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 139967 by I want to learn more last updated on 02/May/21

Answered by mr W last updated on 02/May/21

Commented by mr W last updated on 02/May/21

$$\frac{AF}{AD}=\frac{AB}{AH}=\frac{3k}{3k+\sqrt{2}k+\sqrt{2}k}=\frac{3}{3+2\sqrt{2}}$$$$\frac{ED}{AD}=\frac{BC}{AC}=\frac{\sqrt{2}k}{3k+\sqrt{2}k}=\frac{\sqrt{2}}{3+\sqrt{2}}$$$$\frac{x}{y}=\frac{AF}{ED}=\frac{3}{3+2\sqrt{2}}\times \frac{3+\sqrt{2}}{\sqrt{2}}=\frac{15}{\sqrt{2}}-9\approx 1.6066$$

Commented by I want to learn more last updated on 02/May/21

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.\mathrm{Thanks}.$$

Commented by Tawa11 last updated on 14/Sep/21

$$\mathrm{nice}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com