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Question Number 199234 by mr W last updated on 30/Oct/23

Commented by mr W last updated on 30/Oct/23

$$thesidelengthofthesquaresis1.$$$$findtheareaoftheisoscelesright$$$$angledtriangle.$$

Answered by deleteduser1 last updated on 30/Oct/23

$$GF=\sqrt{1^2+2^2}=\sqrt{5}$$$$\frac{sin45}{1}=\frac{sinAFE}{AE}=\frac{\frac{2\sqrt{5}}{5}}{AE}\Rightarrow AE=\frac{\frac{2\sqrt{5}}{5}}{\frac{\sqrt{2}}{2}}=\frac{2\sqrt{10}}{5}$$$$sinAFE=sin\theta ;sinFEA=sin\beta$$$$sinFEA=sin(135-\theta )=sin135cos\theta -sin\theta cos135$$$$\frac{\sqrt{10}}{10}+\frac{\sqrt{2}}{2}\times \frac{2\sqrt{5}}{5}=\frac{3\sqrt{10}}{10}$$$$sin\left(DEC\right)=sin(90-\beta )=cos\beta =\sqrt{1-\frac{90}{100}}=\frac{\sqrt{10}}{10}$$$$\frac{sin\left(DEC\right)}{DC}=\frac{sin90}{3}\Rightarrow DC=\frac{3\sqrt{10}}{10}$$$$EC=\sqrt{{ED}^2-{CD}^2}=\sqrt{9-\frac{9}{10}}=\frac{\sqrt{810}}{10}=\frac{9\sqrt{10}}{10}$$$$\Rightarrow AC=\frac{13\sqrt{10}}{10}\Rightarrow area=\frac{1}{2}{\left(\frac{13\sqrt{10}}{10}\right)}^2=8.45$$

Commented by mr W last updated on 30/Oct/23

$$thankssir!$$

Commented by deleteduser1 last updated on 30/Oct/23

Answered by mr W last updated on 30/Oct/23

Commented by mr W last updated on 30/Oct/23

$$\alpha +\beta =45°$$$$CB=\sqrt{2}\times DB=\sqrt{2}\times \sqrt{5}=\sqrt{10}$$$$BA=1\times \frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}$$$$CA=\sqrt{10}+\frac{3\sqrt{10}}{10}=\frac{13\sqrt{10}}{10}$$$$area=\frac{1}{2}\times {\left(\frac{13\sqrt{10}}{10}\right)}^2=\frac{169}{20}=8.45$$

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