Category: Geometry

Question Number 199672 by Mingma last updated on 07/Nov/23 Answered by mr W last updated on 07/Nov/23 Commented by mr W last updated on 07/Nov/23…

Question Number 199510 by universe last updated on 04/Nov/23 Answered by mr W last updated on 06/Nov/23 Commented by mr W last updated on 06/Nov/23…

Question Number 199413 by universe last updated on 03/Nov/23 Answered by ajfour last updated on 03/Nov/23 $$b\cos \alpha =R$$$${\{R\tan \alpha +\left(\frac{a}{b}\right)R\}}^2+{\left\{\left(\frac{a}{b}\right)R\tan \alpha \right\}}^2=R^2$$$$say\tan \alpha =t$$$$\Rightarrow\:{t}^{\mathrm{2}}…

Question Number 199358 by ajfour last updated on 01/Nov/23 Commented by ajfour last updated on 01/Nov/23 $$FindR.$$ Answered by mr W last updated…

Question Number 199286 by ajfour last updated on 31/Oct/23 Answered by ajfour last updated on 31/Oct/23 $$Letheightof//gm=h$$$$AB=1$$$$2r+\sqrt{1-4r^2}=p$$$$\frac{{r}}{\mathrm{2}}\left(\mathrm{1}+{p}+\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }\right)=\frac{{p}}{\mathrm{2}}…