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 185674 by ajfour last updated on 25/Jan/23

Commented by ajfour last updated on 25/Jan/23

Find the radii (equal) of the   circles, given outer figure is  a square of side a.

$${Find}\:{the}\:{radii}\:\left({equal}\right)\:{of}\:{the}\: \\ $$$${circles},\:{given}\:{outer}\:{figure}\:{is} \\ $$$${a}\:{square}\:{of}\:{side}\:{a}. \\ $$

Answered by mr W last updated on 25/Jan/23

Commented by mr W last updated on 25/Jan/23

x=a−2r−(r/(tan 22.5°))=a−2r−(r/( (√2)−1))  y=a−r  (x+y)^2 =(x+r)^2 +(y+r)^2   xy=(x+y)r+r^2   (a−2r−(r/( (√2)−1)))(a−r)=(a−2r−(r/( (√2)−1))+a−r)r+r^2   let λ=(a/r)  λ^2 −(6+(√2))λ+6+2(√2)=0  λ=((6+(√2)+(√(14+4(√2))))/2)  ⇒(r/a)=(2/(6+(√2)+(√(14+4(√2)))))≈0.169

$${x}={a}−\mathrm{2}{r}−\frac{{r}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°}={a}−\mathrm{2}{r}−\frac{{r}}{\:\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$${y}={a}−{r} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} =\left({x}+{r}\right)^{\mathrm{2}} +\left({y}+{r}\right)^{\mathrm{2}} \\ $$$${xy}=\left({x}+{y}\right){r}+{r}^{\mathrm{2}} \\ $$$$\left({a}−\mathrm{2}{r}−\frac{{r}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\right)\left({a}−{r}\right)=\left({a}−\mathrm{2}{r}−\frac{{r}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}+{a}−{r}\right){r}+{r}^{\mathrm{2}} \\ $$$${let}\:\lambda=\frac{{a}}{{r}} \\ $$$$\lambda^{\mathrm{2}} −\left(\mathrm{6}+\sqrt{\mathrm{2}}\right)\lambda+\mathrm{6}+\mathrm{2}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$$\lambda=\frac{\mathrm{6}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{14}+\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{r}}{{a}}=\frac{\mathrm{2}}{\mathrm{6}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{14}+\mathrm{4}\sqrt{\mathrm{2}}}}\approx\mathrm{0}.\mathrm{169} \\ $$

Commented by ajfour last updated on 27/Jan/23

Thank you sir!

$${Thank}\:{you}\:{sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com