Mensuration Questions

Question Number 105883 by bemath last updated on 01/Aug/20

Answered by mr W last updated on 01/Aug/20

$${AG}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\mathrm{2}×\mathrm{3}×\mathrm{6}×\mathrm{cos}\:\mathrm{120}°}=\mathrm{3}\sqrt{\mathrm{7}} \\$$$$\frac{\mathrm{sin}\:\angle{AGF}}{\mathrm{3}}=\frac{\mathrm{sin}\:\mathrm{120}°}{\mathrm{3}\sqrt{\mathrm{7}}} \\$$$$\mathrm{sin}\:\angle{AGF}=\frac{\sqrt{\mathrm{21}}}{\mathrm{14}} \\$$$$\mathrm{cos}\:\angle{AGF}=\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{14}} \\$$$${HG}×\frac{\sqrt{\mathrm{21}}}{\mathrm{14}}=\left(\mathrm{3}−{HG}×\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{14}}\right)\mathrm{tan}\:\mathrm{60}° \\$$$$\Rightarrow{HG}=\sqrt{\mathrm{7}} \\$$$${x}={FH}=\mathrm{3}\sqrt{\mathrm{7}}−\sqrt{\mathrm{7}}=\mathrm{2}\sqrt{\mathrm{7}} \\$$

Commented by bemath last updated on 01/Aug/20

$${thank}\:{you}\:{sir} \\$$

Answered by mr W last updated on 01/Aug/20

Commented by mr W last updated on 01/Aug/20

$${FL}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\$$$${LG}=\mathrm{3}+\mathrm{3}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{15}}{\mathrm{2}} \\$$$${FG}=\sqrt{\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{15}}{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{7}} \\$$$$\frac{{HG}}{{FG}}=\frac{{BG}}{{KG}}=\frac{\mathrm{1}}{\mathrm{3}} \\$$$$\Rightarrow{HG}=\frac{\mathrm{1}}{\mathrm{3}}{FG} \\$$$${x}={FH}=\frac{\mathrm{2}}{\mathrm{3}}{FG}=\mathrm{2}\sqrt{\mathrm{7}} \\$$

Answered by Coronavirus last updated on 01/Aug/20

$$\mathrm{en}\:\mathrm{appliquant}\:\mathrm{le}\:\mathrm{theoreme}\:\mathrm{d}'\:\mathrm{Alkashi}\:\mathrm{au}\:\mathrm{triangle}\:\mathrm{A}{GF} \\$$$${alors}\:{FG}^{\mathrm{2}} =\mathrm{FA}^{\mathrm{2}} +\mathrm{A}{G}^{\mathrm{2}} +\mathrm{2}{F}\overset{\rightarrow} {{A}}.{A}\overset{\rightarrow} {{G}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{6}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{3}\right)\left(\mathrm{6}\right)\mathrm{cos}\left({F}\overset{\rightarrow} {{A}};{A}\overset{\rightarrow} {{G}}\right)\:\:{car}\:\begin{cases}{{AG}={AB}+{BG}}\\{{BG}={FG}={ED}=\mathrm{3}}\end{cases} \\$$$$\mathrm{or}\:\left(\mathrm{F}\overset{\rightarrow} {{A}}\:;\:{A}\overset{\rightarrow} {{G}}\right)=\:\frac{\pi}{\mathrm{3}}\Rightarrow\mathrm{cos}\left(\mathrm{F}\overset{\rightarrow} {{A}}\:;\:{A}\overset{\rightarrow} {{G}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\$$$${aussi}\:\:{FG}^{\mathrm{2}} =\mathrm{9}+\mathrm{36}+\mathrm{18}=\mathrm{63} \\$$$${donc}\:\:{FG}=\sqrt{\mathrm{63}} \\$$