Question-51526

Question Number 51526 by ajfour last updated on 27/Dec/18

Commented by ajfour last updated on 27/Dec/18

F o r t h e a r e a i n b l a c k a n d y e l l o w

t o b e t h e s a m e, f i n d r i n t e r m s o f R .

Answered by mr W last updated on 28/Dec/18

\cos θ = \frac{R - r}{R + r} = \frac{1 - \frac{r}{R}}{1 + \frac{r}{R}} = \frac{1 - λ}{1 + λ}

\sin θ = \frac{\sqrt{{(1 + λ)}^{2} - {(1 - λ)}^{2}}}{1 + λ} = \frac{2 \sqrt{λ}}{1 + λ}

A_{b l a c k} = \frac{(R + r) (R + r) \sin θ}{2} - \frac{R^{2} θ}{2} - \frac{r^{2}}{2} (π - θ)

A_{b l a c k} = \frac{{(R + r)}^{2} \sin θ}{2} - \frac{(R^{2} - r^{2}) θ}{2} - \frac{π r^{2}}{2} = π r^{2}

{(R + r)}^{2} \sin θ - (R^{2} - r^{2}) θ = 3 π r^{2}

\Rightarrow 2 (1 + λ) \sqrt{λ} - (1 - λ^{2}) \cos^{- 1} \frac{1 - λ}{1 + λ} = 3 π λ^{2}

\Rightarrow λ = 0.0886

\Rightarrow r = 0.0886 R

Commented by mr W last updated on 28/Dec/18

Commented by ajfour last updated on 28/Dec/18

y e s S i r, i h a d n o t c a c u l a t e d b u t

h a d a r r i v e d a t t h e s a m e e q u a t i o n .

Facebook Tweet Pin