Question-162473

Question Number 162473 by amin96 last updated on 29/Dec/21

Answered by mr W last updated on 29/Dec/21

Commented by amin96 last updated on 29/Dec/21

Thanks alot sir mr W

Commented by mr W last updated on 29/Dec/21

\cos α = \frac{2}{6} = \frac{1}{3}

\tan \frac{α}{2} = \sqrt{\frac{1 - \cos α}{1 + \cos α}} = \sqrt{\frac{3 - 1}{3 + 1}} = \frac{1}{\sqrt{2}}

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

A B = \frac{a}{\tan \frac{α}{2}} = \sqrt{2} a

O A = \frac{3}{\tan α} = \frac{3}{2 \sqrt{2}} = \frac{3 \sqrt{2}}{4}

O B = \frac{3 \sqrt{2}}{4} + \sqrt{2} a

O C = 3 - a

{(3 - a)}^{2} = a^{2} + {(\frac{3 \sqrt{2}}{4} + \sqrt{2} a)}^{2}

2 a^{2} + 9 a - \frac{63}{8} = 0

a = \frac{- 9 + 12}{4} = \frac{3}{4}

A_{g r e e n} = 2 \times π {(\frac{3}{4})}^{2} = \frac{9 π}{8}

Commented by Tawa11 last updated on 30/Dec/21

Great sir

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