Question Number 208493 by Tawa11 last updated on 17/Jun/24 Answered by A5T last updated on 17/Jun/24 $${a}\left(\mathrm{3}{a}\right)={x}^{\mathrm{2}} \Rightarrow{x}={a}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{4}{a}\right)^{\mathrm{2}} =\mathrm{2}\left({a}\sqrt{\mathrm{3}}+{y}\right)^{\mathrm{2}} =\mathrm{16}{a}^{\mathrm{2}} \Rightarrow{r}={y}=\mathrm{2}{a}\sqrt{\mathrm{2}}−{a}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{r}={y}={a}\left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)…
Question Number 208468 by Tawa11 last updated on 16/Jun/24 Answered by A5T last updated on 17/Jun/24 $$\frac{{OD}×{OC}}{{AO}×{OB}}=\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\mathrm{9}}{\mathrm{25}}\Rightarrow{k}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\frac{{AB}}{{DC}}=\frac{\mathrm{5}}{\mathrm{3}};\:{AB}=\mathrm{5}{x},{DC}=\mathrm{3}{x} \\ $$$$\left[{ABCD}\right]=\frac{\left(\mathrm{8}{x}\right){h}}{\mathrm{2}}=\mathrm{4}{xh} \\ $$$$\left[{OBC}\right]=\left[{OAD}\right]=\frac{\mathrm{5}{xh}}{\mathrm{2}}−\mathrm{25}=\frac{\mathrm{3}{xh}}{\mathrm{2}}−\mathrm{9}\Rightarrow{xh}=\mathrm{16} \\…
Question Number 208466 by Tawa11 last updated on 16/Jun/24 Answered by A5T last updated on 17/Jun/24 Commented by A5T last updated on 17/Jun/24 $$\mathrm{4}{x}+\mathrm{3}{x}+\mathrm{2}{x}=\mathrm{180}\Rightarrow{x}=\mathrm{20}° \\…
Question Number 208465 by Tawa11 last updated on 16/Jun/24 Answered by A5T last updated on 17/Jun/24 $$\frac{\mathrm{3}\sqrt{\mathrm{2}}}{{s}}=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{10}}\Rightarrow{s}=\frac{\mathrm{10}\sqrt{\mathrm{10}}}{\mathrm{5}}=\mathrm{2}\sqrt{\mathrm{10}}\Rightarrow{s}^{\mathrm{2}} =\mathrm{40} \\ $$ Commented by Tawa11 last updated…
Question Number 208467 by Tawa11 last updated on 16/Jun/24 Answered by A5T last updated on 17/Jun/24 Commented by A5T last updated on 17/Jun/24 $${sin}\theta=\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\:\mathrm{10}}\Rightarrow{cos}\theta=\frac{\sqrt{\mathrm{10}}}{\mathrm{10}} \\…
Question Number 208455 by efronzo1 last updated on 16/Jun/24 Commented by mr W last updated on 16/Jun/24 $${the}\:{green}\:{one}\:{even}\:{doesn}'{t}\:{need}\:{to}\:{be}\: \\ $$$${a}\:{semicircle}. \\ $$ Commented by mr…
Question Number 208385 by efronzo1 last updated on 14/Jun/24 Commented by efronzo1 last updated on 15/Jun/24 Commented by efronzo1 last updated on 15/Jun/24 $$\mathrm{i}'\mathrm{m}\:\mathrm{stuck}\:\mathrm{this}\:\mathrm{step} \\…
Question Number 208344 by mr W last updated on 13/Jun/24 Commented by mr W last updated on 13/Jun/24 $${find}\:{green}\:{area}=? \\ $$ Commented by naka3546 last…
Question Number 208288 by efronzo1 last updated on 10/Jun/24 Answered by A5T last updated on 10/Jun/24 $$\mathrm{8}\left(\mathrm{8}+{x}\right)=\left(\mathrm{2}{r}\right)\left(\mathrm{2}{r}+\mathrm{2}{R}\right)=\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}{rR} \\ $$$$\Rightarrow{x}=\frac{{r}^{\mathrm{2}} +{rR}−\mathrm{16}}{\mathrm{2}}…\left({i}\right) \\ $$$$\left(\mathrm{8}+{x}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +\left(\mathrm{2}{r}+{R}\right)^{\mathrm{2}}…
Question Number 208158 by necx122 last updated on 06/Jun/24 $${X},\:{Y}\:{and}\:{Z}\:{are}\:{points}\:{on}\:{the}\:{sides}\:{AB}, \\ $$$${BC}\:{and}\:{AC}\:{of}\:{the}\:{triangle}\:{ABC},\:{such} \\ $$$${that}\:{AX}:{XB}\:=\mathrm{4}:\mathrm{3},\:{BY}:{YC}=\mathrm{2}:\mathrm{3},\: \\ $$$${CZ}:{ZA}=\mathrm{2}:\mathrm{1}.\:{Find}\:{the}\:{ratio}\:{of}\:{the}\:{area} \\ $$$${of}\:{the}\:{triangle}\:{XYZ}\:{to}\:{that}\:{of}\:{triangle} \\ $$$${ABC}. \\ $$ Answered by mr…