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 208493 by Tawa11 last updated on 17/Jun/24

Answered by A5T last updated on 17/Jun/24

$$a\left(3a\right)=x^2\Rightarrow x=a\sqrt{3}$$$${\left(4a\right)}^2=2{(a\sqrt{3}+y)}^2=16a^2\Rightarrow r=y=2a\sqrt{2}-a\sqrt{3}$$$$\Rightarrow r=y=a(2\sqrt{2}-\sqrt{3})$$$$a^2{(2\sqrt{2}-\sqrt{3})}^2=a^2{(2\sqrt{2}-\sqrt{3})}^2+4a^2-4a^2(2\sqrt{2}-\sqrt{3})cos\theta$$$$\Rightarrow \theta ={cos}^{-1}\left(\frac{1}{2\sqrt{2}-\sqrt{3}}\right)\approx 24.2°$$

Commented by Tawa11 last updated on 17/Jun/24

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

Commented by Tawa11 last updated on 17/Jun/24

$$\mathrm{Am}\mathrm{still}\mathrm{trying}\mathrm{to}\mathrm{understand}\mathrm{your}\mathrm{workings}\mathrm{sir}.$$$$\mathrm{It}\mathrm{will}\mathrm{be}\mathrm{appreciated}\mathrm{if}\mathrm{you}\mathrm{can}\mathrm{expanciate}\mathrm{sir}.$$

Commented by A5T last updated on 17/Jun/24

Commented by A5T last updated on 17/Jun/24

$${}^{\prime }Powerofa{point}^{\prime }willgive{x}^2=a(a+2a)=a\left(3a\right)$$$$\Rightarrow x=a\sqrt{3};$$$$y=randPythagorastheoremgives:$$$${(a+2a+a)}^2={(x+y)}^2+{(x+y)}^2=2{(x+y)}^2$$$$\Rightarrow {\left(4a\right)}^2=2{(a\sqrt{3}+y)}^2\Rightarrow r=y=a(2\sqrt{2}-\sqrt{3})$$$$Cosineruleonthetrianglewithsidesr,r,2a$$$$givesthelastequation.$$

Commented by Tawa11 last updated on 17/Jun/24

$$\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}.$$

Answered by mr W last updated on 17/Jun/24

Commented by mr W last updated on 17/Jun/24

$$AE=2a$$$$DE=\sqrt{r^2-a^2}$$$$AD=\sqrt{2}r$$$$\sqrt{2}r+\sqrt{r^2-a^2}=2a$$$$r^2-a^2={(2a-\sqrt{2}r)}^2=4a^2+2r^2-4\sqrt{2}ar$$$$r^2-4\sqrt{2}ar+5a^2=0$$$$r=(2\sqrt{2}-\sqrt{3})a$$$$\cos \theta =\frac{a}{(2\sqrt{2}-\sqrt{3})a}=\frac{2\sqrt{2}+\sqrt{3}}{5}$$$$\Rightarrow \theta ={\cos }^{-1}\frac{2\sqrt{2}+\sqrt{3}}{5}\approx 24.203°$$

Commented by Tawa11 last updated on 17/Jun/24

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com