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Question Number 89361 by 675480065 last updated on 17/Apr/20

Answered by mr W last updated on 17/Apr/20

Commented by mr W last updated on 17/Apr/20

$$\sin \alpha =\frac{s}{6}$$$$\frac{\sqrt{2}s}{\sin 45°}=\frac{4}{\sin (45°+\alpha +45°)}=\frac{4}{\cos \alpha }$$$$2s=\frac{4}{\cos \alpha }$$$$\Rightarrow \cos \alpha =\frac{2}{s}$$$${\left(\frac{s}{6}\right)}^2+{\left(\frac{2}{s}\right)}^2=1$$$$s^4-36s^2+144=0$$$$areaofsquare=s^2=18-6\sqrt{5}\approx 4.58$$

Commented by 675480065 last updated on 17/Apr/20

$$\mathrm{thnks}\mathrm{but}\mathrm{i}\mathrm{cant}\mathrm{go}\mathrm{further}$$

Commented by mr W last updated on 17/Apr/20

$$thenyouhaveselectedaquestion$$$$whichisnotsuitableforyourlevel$$$$yet.weneedheretoapplysineslaw,$$$$trigonometry,quadraticequationetc.$$

Commented by Tony Lin last updated on 17/Apr/20

$$thiscanbeevaluatedwithouttrigometry$$

Commented by Tony Lin last updated on 17/Apr/20

Commented by Tony Lin last updated on 17/Apr/20

$$△DBE\sim △CDF$$$$\Rightarrow \frac{\sqrt{6^2-s^2}}{6}=\frac{2\sqrt{2}}{\sqrt{2}s}$$$$12=s\sqrt{36-s^2}$$$$144=s^2(36-s^2)$$$$s^4-36s^2+144=0$$$$s^2=18-6\sqrt{5}$$

Commented by mr W last updated on 18/Apr/20

$$nicesolutionsir!$$

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