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Category: Geometry

Question-41787

Question Number 41787 by Tawa1 last updated on 12/Aug/18 Answered by tanmay.chaudhury50@gmail.com last updated on 12/Aug/18 $${grey}\:{area}={ya}\:+\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right){a} \\ $$$$=\frac{\mathrm{2}{ya}+{xa}−{ya}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}\right){a} \\ $$ Terms of…

Question-41766

Question Number 41766 by ajfour last updated on 12/Aug/18 Answered by MrW3 last updated on 18/Aug/18 $${let}'{s}\:{say}\:\mathrm{0}<{b}\leqslant{a}\leqslant\mathrm{2}{R} \\ $$$$ \\ $$$${B}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${O}\left(\frac{{b}}{\mathrm{2}},\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}\right)…

Question-172723

Question Number 172723 by mnjuly1970 last updated on 30/Jun/22 Answered by mr W last updated on 30/Jun/22 $$\mathrm{tan}\:\alpha=\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}} \\ $$$$\left(\mathrm{3}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}−\mathrm{1}\right)\left(\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{5}{r}+\frac{{r}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}=\mathrm{4}…

Question-41620

Question Number 41620 by Tawa1 last updated on 10/Aug/18 Commented by Tawa1 last updated on 10/Aug/18 $$\mathrm{If}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{small}\:\mathrm{circles}\:\mathrm{in}\:\mathrm{the}\:\mathrm{diagram}\:\mathrm{has}\:\mathrm{the}\:\mathrm{radius}\:=\:\mathrm{r}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{large}\:\mathrm{circle}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{r} \\ $$ Answered by MrW3 last…