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Question Number 208344 by mr W last updated on 13/Jun/24

Commented by mr W last updated on 13/Jun/24

$$findgreenarea=?$$

Commented by naka3546 last updated on 13/Jun/24

$$\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{square}?$$

Commented by mr W last updated on 13/Jun/24

$$yes$$

Commented by Tawa11 last updated on 21/Jun/24

$$\mathrm{Good}\mathrm{one}\mathrm{sir}$$

Answered by A5T last updated on 13/Jun/24

Commented by A5T last updated on 13/Jun/24

$$\frac{sin45}{y}=\frac{sin90}{4}\Rightarrow y=2\sqrt{2}\Rightarrow BC=3y=6\sqrt{2}\Rightarrow BH=2\sqrt{10}$$$$\frac{sinHBC}{4}=\frac{sin45}{2\sqrt{10}}=\frac{\frac{\sqrt{2}}{2}}{\frac{2\sqrt{10}}{1}}=\frac{\sqrt{5}}{20}\Rightarrow sinHBC=\frac{\sqrt{5}}{5}$$$$\Rightarrow cosHBC=\frac{2\sqrt{5}}{5};\frac{\frac{2\sqrt{5}}{5}}{6\sqrt{2}}=\frac{1}{BF}\Rightarrow BF=3\sqrt{10}$$$$\frac{sinGHF}{6\sqrt{2}}=\frac{\frac{\sqrt{5}}{5}}{4}\Rightarrow sinGHF=\frac{\frac{6\sqrt{10}}{5}}{4}=\frac{3\sqrt{10}}{10}$$$$\frac{AE}{EB}\times \frac{BF}{FH}\times \frac{HG}{GA}=1\Rightarrow \frac{AE}{EB=6\sqrt{2}-AE}=\frac{FH}{BF}=\frac{1}{3}$$$$\Rightarrow 4AE=6\sqrt{2}\Rightarrow AE=\frac{3\sqrt{2}}{2}$$$$\left[green\right]=\frac{1}{2}[\frac{3\sqrt{2}}{2}\times 2\sqrt{2}+4\times \sqrt{10}\times \frac{3\sqrt{10}}{10}+6\sqrt{2}\times 2\sqrt{2}]$$$$=\frac{1}{2}[6+12+24]=21$$

Answered by MM42 last updated on 13/Jun/24

$$B=\frac{1}{2}{S}_{FGC}$$$$\mathrm{\Delta }AGE\sim \mathrm{\Delta }FGC\Rightarrow \frac{A}{B}=\frac{1}{2}$$$$C=2S_{FCH}=2B$$$$C=4A\mathrm{\&}B=2A$$$$2a^2=144\Rightarrow a=6\sqrt{2}$$$$C=\frac{1}{2}\times 6\sqrt{2}\times 4\times \frac{\sqrt{2}}{2}=12$$$$\Rightarrow S=A+B+C=21✓$$$$$$

Commented by MM42 last updated on 13/Jun/24

Answered by mr W last updated on 13/Jun/24

Commented by mr W last updated on 13/Jun/24

$$a=sidelengthofsquare$$$$S=areaofsquare$$$$\sqrt{2}a=3\times 4\Rightarrow a=\frac{12}{\sqrt{2}}$$$$S=a^2={\left(\frac{12}{\sqrt{2}}\right)}^2=72$$$$A=\frac{1}{3}\times \frac{S}{2}=\frac{S}{6}$$$$GC=2\times AG$$$$\frac{B+B}{C}=2^2\Rightarrow B=2C$$$$A+B=B+B+2C\Rightarrow A=B+2C=2B$$$$\Rightarrow B=\frac{A}{2}=\frac{S}{12}\Rightarrow C=\frac{B}{2}=\frac{S}{24}$$$$greenarea=A+B+C$$$$=(\frac{1}{6}+\frac{1}{12}+\frac{1}{24})S=\frac{7}{24}S=21✓$$

Answered by cherokeesay last updated on 13/Jun/24

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