Question Number 209398 by SonGoku last updated on 09/Jul/24 $$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{determine}\:\mathrm{the}\:\mathrm{points}\:\mathrm{A}\left(\mathrm{x}_{\mathrm{1}} ,\:\mathrm{y}_{\mathrm{1}} \right)\:\mathrm{and} \\ $$$$\mathrm{B}\left(\mathrm{x}_{\mathrm{2}} ,\:\mathrm{y}_{\mathrm{2}} \right)\:\mathrm{knowing}\:\mathrm{that}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{them}\:\mathrm{is} \\ $$$$\mathrm{2}\sqrt{\mathrm{29}}? \\ $$ Answered by Frix last updated…
Question Number 209389 by a.lgnaoui last updated on 08/Jul/24 $$\mathrm{Cercle}\:\mathrm{C}\:\:\:\mathrm{de}\:\mathrm{rayon}\:\boldsymbol{\mathrm{R}}=\mathrm{5} \\ $$$$\mathrm{petits}\:\mathrm{cercles}\:\mathrm{de}\:\mathrm{meme}\:\mathrm{rayon}\:\boldsymbol{\mathrm{r}} \\ $$$$\mathrm{Determiner}\:\mathrm{Surface}\:\left(\boldsymbol{\mathrm{ABCDEF}}\right)? \\ $$ Commented by a.lgnaoui last updated on 08/Jul/24 Terms of…
Question Number 209307 by efronzo1 last updated on 06/Jul/24 Commented by mr W last updated on 06/Jul/24 $$\mid{QR}\mid<\frac{\mathrm{8}}{\mathrm{3}}−\mathrm{1}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\frac{{QR}}{{PR}}=\frac{\mathrm{2}}{\mathrm{1}}\:\Rightarrow\frac{{PQ}}{{QR}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mid{PQ}\mid=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}×{QR}<\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}×\frac{\mathrm{5}}{\mathrm{3}}=\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{6}}<\sqrt{\mathrm{5}} \\ $$$$\Rightarrow{impossible}\:{that}\:\mid{PQ}\mid=\sqrt{\mathrm{5}}\:! \\…
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Question Number 208327 by efronzo1 last updated on 12/Jun/24 Answered by A5T last updated on 12/Jun/24 $${DE}=\mathrm{5}{x},{DF}=\mathrm{12}{x} \\ $$$${Let}\:{the}\:{perpendicular}\:{from}\:{A}\:{to}\:{BC}\:{meet}\:{it}\:{at}\:{H}; \\ $$$${AH}×{BC}=\mathrm{3}×\mathrm{4}=\mathrm{12}\Rightarrow{AH}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\Rightarrow{BH}=\sqrt{\mathrm{9}−\frac{\mathrm{144}}{\mathrm{25}}}=\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\frac{{AD}}{{AB}}=\frac{{DE}}{{BC}}={x}\Rightarrow{AD}=\mathrm{3}{x}\Rightarrow{AE}=\mathrm{4}{x}…