Question Number 219784 by Spillover last updated on 01/May/25
Answered by Spillover last updated on 02/May/25
Answered by JV2BTC last updated on 02/May/25
$${why}\:{not} \\ $$$$\frac{\mathrm{2}}{{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{8}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\frac{\mathrm{128}}{\mathrm{17}} \\ $$$$\therefore\:{x}=\frac{\mathrm{8}\sqrt{\mathrm{34}}}{\mathrm{17}}\approx\mathrm{2}.\mathrm{74} \\ $$
Commented by Spillover last updated on 02/May/25
$${correct}.{thank}\:{you} \\ $$
Answered by mr W last updated on 02/May/25
Commented by Spillover last updated on 02/May/25
$${correct}.{thank}\:{you} \\ $$
Commented by mr W last updated on 02/May/25
$$\left(\mathrm{5}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{8}}{\mathrm{5}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\frac{{r}}{\mathrm{5}−{r}}=\frac{\frac{\mathrm{8}}{\mathrm{5}}}{\mathrm{5}−\frac{\mathrm{8}}{\mathrm{5}}}=\frac{\mathrm{8}}{\mathrm{17}}=\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{5}}{\:\sqrt{\mathrm{34}}} \\ $$$${x}=\mathrm{2}{r}\:\mathrm{cos}\:\theta=\mathrm{2}×\frac{\mathrm{8}}{\mathrm{5}}×\frac{\mathrm{5}}{\:\sqrt{\mathrm{34}}}=\frac{\mathrm{8}\sqrt{\mathrm{34}}}{\:\mathrm{17}} \\ $$
Answered by Spillover last updated on 02/May/25
Answered by Spillover last updated on 02/May/25
Answered by Spillover last updated on 02/May/25
Answered by Spillover last updated on 02/May/25
Answered by mr W last updated on 02/May/25
$$\frac{{AD}}{\mathrm{2}}=\frac{\sqrt{\mathrm{10}^{\mathrm{2}} −{AD}^{\mathrm{2}} }}{\mathrm{8}}\:\Rightarrow{AD}=\frac{\mathrm{10}}{\:\sqrt{\mathrm{17}}} \\ $$$$\mathrm{3}×\left(\frac{\mathrm{10}}{\:\sqrt{\mathrm{17}}}\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{5}^{\mathrm{2}} =\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{2}×\mathrm{3}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{8}\sqrt{\mathrm{34}}}{\mathrm{17}} \\ $$
Commented by Spillover last updated on 02/May/25
$${good} \\ $$