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Question Number 161257 by amin96 last updated on 15/Dec/21

Commented by amin96 last updated on 15/Dec/21

$$yes.solution?$$

Answered by som(math1967) last updated on 15/Dec/21

$$let\mathrm{\angle }ACE=\mathrm{\angle }ABO=\theta$$$$\therefore \mathrm{\angle }FCL=90-\theta$$$$radiusofsemicircle=r$$$$tan\theta =\frac{6}{8}=\frac{3}{4}$$$$from△AEC$$$$AE=rtan\theta$$$$\therefore ar△AEC\left(blue\right)=\frac{1}{2}\times r\times rtan\theta$$$$=\frac{r^{2}tan\theta }{2}$$$$from△CFD$$$$FL=rsin(90-\theta )=rcos\theta$$$$CL=rcos(90-\theta )=rsin\theta$$$$ar△CFD\left(Yellow\right)=r^{2}sin\theta cos\theta$$$$\therefore \frac{\mathit{B}\mathit{L}\mathit{U}\mathit{E}}{\mathit{Y}\mathit{E}\mathit{L}\mathit{L}\mathit{O}\mathit{W}}=\frac{{\mathit{r}}^2\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}{2{\mathit{r}}^2\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }}=\frac{{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{\theta }}{2}$$$$\frac{1}{2}\times \frac{25}{16}=\frac{25}{32}$$$$$$

Commented by som(math1967) last updated on 15/Dec/21

Commented by Tawa11 last updated on 15/Dec/21

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

Answered by Rasheed.Sindhi last updated on 15/Dec/21

Commented by Rasheed.Sindhi last updated on 15/Dec/21

$$\mathbb{W}\mathrm{ithout}\mathbb{R}\mathrm{eferencing}\mathbb{A}\mathrm{ny}\mathbb{A}\mathrm{ngle}$$$$\bullet \mathrm{A}\mathrm{\&}\mathrm{E}\mathrm{are}\mathrm{y}\mathrm{\&}\mathrm{x}\mathrm{intercepts}\mathrm{of}\mathrm{y}=-\frac{3}{4}\mathrm{x}+6$$$$\mathrm{A}=(0,-\frac{3}{4}\cdot 0+6)=(0,6)$$$$\mathrm{OA}=6$$$$\mathrm{B}=(\mathrm{x},0)=(8,0)[\because 0=-\frac{3}{4}\mathrm{x}+6\Rightarrow \mathrm{x}=8]$$$$\mathrm{OE}=8$$$$\mathrm{AE}=\sqrt{{\mathrm{OA}}^2+{\mathrm{OE}}^2}=\sqrt{6^2+8^2}=10$$$$\bullet \mathrm{B}=(\mathrm{r},\mathrm{r}):\mathrm{y}=-\frac{3}{4}\mathrm{x}+6\Rightarrow \mathrm{r}=-\frac{3}{4}\mathrm{r}+6$$$$\Rightarrow \mathrm{r}=\frac{24}{7}$$$$\bullet ▴\mathrm{AFB}:$$$$\Rightarrow \mathrm{r}=\frac{24}{7}\Rightarrow \mathrm{BF}=\frac{24}{7}$$$$\mathrm{AF}=6-\mathrm{r}=6-\frac{24}{7}=\frac{18}{7}$$$$▴\mathrm{AFB}=\frac{1}{2}\cdot \mathrm{BF}\cdot \mathrm{AF}=\frac{1}{2}\cdot \frac{24}{7}\cdot \frac{18}{7}=\frac{216}{49}$$$$\bullet △\mathrm{AOE}\sim △\mathrm{GCB}$$$$\frac{\mathrm{AE}}{\mathrm{BG}}=\frac{\mathrm{OA}}{\mathrm{BC}}\Rightarrow \frac{10}{\frac{24}{7}}=\frac{6}{\mathrm{BC}}\Rightarrow \mathrm{BC}=\frac{\frac{24}{7}\cdot 6}{10}$$$$=\frac{24}{7}\cdot 6\cdot \frac{1}{10}=\frac{72}{35}\Rightarrow \mathrm{BC}=\frac{72}{35}$$$$\mathrm{GC}=\sqrt{{\mathrm{BG}}^2-{\mathrm{BC}}^2}=\sqrt{{\left(\frac{24}{7}\right)}^2-{\left(\frac{72}{35}\right)}^2}$$$$=\frac{96}{35}$$$$\bullet ▴\mathrm{BGC}=\frac{1}{2}\cdot \mathrm{BC}\cdot \mathrm{GC}=\frac{1}{2}\cdot \frac{72}{35}\cdot \frac{96}{35}$$$$▴\mathrm{BGD}=2\cdot ▴\mathrm{BGC}=\frac{72\cdot 96}{{35}^2}$$$$\bullet \frac{Blue}{Yellow}=\frac{▴\mathrm{AFB}}{▴\mathrm{BGD}}=\frac{\frac{216}{49}}{\frac{72\cdot 96}{{35}^2}}=\frac{216}{49}\cdot \frac{{35}^2}{72\cdot 96}$$$$=\frac{25}{32}$$

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