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 20599 by Tinkutara last updated on 28/Aug/17

$$\mathrm{In}\mathrm{a}\mathrm{rectangle}ABCD,E\mathrm{is}\mathrm{the}\mathrm{midpoint}$$$$\mathrm{of}AB;F\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{on}AC\mathrm{such}\mathrm{that}BF$$$$\mathrm{is}\mathrm{perpendicular}\mathrm{to}AC;\mathrm{and}FE$$$$\mathrm{perpendicular}\mathrm{to}BD.\mathrm{Suppose}BC=8\sqrt{3}.$$$$\mathrm{Find}AB.$$

Answered by ajfour last updated on 29/Aug/17

Commented by Tinkutara last updated on 29/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Commented by ajfour last updated on 29/Aug/17

$$EF=AE=BE(radiiofcircle$$$$withEascentre)$$$$so\mathrm{\angle }AFE=\theta$$$$\mathrm{\angle }DBF=90°-2\theta ,so\mathrm{\angle }EBF=2\theta$$$$now\mathrm{\angle }AFE+\mathrm{\angle }EFB=90°$$$$so\theta +2\theta =90°$$$$\Rightarrow \tan \theta =\frac{BC}{AB}=\frac{8\sqrt{3}}{AB}=\frac{1}{\sqrt{3}}$$$$or\mathit{A}\mathit{B}=24.$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com