Category: Geometry

Question Number 219117 by Spillover last updated on 19/Apr/25 Answered by A5T last updated on 20/Apr/25 $${(\mathrm{r}+1)}^2=1^2+{(3-\mathrm{r}-\sqrt{3})}^2$$$$\Rightarrow 4\mathrm{r}=6+\mathrm{r}\sqrt{3}-3\sqrt{3}$$$$\mathrm{r}=\frac{6-3\sqrt{3}}{4-\sqrt{3}}$$…

Question Number 218956 by Spillover last updated on 17/Apr/25 Answered by mr W last updated on 17/Apr/25 Commented by mr W last updated on 18/Apr/25…

Question Number 218953 by Spillover last updated on 17/Apr/25 Answered by mr W last updated on 18/Apr/25 $$eqn.ofAC:$$$$\frac{x}{20}+\frac{y}{15}=1$$$$G(\frac{20}{3},\frac{15}{3})$$$${r}=\frac{\mid\frac{\mathrm{1}}{\mathrm{20}}×\frac{\mathrm{20}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{15}}×\frac{\mathrm{15}}{\mathrm{3}}−\mathrm{1}\mid}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{20}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{15}^{\mathrm{2}}…

Question Number 218890 by Spillover last updated on 17/Apr/25 Answered by mr W last updated on 17/Apr/25 $$\frac{1}{1}\times \frac{BF}{FD}\times \frac{3}{4}=1\Rightarrow \frac{BF}{FD}=\frac{4}{3}$$$$\Rightarrow {\mathrm{\Delta }}_{ADF}=\frac{3}{7}\times {\mathrm{\Delta }}_{ADB}=\frac{3}{7}\times \frac{3\times 3}{2}=\frac{27}{14}$$$$\left[{CDFE}\right]=\Delta_{{ACF}} −\Delta_{{ADF}}…

Question Number 218853 by Lekhraj last updated on 16/Apr/25 Answered by mr W last updated on 17/Apr/25 Commented by mr W last updated on 17/Apr/25…