Question-156940

Question Number 156940 by amin96 last updated on 17/Oct/21

Answered by mr W last updated on 17/Oct/21

Commented by mr W last updated on 17/Oct/21

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

A_{g r e e n} = \frac{π {(2 r)}^{2}}{4} - \frac{{(2 r)}^{2}}{2} - [\frac{π r^{2}}{4} - \frac{r^{2}}{2}]

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

θ = 60 °

O A = \frac{r}{\tan \frac{θ}{2}}

O B = 2 r

A B = b = 10

{O B}^{2} = {(2 r)}^{2} = {(\frac{r}{\tan \frac{θ}{2}})}^{2} + b^{2} - 2 b \times \frac{r}{\tan \frac{θ}{2}} \cos θ

(4 - \frac{1}{\tan^{2} 30}) r^{2} + 20 \times \frac{r}{\tan 30} \cos 60 - 100 = 0

r^{2} + 10 \sqrt{3} r - 100 = 0

r = 5 (\sqrt{7} - \sqrt{3})

A_{g r e e n} = \frac{75 (π - 2) (5 - \sqrt{21})}{2} \approx 17.8698

Commented by Tawa11 last updated on 17/Oct/21

Great sir

Commented by amin96 last updated on 17/Oct/21

c o o l s i r . b r a v o

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