Question Number 209960 by a.lgnaoui last updated on 27/Jul/24
$$\mathrm{Determiner}\:\:\mathrm{Aire}\:\left(\boldsymbol{\mathrm{ABH}}\right) \\ $$$$\:\:\:\mathrm{AH}\bot\mathrm{CE} \\ $$
Commented by a.lgnaoui last updated on 27/Jul/24
Answered by mr W last updated on 27/Jul/24
Commented by mr W last updated on 27/Jul/24
$${BD}=\frac{\mathrm{3}}{\mathrm{tan}\:\mathrm{2}\alpha}={CD} \\ $$$${AD}=\frac{\mathrm{3}}{\mathrm{sin}\:\mathrm{2}\alpha} \\ $$$${AC}=\mathrm{5}\:\mathrm{cos}\:\alpha=\frac{\mathrm{3}}{\mathrm{sin}\:\mathrm{2}\alpha}+\frac{\mathrm{3}}{\mathrm{tan}\:\mathrm{2}\alpha} \\ $$$$\mathrm{5}\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\mathrm{2}\alpha=\mathrm{3}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha\right) \\ $$$$\mathrm{5}\:\mathrm{sin}\:\mathrm{2}\alpha=\mathrm{6}\:\mathrm{cos}\:\alpha \\ $$$$\mathrm{10}\:\mathrm{sin}\:\alpha=\mathrm{6} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${A}_{\Delta{ABH}} =\frac{\mathrm{3}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\mathrm{2}}=\frac{\mathrm{9}×\mathrm{3}×\mathrm{4}}{\mathrm{2}×\mathrm{5}×\mathrm{5}}=\frac{\mathrm{54}}{\mathrm{25}} \\ $$
Answered by A5T last updated on 27/Jul/24
Commented by A5T last updated on 27/Jul/24
![AB=AE=3;CE=5⇒AC=(√(5^2 −3^2 ))=4 CE×CH=CA^2 =5CH⇒CH=((16)/5)=CE−EH ⇒BH=EH=5−((16)/5)=(9/5) ⇒[ABH]=(1/2)×(9/5)×(√(3^2 −(9^2 /5^2 )))=((54)/(25))=2.16](https://www.tinkutara.com/question/Q209971.png)
$${AB}={AE}=\mathrm{3};{CE}=\mathrm{5}\Rightarrow{AC}=\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }=\mathrm{4} \\ $$$${CE}×{CH}={CA}^{\mathrm{2}} =\mathrm{5}{CH}\Rightarrow{CH}=\frac{\mathrm{16}}{\mathrm{5}}={CE}−{EH} \\ $$$$\Rightarrow{BH}={EH}=\mathrm{5}−\frac{\mathrm{16}}{\mathrm{5}}=\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\Rightarrow\left[{ABH}\right]=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{9}}{\mathrm{5}}×\sqrt{\mathrm{3}^{\mathrm{2}} −\frac{\mathrm{9}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} }}=\frac{\mathrm{54}}{\mathrm{25}}=\mathrm{2}.\mathrm{16} \\ $$