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Question Number 205880 by cortano12 last updated on 01/Apr/24

Commented by cortano12 last updated on 01/Apr/24

$$\underbrace{}$$

Answered by mr W last updated on 02/Apr/24

Commented by mr W last updated on 02/Apr/24

$${\mathrm{\Delta }}_{BCE}=\frac{square}{4}=\frac{{12}^2}{4}=36$$$$\frac{EF}{FB}=\frac{EC}{AB}=\frac{1}{2}\Rightarrow \frac{EF}{EB}=\frac{1}{3}$$$$A_1=\frac{1}{3}{\mathrm{\Delta }}_{BCE}$$$$\mathrm{\Delta }BGH\sim \mathrm{\Delta }BCE$$$$\frac{BH}{BE}=\frac{1}{\sqrt{1^2+2^2}}=\frac{1}{\sqrt{5}}$$$$A_2={\left(\frac{1}{\sqrt{5}}\right)}^2{\mathrm{\Delta }}_{BCE}=\frac{1}{5}{\mathrm{\Delta }}_{BCE}$$$$shaded{A}_3=(1-\frac{1}{3}-\frac{1}{5}){\mathrm{\Delta }}_{BCE}=\frac{7}{15}{\mathrm{\Delta }}_{BCE}$$$$=\frac{7}{15}\times 36=\frac{84}{5}=16.8✓$$

Answered by A5T last updated on 01/Apr/24

Commented by A5T last updated on 01/Apr/24

$$\frac{sin(90-\theta )}{y=FH}=\frac{sin(90-\beta )}{BF};\frac{sin\beta }{BF}=\frac{sin\theta }{x}$$$$x=\frac{BFsin\theta }{sin\beta };\frac{1}{y}=\frac{sin(90-\beta )}{BFsin(90-\theta )}\Rightarrow \frac{x}{y}=\frac{tan\theta }{tan\beta }=\frac{\frac{2}{1}}{\frac{1}{2}}=4$$$$\frac{\frac{1}{2}\times 4y\times 2z}{\frac{1}{2}\times 5y\times 3z}=\frac{8}{15}=\frac{\left[AGF\right]}{\left[ACH\right]}\Rightarrow \frac{8}{15}\times 36=\left[AGF\right]=19.2$$$$\Rightarrow \left[CGFH\right]=\left[ACH\right]-\left[AGF\right]=36-19.2=16.8$$

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