Question Number 205861 by mr W last updated on 01/Apr/24 Answered by A5T last updated on 01/Apr/24 $$\mathrm{9}×\mathrm{9}=\left(\mathrm{2}{r}−{x}\right){x}\Rightarrow\mathrm{2}{r}−{x}=\frac{\mathrm{81}}{{x}} \\ $$$$\mathrm{21}×\mathrm{21}=\left(\mathrm{2}{r}−\mathrm{16}−{x}\right)\left(\mathrm{16}+{x}\right)=\left(\frac{\mathrm{81}}{{x}}−\mathrm{16}\right)\left(\mathrm{16}+{x}\right) \\ $$$$\Rightarrow{x}=\mathrm{2}\Rightarrow{r}=\mathrm{21}.\mathrm{25} \\ $$ Commented…
Question Number 205734 by mr W last updated on 28/Mar/24 Answered by A5T last updated on 28/Mar/24 Commented by A5T last updated on 28/Mar/24 $$\frac{\mathrm{48}×\mathrm{144}×\mathrm{13}}{\mathrm{4}{R}}=\frac{\mathrm{48}×\mathrm{36}}{\mathrm{2}}\Rightarrow{R}=\mathrm{26}…
Question Number 205660 by marie last updated on 26/Mar/24 Answered by Skabetix last updated on 26/Mar/24 $$\mathrm{E}{xercice}\:{n}°\mathrm{4} \\ $$$$\left.\mathrm{1}\right)\:{D}'{apres}\:{le}\:{theoreme}\:{de}\:{Pythagore}\:: \\ $$$${BC}^{\mathrm{2}} ={AC}^{\mathrm{2}} +{AB}^{\mathrm{2}} \\ $$$$\Leftrightarrow{BC}^{\mathrm{2}}…
Question Number 205657 by mnjuly1970 last updated on 26/Mar/24 Answered by mr W last updated on 26/Mar/24 $${small}\:{square}:\:{a}×{a}=\mathrm{6}×\mathrm{6} \\ $$$${big}\:{square}:\:{b}×{b} \\ $$$${area}\:{of}\:{triangle}: \\ $$$${A}_{\mathrm{1}} +{A}_{\mathrm{2}}…
Question Number 205639 by mr W last updated on 26/Mar/24 Answered by mahdipoor last updated on 26/Mar/24 $${Trapezium}−\mathrm{2}\:{Right}\:{triangle}={area}\:\Rightarrow \\ $$$$\left[\frac{\left({r}\right)+\left(\mathrm{4}+{r}\right)}{\mathrm{2}}×\mathrm{4}\right]−\left[\frac{\left(\mathrm{4}+{r}\right)\left({r}\right)}{\mathrm{2}}+\frac{\left(\mathrm{4}−{r}\right)\left({r}\right)}{\mathrm{2}}\right]=\mathrm{8} \\ $$ Commented by mr…
Question Number 205614 by mr W last updated on 25/Mar/24 Answered by A5T last updated on 25/Mar/24 Commented by A5T last updated on 25/Mar/24 $$\frac{{sin}\left(\angle{CFB}\right)}{\mathrm{1}}=\frac{{sin}\mathrm{90}=\mathrm{1}}{\:\sqrt{\mathrm{0}.\mathrm{5}^{\mathrm{2}}…
Question Number 205562 by mr W last updated on 24/Mar/24 Answered by A5T last updated on 24/Mar/24 Commented by A5T last updated on 24/Mar/24 $${CE}=\sqrt{\mathrm{2}{r}^{\mathrm{2}}…
Question Number 205530 by cherokeesay last updated on 23/Mar/24 Answered by mr W last updated on 24/Mar/24 $${P}\left({a}\:\mathrm{cos}\:\theta,\:{b}\:\mathrm{sin}\:\theta\right) \\ $$$$\mathrm{tan}\:\varphi=−\frac{{dy}}{{dx}}=\frac{{b}}{{a}\:\mathrm{tan}\:\theta} \\ $$$${r}={a}\:\mathrm{cos}\:\theta−{r}\:\mathrm{sin}\:\varphi\:\Rightarrow{r}\:\mathrm{sin}\:\varphi={a}\:\mathrm{cos}\:\theta−{r} \\ $$$${r}={b}\:\mathrm{sin}\:\theta−{r}\:\mathrm{cos}\:\varphi\:\Rightarrow{r}\:\mathrm{cos}\:\varphi={b}\:\mathrm{sin}\:\theta−{r} \\…
Question Number 205507 by mr W last updated on 23/Mar/24 Commented by mr W last updated on 23/Mar/24 $${an}\:{unsolved}\:{old}\:{question} \\ $$ Answered by mr W…
Question Number 205428 by mr W last updated on 21/Mar/24 Answered by Rasheed.Sindhi last updated on 21/Mar/24 Commented by Rasheed.Sindhi last updated on 21/Mar/24 $${White}\:{area}\:{of}\:\Box{ABCD}\:{is}\:{composed}…