Question Number 185893 by mr W last updated on 29/Jan/23
Commented by mr W last updated on 29/Jan/23
$${length}\:{of}\:{common}\:{tangent}\:{line}\:{x}=? \\ $$
Commented by som(math1967) last updated on 29/Jan/23
$${x}=\sqrt{\mathrm{7}}\:? \\ $$
Commented by mr W last updated on 29/Jan/23
$${yes},\:{thanks}! \\ $$
Answered by a.lgnaoui last updated on 29/Jan/23
Answered by som(math1967) last updated on 29/Jan/23
Commented by som(math1967) last updated on 29/Jan/23
$${AB}={diagonal}\:{of}\:{square} \\ $$$$={dist}\:{between}\:{two}\:{centre}=\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${BC}=\mathrm{2}+\mathrm{3}=\mathrm{5} \\ $$$$\therefore{AC}={DE}={x}=\sqrt{\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }=\sqrt{\mathrm{7}} \\ $$$$ \\ $$
Answered by a.lgnaoui last updated on 29/Jan/23
$${Triangle}\:{AEF}\:{rectangle}\:{en}\:{E} \\ $$$${EF}^{\mathrm{2}} ={AF}^{\mathrm{2}} −\:\mathrm{4} \\ $$$${Truangke}\:{DFE}\:{rectangle}\:{en}\:{F} \\ $$$${EF}^{\mathrm{2}} ={DE}^{\mathrm{2}} −\mathrm{9} \\ $$$${AD}^{\mathrm{2}} ={AE}^{\mathrm{2}} +{DE}^{\mathrm{2}} −{AE}×{DE}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\alpha\right)\left({i}\right) \\ $$$$\left(\mathrm{4}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} =\mathrm{4}+{DE}^{\mathrm{2}} +\mathrm{4}{DE}\mathrm{sin}\:\alpha \\ $$$$\bigtriangleup{EFD}\:\:\:\: \\ $$$${EF}=\frac{\mathrm{3}}{\mathrm{tan}\:\alpha} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{3}}{{DE}}\:\:\:\:\:{DE}^{\mathrm{2}} =\mathrm{9}+{EF}^{\mathrm{2}} \\ $$$$\left({i}\right)\:\:\Rightarrow\mathrm{32}=\mathrm{4}+\mathrm{9}+{EF}^{\mathrm{2}} +\mathrm{4}\frac{\mathrm{3}}{\mathrm{sin}\:\alpha}\mathrm{sin}\:\alpha \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{32}=\mathrm{25}+{EF}^{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{E}}{F}=\sqrt{\mathrm{7}} \\ $$$$ \\ $$
Answered by ajfour last updated on 29/Jan/23
Commented by mr W last updated on 29/Jan/23
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