Question Number 219116 by Spillover last updated on 19/Apr/25 Answered by A5T last updated on 20/Apr/25 $$\mathrm{tan30}°=\frac{\mathrm{r}}{\mathrm{x}}\Rightarrow\mathrm{x}=\mathrm{r}\sqrt{\mathrm{3}} \\ $$$$\mathrm{k}=\mathrm{sin15}°=\frac{\mathrm{R}}{\left(\mathrm{2r}+\mathrm{2r}\sqrt{\mathrm{3}}\right)−\mathrm{R}} \\ $$$$\Rightarrow\mathrm{R}+\mathrm{Rk}=\mathrm{kr}\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\right)\Rightarrow\frac{\mathrm{r}}{\mathrm{R}}=\frac{\mathrm{1}+\mathrm{k}}{\mathrm{k}\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\right)} \\ $$$$\mathrm{sin30}°=\mathrm{2sin15}°\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{15}°} \\…
Question Number 219120 by Spillover last updated on 19/Apr/25 Answered by mr W last updated on 20/Apr/25 Commented by mr W last updated on 20/Apr/25…
Question Number 219118 by Spillover last updated on 19/Apr/25 Answered by A5T last updated on 20/Apr/25 $$\mathrm{R}^{\mathrm{2}} =\left(\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \Rightarrow\mathrm{R}=\frac{\mathrm{a}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{L}=\mathrm{R}−\frac{\mathrm{a}}{\mathrm{2}}=\frac{\mathrm{a}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{a}}{\mathrm{L}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\…
Question Number 219117 by Spillover last updated on 19/Apr/25 Answered by A5T last updated on 20/Apr/25 $$\left(\mathrm{r}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{r}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4r}=\mathrm{6}+\mathrm{r}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\mathrm{r}=\frac{\mathrm{6}−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}−\sqrt{\mathrm{3}}} \\ $$…
Question Number 219119 by Spillover last updated on 19/Apr/25 Answered by A5T last updated on 20/Apr/25 $$\mathrm{R}=\mathrm{6} \\ $$$$\mathrm{Area}=\mathrm{4}×\left[\mathrm{2}\left(\frac{\mathrm{6}^{\mathrm{2}} \pi}{\mathrm{4}}−\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}\right)\right]=\mathrm{8}\left[\frac{\mathrm{36}\pi−\mathrm{72}}{\mathrm{4}}\right]=\mathrm{72}\left(\pi−\mathrm{2}\right) \\ $$ Answered by…
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 218957 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 17/Apr/25…
Question Number 218953 by Spillover last updated on 17/Apr/25 Answered by mr W last updated on 18/Apr/25 $${eqn}.\:{of}\:{AC}: \\ $$$$\frac{{x}}{\mathrm{20}}+\frac{{y}}{\mathrm{15}}=\mathrm{1} \\ $$$${G}\left(\frac{\mathrm{20}}{\mathrm{3}},\:\frac{\mathrm{15}}{\mathrm{3}}\right) \\ $$$${r}=\frac{\mid\frac{\mathrm{1}}{\mathrm{20}}×\frac{\mathrm{20}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{15}}×\frac{\mathrm{15}}{\mathrm{3}}−\mathrm{1}\mid}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{20}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{15}^{\mathrm{2}}…
Question Number 218954 by Spillover last updated on 17/Apr/25 Commented by AntonCWX8 last updated on 17/Apr/25 $${All}\:{are}\:{from}\:{Facebook}… \\ $$ Answered by Spillover last updated on…
Question Number 218890 by Spillover last updated on 17/Apr/25 Answered by mr W last updated on 17/Apr/25 $$\frac{\mathrm{1}}{\mathrm{1}}×\frac{{BF}}{{FD}}×\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{1}\:\Rightarrow\frac{{BF}}{{FD}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow\Delta_{{ADF}} =\frac{\mathrm{3}}{\mathrm{7}}×\Delta_{{ADB}} =\frac{\mathrm{3}}{\mathrm{7}}×\frac{\mathrm{3}×\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{27}}{\mathrm{14}} \\ $$$$\left[{CDFE}\right]=\Delta_{{ACF}} −\Delta_{{ADF}}…