Question Number 189899 by Mingma last updated on 23/Mar/23
Answered by HeferH last updated on 23/Mar/23
Commented by HeferH last updated on 24/Mar/23
$$\left(\mathrm{6}−\mathrm{r}\right)+\left(\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{r}\right)=\mathrm{12} \\ $$$$\:\mathrm{6}+\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{12}\:=\:\mathrm{2r} \\ $$$$\:\mathrm{r}\:=\:\frac{\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{6}}{\mathrm{2}}\:=\:\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3} \\ $$$$\:\frac{\mathrm{x}}{\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}}=\:\frac{\mathrm{6}\sqrt{\mathrm{3}}−\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\right)}{\:\sqrt{\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{36}}} \\ $$$$\:\mathrm{x}\:=\:\frac{\left(\mathrm{3}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{9}}{\:\sqrt{\mathrm{72}−\mathrm{18}\sqrt{\mathrm{3}}}}\:=\:\frac{\mathrm{18}}{\:\sqrt{\mathrm{8}\centerdot\mathrm{9}−\mathrm{2}\centerdot\mathrm{9}\sqrt{\mathrm{3}}}}=\frac{\mathrm{18}}{\mathrm{3}\sqrt{\mathrm{8}−\mathrm{2}\sqrt{\mathrm{3}}}} \\ $$$$\:\mathrm{x}\:=\:\frac{\mathrm{6}}{\:\sqrt{\mathrm{8}−\mathrm{2}\sqrt{\mathrm{3}}}}\:\approx\:\mathrm{2}.\mathrm{81} \\ $$
Commented by Rupesh123 last updated on 24/Mar/23
Excellent!