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Category: Geometry

Question-188965

Question Number 188965 by normans last updated on 09/Mar/23 Answered by mr W last updated on 09/Mar/23 $${say}\:{edge}\:{length}\:{of}\:{cube}\:{is}\:\mathrm{1}. \\ $$$${base}\:{of}\:{isosceles}\:{triangle}\:{is}\:\sqrt{\mathrm{2}}. \\ $$$${length}\:{of}\:{its}\:{legs}\:{is}\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}. \\…

find-the-equations-of-two-circles-which-thier-center-both-2-2-and-tangent-with-circle-x-2-y-2-8x-10y-5-0-

Question Number 123393 by malwan last updated on 25/Nov/20 $${find}\:{the}\:{equations}\:{of}\:\left[{two}\right]\:{circles} \\ $$$${which}\:{thier}\:{center}\:{both}\:\left(\mathrm{2},−\mathrm{2}\right) \\ $$$${and}\:{tangent}\:{with}\:{circle} \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:−\mathrm{8}{x}\:+\:\mathrm{10}{y}\:+\:\mathrm{5}\:=\:\mathrm{0} \\ $$ Commented by malwan last updated…

Question-188845

Question Number 188845 by ajfour last updated on 07/Mar/23 Answered by a.lgnaoui last updated on 07/Mar/23 $$\bigtriangleup{OAB}\:\:\:\left({recrangle}\:{en}\:{O}\right) \\ $$$${AB}={a}+\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\Rightarrow{AH}=\frac{{a}}{\mathrm{2}}+\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${OA}={OB}\:={R} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{4}}=\frac{{AH}}{{R}}\Rightarrow{R}=\frac{\mathrm{2}{AH}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{R}}=\mathrm{1}+\sqrt{\mathrm{2}}…

Question-188755

Question Number 188755 by normans last updated on 06/Mar/23 Commented by mr W last updated on 06/Mar/23 $${with}\:{given}\:{data}\:{the}\:{area}\:{of}\:\Delta{CFE} \\ $$$${is}\:{not}\:{uniquely}\:{defined}!\: \\ $$$${it}'{s}\:{dependent}\:{from}\:{the}\:{angle}\:\angle{A}. \\ $$ Commented…