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 Number 206554 by cortano21 last updated on 18/Apr/24 Answered by starsouf last updated on 18/Apr/24 $$ \\ $$ Answered by mr W last updated…
Question Number 206572 by mr W last updated on 18/Apr/24 Commented by mr W last updated on 18/Apr/24 $${radius}\:{of}\:{circles}\:{is}\:\mathrm{1}. \\ $$$${find}\:{AB}=? \\ $$ Answered by…
Question Number 206571 by mr W last updated on 19/Apr/24 Answered by cortano21 last updated on 19/Apr/24 $$\:\:\underline{\:} \\ $$ Commented by mr W last…
Question Number 206491 by cortano21 last updated on 16/Apr/24 Commented by cortano21 last updated on 16/Apr/24 $$\:\:\underbrace{\:} \\ $$ Answered by mr W last updated…
Question Number 206466 by cortano21 last updated on 15/Apr/24 Commented by cortano21 last updated on 15/Apr/24 $$\:\: \\ $$ Answered by mr W last updated…
Question Number 206430 by cortano21 last updated on 14/Apr/24 Answered by mr W last updated on 14/Apr/24 $${AB}={a},\:{say} \\ $$$$\mathrm{7}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} −\mathrm{2}×\mathrm{8}{a}\:\mathrm{cos}\:\mathrm{60}° \\ $$$${a}^{\mathrm{2}}…
Question Number 206396 by cortano21 last updated on 13/Apr/24 Answered by mr W last updated on 13/Apr/24 Commented by mr W last updated on 13/Apr/24…
Question Number 206321 by cortano21 last updated on 12/Apr/24 Answered by TonyCWX08 last updated on 12/Apr/24 $${By}\:\Delta{BCD}\:\backsim\:\Delta{CDE} \\ $$$$\frac{{x}}{\:\sqrt{\mathrm{14}}}=\frac{\mathrm{7}\sqrt{\mathrm{14}}}{{x}} \\ $$$${x}^{\mathrm{2}} =\mathrm{98} \\ $$$$ \\…
Question Number 206292 by cortano21 last updated on 11/Apr/24 Answered by A5T last updated on 11/Apr/24 $${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}}…