Question-72930

Question Number 72930 by TawaTawa last updated on 05/Nov/19

Answered by ajfour last updated on 05/Nov/19

Commented by ajfour last updated on 05/Nov/19

R \sin α = R - 2 r

E F = \frac{r}{\tan (\frac{π}{4} - \frac{α}{2})} + r = 2 R \sin α

\Rightarrow l e t \frac{r}{R} = x

\Rightarrow \sin α = 1 - 2 x

l e t \frac{α}{2} = θ, \tan θ = m

\Rightarrow \frac{2 m}{1 + m^{2}} = 1 - 2 x

\Rightarrow m^{2} - \frac{2 m}{(1 - 2 x)} + 1 = 0

a n d x (\frac{1 + m}{1 - m} + 1) = 2 (1 - 2 x)

\Rightarrow x = (1 - 2 x) (1 - m)

o r m = 1 - \frac{x}{1 - 2 x} = \frac{1 - 3 x}{1 - 2 x}

\Rightarrow {(\frac{1 - 3 x}{1 - 2 x})}^{2} - \frac{2 (1 - 3 x)}{{(1 - 2 x)}^{2}} + 1 = 0

\Rightarrow {(1 - 3 x)}^{2} - 2 (1 - 3 x) + {(1 - 2 x)}^{2} = 0

13 x^{2} - 4 x = 0

\Rightarrow x = \frac{4}{13}

\frac{s h a d e d a r e a}{a r e a o f s e m i c i r c l e} = \frac{2 π r^{2}}{π R^{2} / 2}

= 4 x^{2} = \frac{64}{169} \cdot

Commented by TawaTawa last updated on 05/Nov/19

God bless you sir . Thanks for your time