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Category: Geometry

Question-206615

Question Number 206615 by cortano21 last updated on 20/Apr/24 Answered by mr W last updated on 20/Apr/24 $$\frac{{R}−{r}}{{R}+{r}}=\mathrm{sin}\:\mathrm{22}.\mathrm{5}°=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}{R}\right){R}=\left(\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\right){r} \\ $$$$\frac{{r}}{{R}}=\frac{\mathrm{2}−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}\approx\mathrm{0}.\mathrm{4465} \\ $$ Terms…

Question-206292

Question Number 206292 by cortano21 last updated on 11/Apr/24 Answered by A5T last updated on 11/Apr/24 $${Let}\:{AE}={x};{BE}={y};{BF}={v};{FC}={w} \\ $$$${S}+\mathrm{39}=\frac{\left(\mathrm{2}{v}+{w}\right)\left({x}+{y}\right)}{\mathrm{2}}=\frac{\mathrm{2}{v}\left({x}+{y}\right)}{\mathrm{2}}+\mathrm{15}\Rightarrow{S}={vx} \\ $$$${wx}=\mathrm{54}−{S}=\mathrm{30}−{yw}\Rightarrow{yw}={S}−\mathrm{24} \\ $$$$\frac{\left[{BFD}\right]}{\left[{DFC}\right]}=\frac{{v}}{{w}}\Rightarrow\left[{BFD}\right]=\frac{\mathrm{15}{v}}{{w}} \\ $$$$\frac{\left[{EDB}\right]}{\left[{ADE}\left[\right.\right.}=\frac{{y}}{{x}}\Rightarrow{EDB}=\frac{\mathrm{27}{y}}{{x}}…