Question-45366

Question Number 45366 by behi83417@gmail.com last updated on 12/Oct/18

Commented by behi83417@gmail.com last updated on 12/Oct/18

s h a d e d a r e a = ?

Commented by ajfour last updated on 12/Oct/18

Answered by ajfour last updated on 12/Oct/18

x = \frac{r^{2}}{2} θ = \frac{9}{2} (180 ° - 20 °) \times \frac{π}{180 °}

= \frac{9}{2} \times \frac{8 π}{9} = 4 π

(s i n c e ∠ A C B = 150 ° - 130 ° = 20 °) .

Commented by behi83417@gmail.com last updated on 12/Oct/18

t h a n k s a l o t s i r A j f o u r .

Answered by MrW3 last updated on 12/Oct/18

= F o r a n y t h r e e c i r c l e s =

R a d i u s o f c i r c l e :

b l a c k = r_{1}

g r e e n = r_{2}

r e d = r_{2}

{(r_{1} - r_{2})}^{2} = {(r_{1} - r_{3})}^{2} + {(r_{2} + r_{3})}^{2} - 2 (r_{1} - r_{3}) (r_{2} + r_{3}) \cos θ

\cos θ = \frac{{(r_{1} - r_{3})}^{2} + {(r_{2} + r_{3})}^{2} - {(r_{1} - r_{2})}^{2}}{2 (r_{1} - r_{3}) (r_{2} + r_{3})}

\cos θ = \frac{r_{3}^{2} - r_{1} r_{3} + r_{2} r_{3} + r_{1} r_{2}}{(r_{1} - r_{3}) (r_{2} + r_{3})}

\cos θ = \frac{r_{2} (r_{1} + r_{3}) - r_{3} (r_{1} - r_{3})}{(r_{1} - r_{3}) (r_{2} + r_{3})}

θ = \cos^{- 1} {\frac{r_{2} (r_{1} + r_{3}) - r_{3} (r_{1} - r_{3})}{(r_{1} - r_{3}) (r_{2} + r_{3})}}

A_{s h a d e} = \frac{r_{3}^{2} (π - θ)}{2} = \frac{r_{3}^{2}}{2} [π - \cos^{- 1} {\frac{r_{2} (r_{1} + r_{3}) - r_{3} (r_{1} - r_{3})}{(r_{1} - r_{3}) (r_{2} + r_{3})}}]

Commented by behi83417@gmail.com last updated on 12/Oct/18

t h a n k y o u s o m u c h s i r .

h o w c a n w e f i n d r_{1} a n d r_{2} ?

Commented by MrW3 last updated on 12/Oct/18

Commented by MrW3 last updated on 12/Oct/18

t h i s s h o w s t h e q u e s t i o n I s o l v e d . i t

i s n o t d i r e c t l y y o u r q u e s t i o n .

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