Question Number 209241 by mr W last updated on 05/Jul/24 Answered by A5T last updated on 05/Jul/24 Commented by A5T last updated on 05/Jul/24 $$\sqrt{{a}^{\mathrm{2}}…
Question Number 209137 by mr W last updated on 02/Jul/24 Answered by kapoorshah last updated on 02/Jul/24 $${Math}\:{Booster} \\ $$ Answered by cherokeesay last updated…
Question Number 209162 by dr1001sa last updated on 02/Jul/24 $$ \\ $$Cyclic quadrilateral ABCD is inscribed in circle. Point S is intersection of diagonals AC…
Question Number 208940 by efronzo1 last updated on 28/Jun/24 Answered by kapoorshah last updated on 28/Jun/24 $${r}\:=\:{OD}\:=\:\sqrt{\mathrm{10}} \\ $$$${BE}\:=\:\sqrt{\mathrm{15}} \\ $$$${BOE}\:\sim\:{BCA} \\ $$$$\frac{{BO}}{{BE}}\:=\:\frac{{BC}}{{AB}} \\ $$$$\frac{\sqrt{\mathrm{10}}}{\:\sqrt{\mathrm{15}}}\:=\:\frac{{BC}}{\mathrm{2}\sqrt{\mathrm{10}}}\:\:\Rightarrow\:{BC}\:=\:\frac{\mathrm{20}}{\:\sqrt{\mathrm{15}}}\:=\:\mathrm{5}.\mathrm{164}…
Question Number 208913 by efronzo1 last updated on 27/Jun/24 Answered by A5T last updated on 27/Jun/24 $$\frac{{sin}\beta}{\mathrm{6}}=\frac{\mathrm{1}}{{AC}};\frac{{sin}\left(\mathrm{60}−\beta\right)}{\mathrm{3}}=\frac{\mathrm{1}}{{AC}} \\ $$$$\Rightarrow\frac{{sin}\left(\mathrm{60}−\beta\right)}{{sin}\beta}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\beta\approx\mathrm{40}.\mathrm{8934} \\ $$$$\Rightarrow{sin}\beta\approx\mathrm{0}.\mathrm{654654};\:{we}\:{can}\:{then}\:{find}\:{AC}=\frac{\mathrm{6}}{{sin}\beta} \\ $$$${AC}\approx\mathrm{9}.\mathrm{165151} \\ $$…
Question Number 208915 by Tawa11 last updated on 27/Jun/24 Answered by mr W last updated on 27/Jun/24 Commented by mr W last updated on 27/Jun/24…
Question Number 208896 by efronzo1 last updated on 26/Jun/24 Answered by MM42 last updated on 27/Jun/24 $${s}_{\mathrm{1}} =\mathrm{32}−\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{64}−{x}^{\mathrm{2}} }{dx}\:\:\:\:\:;\:\:{x}=\mathrm{8}{sin}\theta \\ $$$$\Rightarrow=\mathrm{32}−\mathrm{64}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{cos}^{\mathrm{2}}…
Question Number 208828 by efronzo1 last updated on 24/Jun/24 Answered by mr W last updated on 24/Jun/24 $${BM}={MC}={a},\:{say} \\ $$$${AM}={DM}=\mathrm{2}{a} \\ $$$$\frac{{BM}}{\mathrm{6}}=\frac{{MQ}}{\mathrm{4}}\:\Rightarrow{MQ}=\frac{\mathrm{2}{a}}{\mathrm{3}}\:\Rightarrow{QD}=\frac{\mathrm{4}{a}}{\mathrm{3}} \\ $$$$\left(\mathrm{6}+\mathrm{4}\right)^{\mathrm{2}} ={a}^{\mathrm{2}}…
Question Number 208782 by cherokeesay last updated on 22/Jun/24 Answered by mr W last updated on 23/Jun/24 Commented by cherokeesay last updated on 23/Jun/24 $${thank}\:{you}\:{master}\:!…
Question Number 208772 by Jubr last updated on 22/Jun/24 Answered by mr W last updated on 22/Jun/24 Commented by mr W last updated on 22/Jun/24…