Question-163921

Question Number 163921 by ajfour last updated on 11/Jan/22

Answered by mr W last updated on 12/Jan/22

R = r a d i u s o f b i g c i r c l e s

r = r a d i u s o f s m a l l c i r c l e s

s e m i d i a g o n a l = R + \sqrt{2} R = \sqrt{{(R + r)}^{2} - R^{2}} + \sqrt{2} r

(\sqrt{2} + 1) R - \sqrt{2} r = \sqrt{2 R r + r^{2}}

r^{2} - 2 (\sqrt{2} + 3) R r + {(\sqrt{2} + 1)}^{2} R^{2} = 0

\frac{r}{R} = \sqrt{2} + 3 - \sqrt{{(\sqrt{2} + 3)}^{2} - {(\sqrt{2} + 1)}^{2}}

\Rightarrow \frac{r}{R} = \sqrt{2} + 3 - 2 \sqrt{\sqrt{2} + 2}

a r e a o f p a r a l l e l o g r a m :

A_{p} = 2 R \sqrt{{(R + r)}^{2} - R^{2}} = 2 R \sqrt{2 R r + r^{2}}

a r e a o f s q u a r e :

A_{s} = 2 {(R + \sqrt{2} R)}^{2}

r a t i o

\frac{A_{s}}{A_{p}} = \frac{2 {(R + \sqrt{2} R)}^{2}}{2 R \sqrt{2 R r + r^{2}}} = \frac{{(1 + \sqrt{2})}^{2}}{\sqrt{(2 + \frac{r}{R}) (\frac{r}{R})}}

\frac{A_{s}}{A_{p}} = \frac{3 + 2 \sqrt{2}}{\sqrt{(5 + \sqrt{2} - 2 \sqrt{\sqrt{2} + 2}) (3 + \sqrt{2} - 2 \sqrt{\sqrt{2} + 2})}}

\frac{A_{s}}{A_{p}} \approx 4.169

Commented by Tawa11 last updated on 12/Jan/22

Great sir

Answered by ajfour last updated on 12/Jan/22

R + R \sqrt{2} = s \sqrt{2}

R + r + 2 \sqrt{R r} = 2 s

\sqrt{\frac{r}{R}} + 1 = \sqrt{\frac{2 s}{R}}

R = (2 - \sqrt{2}) s

r = R {(\sqrt{\frac{2}{2 - \sqrt{2}}} - 1)}^{2}

= R {(\sqrt{2 + \sqrt{2}} - 1)}^{2}

A_{p} = 4 R \sqrt{{(r + R)}^{2} - R^{2}}

= (4 s^{2}) {(2 - \sqrt{2})}^{2} {{[{(\sqrt{2 + \sqrt{2}} - 1)}^{2} + 1]}^{2} - 1}^{1 / 2}

\frac{A_{p}}{A_{s}} = {(2 - \sqrt{2})}^{2} {{[{(\sqrt{2 + \sqrt{2}} - 1)}^{2} + 1]}^{2} - 1}^{1 / 2}

\dots

Facebook Tweet Pin