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 180300 by mnjuly1970 last updated on 10/Nov/22

Answered by Acem last updated on 11/Nov/22

Commented by Acem last updated on 11/Nov/22

$$Sorryi{haven}^{\prime }tacompass$$

Commented by mr W last updated on 12/Nov/22

$$itisnotgiventhat\mathrm{\angle }A=\mathrm{\angle }C=45°.$$$$so\frac{c}{AC}\ne \frac{r}{2r}.$$

Commented by Acem last updated on 12/Nov/22

$$diameter,diameter...mademethinkthat{it}^{\prime }s$$$$belongtothebiggercircle.$$$$$$$${it}^{\prime }sgoodthatthecomposerwrotethisword$$$$sothatwe{don}^{\prime }tguessitasasquare!$$

Answered by a.lgnaoui last updated on 11/Nov/22

$$△\mathrm{OCD}\mathrm{OD}=\mathrm{OC}\measuredangle \mathrm{COD}=45°$$$$△\mathrm{OAE}\mathrm{et}△\mathrm{BCD}\left(\mathrm{triangles}\mathrm{rectangles}\right)$$$$\mathrm{OB}=\mathrm{BD}=\mathrm{BC}\Rightarrow a+b=c\left(1\right)$$$$\cos 45=\frac{a}{2r}=\frac{a+b}{3r},\frac{a}{2}=\frac{a+b}{3}\Rightarrow \frac{a}{b}=2\left(2\right)$$$$\frac{a}{b}\sqrt{\frac{c}{a+b}}=2$$$$$$

Commented by a.lgnaoui last updated on 11/Nov/22

Commented by Acem last updated on 12/Nov/22

$$Coolaussi!d^{\prime }accord,commentsavez-vousque$$$$ED=r?$$

Commented by mr W last updated on 12/Nov/22

$$\mathrm{\angle }CODcanbedifferentthan45°!$$

Commented by mr W last updated on 12/Nov/22

Answered by mr W last updated on 12/Nov/22

Commented by mr W last updated on 12/Nov/22

$$c=2R\sin \alpha$$$$a+b=\frac{2R}{\tan \alpha }\times \cos \alpha$$$$\frac{c}{a+b}=\frac{2R\sin \alpha \tan \alpha }{2R\cos \alpha }={\tan }^2\alpha$$$$\Rightarrow \sqrt{\frac{c}{a+b}}=\tan \alpha$$$$\sqrt{{(R+r)}^2-R^2}+r=\frac{2R}{\tan \alpha }$$$$\sqrt{2Rr+r^2}=\frac{2R}{\tan \alpha }-r$$$$\Rightarrow r=\frac{2R}{\tan \alpha (\tan \alpha +2)}$$$$a=2r\cos \alpha =\frac{4R\cos \alpha }{\tan \alpha (\tan \alpha +2)}$$$$b=\frac{2R}{\tan \alpha }\times \cos \alpha -\frac{4R\cos \alpha }{\tan \alpha (2+\tan \alpha )}$$$$\Rightarrow b=\frac{2R\cos \alpha \tan \alpha }{\tan \alpha (\tan \alpha +2)}$$$$\frac{a}{b}=\frac{4R\cos \alpha }{\tan \alpha (\tan \alpha +2)}\times \frac{\tan \alpha (\tan \alpha +2)}{2R\cos \alpha \tan \alpha }$$$$\Rightarrow \frac{a}{b}=\frac{2}{\tan \alpha }$$$$\Rightarrow \frac{a}{b}\sqrt{\frac{c}{a+b}}=\frac{2}{\tan \alpha }\times \tan \alpha =2✓$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com