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Question Number 166834 by leicianocosta last updated on 28/Feb/22

Answered by Rasheed.Sindhi last updated on 28/Feb/22

Commented by Rasheed.Sindhi last updated on 28/Feb/22

$$LetAC=BC=a$$$${AC}^2+{BC}^2={AB}^2$$$$a^2+a^2=6^2\Rightarrow a=3\sqrt{2}$$$${CD}^2={AC}^2+{AD}^2-2.AC.AD.\cos A$$$$={\left(3\sqrt{2}\right)}^2+2^2-2.3\sqrt{2}.2.\cos 45$$$$=18+4-12\sqrt{2}.\frac{1}{\sqrt{2}}=10$$$$CD=\sqrt{10}$$$$△BCD:$$$$\frac{BD}{\sin C}=\frac{CD}{\sin B}\Rightarrow \frac{4}{\sin C}=\frac{\sqrt{10}}{\sin 45}$$$$\Rightarrow \sqrt{10}\sin C=4\left(\frac{1}{\sqrt{2}}\right)$$$$\sin C=\frac{4}{\sqrt{10}.\sqrt{2}}=\frac{2}{\sqrt{5}}$$$$C={\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)$$$$D=180-(45+{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right))$$$$BDE=D-45$$$$=180-45-{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)-45$$$$=90-{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)$$$$△BDE:$$$$\mathrm{\angle }BDE=90-{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)$$$$BD=4$$$$\mathrm{\angle }B=45$$$$\mathrm{\angle }E=180-(45+90-{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right))$$$$=45+{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)$$$$▴BDE=\frac{{BD}^2.\sin B.\sin D}{2\sin E}$$$$=\frac{4^2\sin 45.\sin (90-{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right))}{2(45+{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right))}$$$$=\frac{\frac{16}{\sqrt{2}}.\cos \left({\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)}{90+{2\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right))}$$

Commented by Tawa11 last updated on 01/Mar/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 28/Feb/22

Commented by mr W last updated on 28/Feb/22

$$AC=3\sqrt{2}$$$${CD}^2={\left(\frac{2}{\sqrt{2}}\right)}^2+{(3\sqrt{2}-\frac{2}{\sqrt{2}})}^2=10\Rightarrow CD=\sqrt{10}$$$$\frac{\sin \alpha }{2}=\frac{\sin 45°}{\sqrt{10}}$$$$\sin \alpha =\frac{1}{\sqrt{5}}\Rightarrow \cos \alpha =\frac{2}{\sqrt{5}}$$$$DF=FE$$$$CE\times \cos \alpha +CE\times \sin \alpha =CD$$$$CE=\frac{\sqrt{10}}{\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{5}}}=\frac{5\sqrt{2}}{3}$$$$BE=3\sqrt{2}-\frac{5\sqrt{2}}{3}=\frac{4\sqrt{2}}{3}$$$$x=\frac{1}{2}\times 4\times \frac{4\sqrt{2}}{3}\times \frac{\sqrt{2}}{2}=\frac{8}{3}$$

Commented by Tawa11 last updated on 01/Mar/22

$$\mathrm{Great}\mathrm{sir}$$

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