Question Number 208743 by Tawa11 last updated on 22/Jun/24 Commented by Tawa11 last updated on 22/Jun/24 $$\mathrm{Area}\:\mathrm{of}\:\mathrm{red}\:\mathrm{circle}. \\ $$ Commented by Tawa11 last updated on…
Question Number 208690 by Tawa11 last updated on 21/Jun/24 Answered by mr W last updated on 21/Jun/24 $$\left(\mathrm{2}{r}−\mathrm{3}\right)×\mathrm{3}=\mathrm{2}^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$${shaded}\:{area}=\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{13}}{\mathrm{6}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{13}}{\mathrm{6}}\right)×\mathrm{2}\approx\mathrm{3}.\mathrm{04} \\ $$…
Question Number 208639 by Tawa11 last updated on 20/Jun/24 Commented by mr W last updated on 20/Jun/24 $${is}\:{it}\:{a}\:{red}\:{square}? \\ $$$${does}\:{one}\:{corner}\:{of}\:{the}\:{square}\:{lie} \\ $$$${on}\:{the}\:{center}\:{of}\:{semicircle}? \\ $$ Commented…
Question Number 208581 by Tawa11 last updated on 18/Jun/24 Commented by mr W last updated on 18/Jun/24 Commented by Tawa11 last updated on 18/Jun/24 $$\mathrm{Thanks}\:\mathrm{sir}.…
Question Number 208493 by Tawa11 last updated on 17/Jun/24 Answered by A5T last updated on 17/Jun/24 $${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…