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Category: Geometry

Question-40867

Question Number 40867 by ajfour last updated on 28/Jul/18 Commented by MrW3 last updated on 28/Jul/18 $${as}\:{shown}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\frac{{a}}{\mathrm{2}{R}} \\ $$$$\Rightarrow\theta=\mathrm{2}\:\mathrm{sin}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{R}} \\ $$$${but}\:{I}\:{think}\:{you}\:{meant}\:{something} \\ $$$${others}. \\…

Question-171765

Question Number 171765 by cherokeesay last updated on 20/Jun/22 Answered by som(math1967) last updated on 21/Jun/22 $${let}\:{OB}={OA}={OM}={r} \\ $$$$\:\therefore{AC}=\sqrt{{r}^{\mathrm{2}} +\frac{{r}^{\mathrm{2}} }{\mathrm{4}}}=\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\bigtriangleup{AOC}\sim\bigtriangleup{OLC} \\ $$$$\:\therefore\frac{{LC}}{{OC}}=\frac{\frac{{r}}{\mathrm{2}}}{\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}}}…

Question-40686

Question Number 40686 by MrW3 last updated on 26/Jul/18 Commented by MrW3 last updated on 26/Jul/18 $${Find}\:{radius}\:{R}\:{of}\:{the}\:{circumsphere}\:{of} \\ $$$${pyramid}. \\ $$$$ \\ $$$${I}\:{got}\:{R}=\frac{\sqrt{\mathrm{3}}{b}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}}…