Question Number 215769 by momoga last updated on 17/Jan/25 In △ABC, it is given that AC⊥CB, CD⊥AB, and CD = 12, AC = BC +…
Question Number 215776 by cherokeesay last updated on 17/Jan/25 Answered by A5T last updated on 18/Jan/25 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{be}\:\mathrm{r} \\ $$$$\mathrm{AC}\parallel\mathrm{OB}\Rightarrow\mathrm{CB}=\sqrt{\mathrm{r}^{\mathrm{2}} +\left(\mathrm{r}−\mathrm{x}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2r}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} −\mathrm{2rx}} \\ $$$$\mathrm{Power}\:\mathrm{point}\:\mathrm{of}\:\mathrm{C}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circle}…
Question Number 215641 by cherokeesay last updated on 12/Jan/25 Answered by mr W last updated on 12/Jan/25 $${AC}=\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{5}} \\ $$$${DC}×{AC}=\mathrm{1}×\left(\mathrm{1}+\mathrm{1}+\mathrm{1}\right)\: \\ $$$$\Rightarrow{DC}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{5}}}=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{5}} \\…
Question Number 215454 by cherokeesay last updated on 07/Jan/25 Answered by mr W last updated on 07/Jan/25 Commented by mr W last updated on 15/Jan/25…
Question Number 215445 by BaliramKumar last updated on 07/Jan/25 Answered by A5T last updated on 07/Jan/25 $$\sqrt{\left(\mathrm{x}+\mathrm{r}\right)^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} }=\sqrt{\left(\mathrm{r}+\mathrm{r}\right)^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} }−\left(\mathrm{r}+\mathrm{x}\right) \\ $$$$\Rightarrow\sqrt{\mathrm{x}\left(\mathrm{x}+\mathrm{2r}\right)}=\mathrm{r}\sqrt{\mathrm{3}}−\mathrm{r}−\mathrm{x} \\ $$$$\Rightarrow\mathrm{4r}−\mathrm{2r}\sqrt{\mathrm{3}}−\mathrm{2x}\sqrt{\mathrm{3}}=\mathrm{0}…
Question Number 215467 by Tawa11 last updated on 07/Jan/25 Commented by MathematicalUser2357 last updated on 08/Jan/25 $${WOT}\:{is}\:{this} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 215400 by ajfour last updated on 05/Jan/25 Commented by ajfour last updated on 05/Jan/25 $${Correction}\:{I}\:{think}-\:\:{Find}\:{b},\:{given}\:{a}. \\ $$ Answered by mr W last updated…
Question Number 215340 by ppch145 last updated on 03/Jan/25 Commented by ppch145 last updated on 04/Jan/25 $$ \\ $$$$\boldsymbol{\mathrm{Yes}}!\:\mathrm{Saving}\:\mathrm{and}\:\mathrm{exporting}\: \\ $$$$\mathrm{ordinary}\:\mathrm{equations}\:\mathrm{always}\: \\ $$$$\mathrm{get}\:\mathrm{done}\:\mathrm{successfully}.\:\mathrm{In}\:\mathrm{fact},\: \\ $$$$\mathrm{some}\:\mathrm{of}\:\mathrm{my}\:\mathrm{drawings}\:\mathrm{get}\:…
Question Number 215313 by AlagaIbile last updated on 02/Jan/25 Answered by mr W last updated on 02/Jan/25 $$\angle{ABD}=\mathrm{135}° \\ $$$$\Rightarrow\angle{CBD}=\mathrm{45}° \\ $$$$\Rightarrow{CD}={CB}=\mathrm{2} \\ $$$$\Rightarrow{AD}=\sqrt{\left(\mathrm{12}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}}…
Question Number 215276 by ppch145 last updated on 02/Jan/25 $$\mathrm{I}\:\mathrm{produced}\:\mathrm{the}\:\mathrm{drawing}\: \\ $$$$\mathrm{shown}\:\mathrm{below}\:\left(\mathrm{Q}.\:\mathrm{215275}\right)\: \\ $$$$\mathrm{using}\:\mathrm{this}\:\mathrm{great}\:\mathrm{app},\:\mathrm{but}\: \\ $$$$\mathrm{unfortunately}\:\mathrm{it}\:\mathrm{refused}\:\mathrm{to}\: \\ $$$$\mathrm{get}\:\mathrm{saved}\:\mathrm{after}\:\mathrm{clicking}\: \\ $$$$“\mathrm{Save}''\:\mathrm{button}\:\mathrm{several}\:\mathrm{times}.\: \\ $$$$\mathrm{Also},\:\mathrm{the}\:“\mathrm{Export}\:\mathrm{As}\:\mathrm{Image}''\: \\ $$$$\mathrm{button}\:\mathrm{is}\:\mathrm{not}\:\mathrm{working}.\:\mathrm{Does}\: \\…