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Question Number 218606 by Spillover last updated on 12/Apr/25

Answered by A5T last updated on 13/Apr/25

Commented by A5T last updated on 13/Apr/25

$$\mathrm{\angle }\mathrm{BAF}=\alpha \Rightarrow \mathrm{\angle }\mathrm{ABF}=\alpha \mathrm{since}\mathrm{BF}=\mathrm{AF}$$$$\mathrm{Similarly}\mathrm{\angle }\mathrm{CAF}=\mathrm{\angle }\mathrm{AFC}=\beta \mathrm{since}\mathrm{AF}=\mathrm{FC}$$$$\mathrm{\angle }\mathrm{ABC}+\mathrm{\angle }\mathrm{BCA}+\mathrm{\angle }\mathrm{CAB}=180°\Rightarrow 2(\alpha +\beta )=180°$$$$\Rightarrow \alpha +\beta =90°=\mathrm{\angle }\mathrm{BAC}$$$$\sin \mathrm{\angle }{\mathrm{BO}}_1{\mathrm{O}}_2=\frac{\sqrt{{(\mathrm{R}+\mathrm{r})}^2-{(\mathrm{R}-\mathrm{r})}^2}}{\mathrm{R}+\mathrm{r}}=\frac{2\sqrt{\mathrm{Rr}}}{\mathrm{R}+\mathrm{r}}$$$$\sin \mathrm{\angle }{\mathrm{AO}}_2\mathrm{C}=\frac{2\sqrt{\mathrm{Rr}}}{\mathrm{R}+\mathrm{r}}$$$$\left[\mathrm{ABC}\right]=\frac{2(\mathrm{R}+\mathrm{r})\sqrt{\mathrm{Rr}}}{2}-\frac{{\mathrm{R}}^2\sqrt{\mathrm{Rr}}}{\mathrm{R}+\mathrm{r}}-\frac{{\mathrm{r}}^2\sqrt{\mathrm{Rr}}}{\mathrm{R}+\mathrm{r}}$$$$\Rightarrow \left[\mathrm{ABC}\right]=\frac{[{(\mathrm{R}+\mathrm{r})}^2-{\mathrm{R}}^2-{\mathrm{r}}^2]\sqrt{\mathrm{Rr}}}{\mathrm{R}+\mathrm{r}}=\frac{2\mathrm{Rr}\sqrt{\mathrm{Rr}}}{\mathrm{R}+\mathrm{r}}$$

Commented by Spillover last updated on 13/Apr/25

$$correct$$

Answered by mr W last updated on 13/Apr/25

Commented by mr W last updated on 13/Apr/25

$$2\alpha +2\beta =180°$$$$\Rightarrow \alpha +\alpha =90°$$$$\Rightarrow \mathrm{\angle }BAC=90°$$$$BC=\sqrt{{(R_1-R_2)}^2-{(R_1-R_2)}^2}=2\sqrt{R_1{R}_2}$$$$h=R_2+\frac{R_2(R_1-R_2)}{R_1+R_2}=\frac{2R_1{R}_2}{R_1+R_2}$$$${\mathrm{\Delta }}_{ABC}=\frac{AB\times h}{2}=\frac{2\sqrt{R_1{R}_2}}{2}\times \frac{2R_1{R}_2}{R_1+R_2}$$$$=\frac{2R_1{R}_2\sqrt{R_1{R}_2}}{R_1+R_2}✓$$

Commented by Spillover last updated on 13/Apr/25

$$correct$$

Answered by Spillover last updated on 13/Apr/25

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