Question Number 212760 by Spillover last updated on 23/Oct/24 Answered by mr W last updated on 23/Oct/24 Commented by mr W last updated on 23/Oct/24…
Question Number 212761 by Spillover last updated on 23/Oct/24 Commented by mr W last updated on 23/Oct/24 $${AB}=\mathrm{2}×{GE}=\mathrm{8}\sqrt{\mathrm{7}} \\ $$ Answered by som(math1967) last updated…
Question Number 212784 by Spillover last updated on 23/Oct/24 Answered by A5T last updated on 24/Oct/24 $${d}^{\mathrm{2}} =\left({r}−{a}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{r}\left({r}−{a}\right){cos}\theta \\ $$$${c}^{\mathrm{2}} =\left({r}−{a}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{r}\left({r}−{a}\right){cos}\theta…
Question Number 212783 by Spillover last updated on 23/Oct/24 Commented by York12 last updated on 25/Oct/24 $$ \\ $$$$ \\ $$$$ \\ $$ Answered by…
Question Number 212733 by cherokeesay last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 Commented by A5T last updated on 22/Oct/24 $$\left({R}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 212721 by Spillover last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 Commented by A5T last updated on 22/Oct/24 $${In}\left[{ABC}\right];\:{tan}\theta=\frac{{R}}{{R}+\mathrm{2}{r}};{AC}=\sqrt{\mathrm{2}{R}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}}…
Question Number 212720 by Spillover last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 $${AB}=\sqrt{\mathrm{2}\left(\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{6};\:{Let}\:\angle{OAB}=\theta \\ $$$$\frac{{sin}\theta}{{r}}=\frac{{sin}\left(\mathrm{180}−\mathrm{2}\theta\right)}{{AB}}\Rightarrow{cos}\theta=\frac{\mathrm{3}}{{r}}\Rightarrow{sin}\mathrm{2}\theta=\frac{\mathrm{6}\sqrt{{r}^{\mathrm{2}} −\mathrm{9}}}{{r}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}}…
Question Number 212723 by ajfour last updated on 22/Oct/24 Commented by ajfour last updated on 22/Oct/24 $${Find}\:\:{R}_{{min}} \\ $$ Answered by mr W last updated…
Question Number 212719 by Spillover last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 Commented by A5T last updated on 22/Oct/24 $${a}=\mathrm{2}\left(\frac{\pi}{\mathrm{6}}−\frac{{sin}\mathrm{60}°}{\mathrm{2}}\right)=\mathrm{2}\left(\frac{\pi}{\mathrm{6}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\…
Question Number 212690 by Spillover last updated on 21/Oct/24 Answered by mr W last updated on 21/Oct/24 $$\mathrm{10}×\left(\mathrm{2}{n}\right)=\left(\mathrm{2}\sqrt{\mathrm{15}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{n}=\mathrm{3} \\ $$$$\Rightarrow{R}={n}+\frac{\mathrm{10}}{\mathrm{2}}=\mathrm{8} \\ $$$${painted}\:{area}\:=\pi\left({R}^{\mathrm{2}} −{n}^{\mathrm{2}}…