrm{\angle }C=90°.$$$$AC=3h,BC=6$$$$A_{△ABC}=9h$$$$eq.ofAD:\frac{x}{5}+\frac{y}{3h}=1$$$$eq.ofBH:\frac{x}{6}+\frac{y}{2h}=1$$$$\Rightarrow y_G(\frac{3}{h}-\frac{5}{3h})=1$$$$y_G=\frac{3h}{4}$$$$eq.ofAE:\frac{x}{3}+\frac{y}{3h}=1$$$$y_F(\frac{1}{h}-\frac{1}{3h})=1$$$$y_F=\frac{3h}{2}$$$$reqdratio=\frac{\frac{3}{2}\left(\frac{3h}{2}\right)-\frac{1}{2}\left(\frac{3h}{4}\right)}{9h}$$$$=\frac{15h}{8}\times \frac{1}{9h}=\frac{5}{24}$$

Commented by Tawa11 last updated on 09/Jan/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 07/Jan/22

Commented by Tawa11 last updated on 09/Jan/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by mr W last updated on 07/Jan/22

$$\frac{DP}{DC}=\frac{AH}{AC}=\frac{1}{3}$$$$DP=\frac{DC}{3}=\frac{1}{3}\times \frac{5BC}{6}=\frac{5BC}{18}=\frac{5\times 6BD}{18}=\frac{5BD}{3}$$$$\frac{BG}{GH}=\frac{BD}{DP}=\frac{3}{5}\Rightarrow \frac{BG}{BH}=\frac{3}{8}$$$$\frac{EQ}{EC}=\frac{AH}{AC}=\frac{1}{3}$$$$EQ=\frac{EC}{3}=\frac{BE}{3}$$$$\frac{FH}{BF}=\frac{EQ}{BE}=\frac{1}{3}\Rightarrow \frac{FH}{BH}=\frac{1}{4}=\frac{2}{8}$$$$\frac{GF}{BH}=\frac{8-3-2}{8}=\frac{3}{8}$$$$\left[AGF\right]=\frac{3}{8}\left[ABH\right]=\frac{3}{8}\times \frac{1}{3}\left[ABC\right]=\frac{\left[ABC\right]}{8}$$$$\left[ADB\right]=\frac{2}{6}\left[ABC\right]=\frac{\left[ABC\right]}{3}$$$$\left[DGFE\right]=\left[ADB\right]-\left[AGF\right]=\frac{\left[ABC\right]}{3}-\frac{\left[ABC\right]}{8}=\frac{5}{24}\left[ABC\right]$$$$\Rightarrow \frac{\left[DGFE\right]}{\left[ABC\right]}=\frac{5}{24}$$

Commented by nurtani last updated on 08/Jan/22

nice Mr W, thank you��

Answered by mr W last updated on 08/Jan/22

$$\underset{―}{\mathit{a}\mathit{l}\mathit{t}\mathit{e}\mathit{r}\mathit{n}\mathit{a}\mathit{t}\mathit{i}\mathit{v}\mathit{e}\mathit{m}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}}$$$$\frac{AG}{AD}=\frac{1\times 6}{1\times 6+2\times 1}=\frac{3}{4}$$$$\frac{AF}{AE}=\frac{1\times 6}{1\times 6+2\times 3}=\frac{1}{2}$$$$\left[AGF\right]=\frac{3}{4}\times \frac{1}{2}\times \left[ADE\right]=\frac{3}{8}\left[ADE\right]$$$$\left[DGFE\right]=\frac{5}{8}\left[ADE\right]=\frac{5}{8}\times \frac{\left[ABC\right]}{3}=\frac{5\left[ABC\right]}{24}$$$$\frac{\left[DGFE\right]}{\left[ABC\right]}=\frac{5}{24}$$

Commented by mr W last updated on 09/Jan/22

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