Category: Geometry

Question Number 208021 by efronzo1 last updated on 02/Jun/24 Answered by A5T last updated on 02/Jun/24 $$\frac{{CD}}{{CD}+{r}}=\frac{{r}}{{r}+{AB}}\Rightarrow\frac{{r}+{AB}}{{r}+{CD}}=\frac{{r}}{{CD}} \\ $$$$\frac{{r}+{AB}}{\mathrm{8}+{r}}=\frac{{r}+{CD}}{\mathrm{3}+{r}}\Rightarrow\frac{{r}+{AB}}{{r}+{CD}}=\frac{\mathrm{8}+{r}}{\mathrm{3}+{r}}=\frac{{r}}{{CD}} \\ $$$$\Rightarrow{CD}=\frac{{r}\left(\mathrm{3}+{r}\right)}{\mathrm{8}+{r}} \\ $$$$\frac{{AB}}{{AB}+{r}}=\frac{{r}}{{CD}+{r}}={AB}\left({CD}+{r}\right)={r}\left({AB}\right)+{r}^{\mathrm{2}} \\ $$$${AB}=\frac{{r}^{\mathrm{2}}…

Question Number 208011 by necx122 last updated on 02/Jun/24 Commented by necx122 last updated on 02/Jun/24 $${If}\:{the}\:{ratio}\:{of}\:<{ARS}\::\:<{BRT}\:=\:\mathrm{9}:\mathrm{4}, \\ $$$${find}\:{the}\:{ratio}\:{of}\:{area}\:{ABC}\::\:{area}\:{ARS} \\ $$ Commented by A5T last…

Question Number 207980 by mnjuly1970 last updated on 01/Jun/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{In}\:\:\:{A}\overset{\Delta} {{B}C}\::\:\:{B}=\:\mathrm{90}^{\:\mathrm{o}} \: \\ $$$$\:\mathrm{BB}'\:\:\bot\:\mathrm{CC}'\:\left(\:\mathrm{BB}'\:{and}\:\mathrm{CC}'\:{are}\:{medians}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\frac{\boldsymbol{{m}}_{\boldsymbol{{c}}} }{\boldsymbol{{m}}_{\boldsymbol{{a}}} \:}\:=\:? \\ $$$$\boldsymbol{{note}}:\:\:\mid\:\mathrm{CC}'\:\mid\:=\:\boldsymbol{{m}}_{\boldsymbol{{c}}} \:\:,\:\mid\mathrm{AA}'\mid=\:\boldsymbol{{m}}_{\boldsymbol{{a}}}…

Question Number 207946 by cherokeesay last updated on 31/May/24 Answered by mr W last updated on 01/Jun/24 Commented by mr W last updated on 01/Jun/24…

Question Number 207764 by mr W last updated on 25/May/24 Answered by som(math1967) last updated on 25/May/24 $$\:{R}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{2}^{\mathrm{2}} +{R}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{R}^{\mathrm{2}} =\mathrm{16}−\mathrm{8}…