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Question Number 200575 by sonukgindia last updated on 20/Nov/23

Answered by deleteduser1 last updated on 20/Nov/23

$$\frac{ra}{2}=\frac{bx}{2}\Rightarrow ra=bx;x=\sqrt{r^2-a^2};b=r-a$$$$ra=bx\Rightarrow ra=(r-a)\left(\sqrt{r^2-a^2}\right)$$$$\Rightarrow 2r=a\underset{-}{+}a\sqrt{5+4\sqrt{2}}\Rightarrow a+2b=\underset{-}{+}a\sqrt{5+4\sqrt{2}}$$$$\Rightarrow a\left(1\stackrel{-}{+}\sqrt{5+4\sqrt{2}}\right)=-2b\Rightarrow \frac{a}{b}=\frac{-2}{1\stackrel{-}{+}\sqrt{5+4\sqrt{2}}}$$$$\Rightarrow \frac{a}{b}\approx 0.883203$$

Commented by mr W last updated on 21/Nov/23

$$howdidyouget?$$$$\Rightarrow 2r=a\underset{-}{+}a\sqrt{5+4\sqrt{2}}$$

Commented by deleteduser1 last updated on 21/Nov/23

$$\frac{r^2{a}^2}{{(r-a)}^2}=r^2-a^2\Rightarrow r^2{a}^2=(r^2-a^2)(r^2+a^2-2ar)$$$$\Rightarrow r^4-2{ar}^3-a^2{r}^2+2a^{3}r-a^4=0$$$$So,wecansolvethepolynomialinr,Iused$$$$wolfram.r=\frac{1}{2}\left(a\underset{-}{+}a\sqrt{5+4\sqrt{2}}\right)$$

Commented by mr W last updated on 21/Nov/23

$$thankssir!ijustwantedtoknowif$$$$youhadawaytoavoidsolving$$$$quadraticequation.$$

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