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Question Number 185324 by cherokeesay last updated on 20/Jan/23

Answered by som(math1967) last updated on 20/Jan/23

AC=(√(2^2 +1^2 ))=(√5)=BC   r×((2+1+(√5))/2)=(1/2)×2×1  ⇒r=(2/(3+(√5)))   R×(((√5)+(√5)+2)/2)=(1/2)×2×2   R=(4/(2((√5)+1)))=(2/( (√5)+1))   1+(r/R)=1+((2/(3+(√5)))/(2/( (√5)+1)))  =1+(((√5)+1)/(3+(√5)))=((4+2(√5))/(3+(√5)))=((2((√5)+2)(3−(√(5))))/(9−5))  =((2((√5)+1))/4)=(((√5)+1)/2)=ϕ

$${AC}=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }=\sqrt{\mathrm{5}}={BC} \\ $$$$\:{r}×\frac{\mathrm{2}+\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{1} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}}{\mathrm{3}+\sqrt{\mathrm{5}}} \\ $$$$\:{R}×\frac{\sqrt{\mathrm{5}}+\sqrt{\mathrm{5}}+\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{2} \\ $$$$\:{R}=\frac{\mathrm{4}}{\mathrm{2}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{1}} \\ $$$$\:\mathrm{1}+\frac{{r}}{{R}}=\mathrm{1}+\frac{\frac{\mathrm{2}}{\mathrm{3}+\sqrt{\mathrm{5}}}}{\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}} \\ $$$$=\mathrm{1}+\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{3}+\sqrt{\mathrm{5}}}=\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{3}+\sqrt{\mathrm{5}}}=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}+\mathrm{2}\right)\left(\mathrm{3}−\sqrt{\left.\mathrm{5}\right)}\right.}{\mathrm{9}−\mathrm{5}} \\ $$$$=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}{\mathrm{4}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}=\varphi \\ $$$$ \\ $$

Commented by som(math1967) last updated on 20/Jan/23

Commented by cherokeesay last updated on 20/Jan/23

Nice ! thank you.

$${Nice}\:!\:{thank}\:{you}. \\ $$

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