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Category: Geometry

Question-208493

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-208468

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-208465

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-208288

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}}…