Category: Geometry

Question Number 201322 by cherokeesay last updated on 04/Dec/23 Answered by AST last updated on 05/Dec/23 $${Let}\:\angle{EDC}=\beta\Rightarrow\angle{ABD}=\mathrm{2}\beta−\mathrm{90} \\ $$$${EC}^{\mathrm{2}} =\mathrm{2}−\mathrm{2}{cos}\beta…\left({i}\right) \\ $$$${BC}^{\mathrm{2}} ={BD}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{BD}\boldsymbol{{D}}{Ccos}\left(\mathrm{2}\beta\right)…\left({ii}\right) \\…

Question Number 201150 by Mingma last updated on 30/Nov/23 Answered by mr W last updated on 02/Dec/23 $${a}={side}\:{length}\:{of}\:{square} \\ $$$$\left(\frac{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }{\mathrm{2}{ax}}\right)^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{2}} +{x}^{\mathrm{2}}…

Question Number 201065 by mr W last updated on 28/Nov/23 Commented by mr W last updated on 28/Nov/23 $${find}\:{the}\:{side}\:{length}\:\boldsymbol{{a}}\:{of}\:{the}\:{square} \\ $$$${in}\:{terms}\:{of}\:\boldsymbol{{r}}. \\ $$ Answered by…

Question Number 201004 by Mingma last updated on 27/Nov/23 Answered by ajfour last updated on 27/Nov/23 Commented by Mingma last updated on 27/Nov/23 Very elegant, sir! Answered…

Question Number 200936 by Mingma last updated on 26/Nov/23 Commented by AST last updated on 26/Nov/23 $${Q}\mathrm{199738} \\ $$ Terms of Service Privacy Policy Contact:…