Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 205614 by mr W last updated on 25/Mar/24

Answered by A5T last updated on 25/Mar/24

Commented by A5T last updated on 25/Mar/24

$$\frac{sin\left(\mathrm{\angle }CFB\right)}{1}=\frac{sin90=1}{\sqrt{0.5^2+1}}=\frac{2}{\sqrt{5}}\Rightarrow cos\left(\mathrm{\angle }CFB\right)=\frac{1}{\sqrt{5}}$$$$sin\left(FEC\right)=sin(135-\mathrm{\angle }CFB)=\frac{3\sqrt{10}}{10}$$$$\frac{sin\left(FEC\right)}{0.5}=\frac{sin\left(EFC\right)}{CE}\Rightarrow CE=\frac{\sqrt{2}}{3}$$$$FE=\sqrt{0.5^2+\frac{2}{9}-\frac{1}{3}}=\frac{\sqrt{5}}{6}$$$$\Rightarrow AE=\sqrt{2}-\frac{\sqrt{2}}{3}=\frac{2\sqrt{2}}{3};BE=\frac{\sqrt{5}}{2}-\frac{\sqrt{5}}{6}=\frac{\sqrt{5}}{3}$$$$AreaofAEB=\frac{AE\times EBsinAEB}{2}=\frac{1}{3}$$$$\Rightarrow \frac{r}{2}\left(\frac{\sqrt{5}+2\sqrt{2}+3}{3}\right)=\frac{1}{3}\Rightarrow r=\frac{2}{3+2\sqrt{2}+\sqrt{5}}\approx 0.248$$

Commented by mr W last updated on 25/Mar/24

��

Answered by mr W last updated on 25/Mar/24

Commented by mr W last updated on 25/Mar/24

$$x+\frac{x}{2}=\sqrt{1^2+1^2}=\sqrt{2}$$$$\Rightarrow x=\frac{2\sqrt{2}}{3}$$$$y+\frac{y}{2}=\sqrt{1^2+0.5^2}=\frac{\sqrt{5}}{2}$$$$\Rightarrow y=\frac{\sqrt{5}}{3}$$$${\mathrm{\Delta }}_{ABC}=\frac{1\times x\times \sin 45°}{2}=\frac{1}{3}$$$${\mathrm{\Delta }}_{ABC}=\frac{(1+x+y)r}{2}=\frac{(3+2\sqrt{2}+\sqrt{5})r}{6}$$$$\frac{(3+2\sqrt{2}+\sqrt{5})r}{6}=\frac{1}{3}$$$$\Rightarrow r=\frac{2}{3+2\sqrt{2}+\sqrt{5}}\approx 0.248$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com