Question Number 153580 by liberty last updated on 08/Sep/21 | ||
Answered by mr W last updated on 08/Sep/21 | ||
Commented by mr W last updated on 08/Sep/21 | ||
$$\frac{{a}−{R}}{{R}}=\frac{{R}}{{b}−{R}} \\ $$$$\Rightarrow\left({a}−{R}\right){b}={aR} \\ $$$$\frac{{r}}{{a}−{r}}=\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$${r}^{\mathrm{2}} {a}^{\mathrm{2}} =\left({a}^{\mathrm{2}} −\mathrm{2}{ar}\right){b}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} \left({a}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \left({a}^{\mathrm{2}} −\mathrm{2}{ar}\right) \\ $$$$\Rightarrow\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right){a}^{\mathrm{2}} −\mathrm{2}{rR}\left({R}−{r}\right){a}−{r}^{\mathrm{2}} {R}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}=\frac{{rR}\left[{R}−{r}+\sqrt{\mathrm{2}{R}\left({R}−{r}\right)}\right]}{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${b}=\frac{{R}}{\mathrm{1}−\frac{{R}}{{a}}}=\frac{{R}}{\mathrm{1}−\frac{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }{{r}\left[{R}−{r}+\sqrt{\mathrm{2}{R}\left({R}−{r}\right)}\right]}} \\ $$$${with}\:{R}=\mathrm{24},\:{r}=\mathrm{21}: \\ $$$$\Rightarrow{a}=\mathrm{56} \\ $$$$\Rightarrow{b}=\mathrm{42} \\ $$$${area}={ab}=\mathrm{2352} \\ $$ | ||
Commented by Tawa11 last updated on 08/Sep/21 | ||
$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$ | ||