Question-175709

Question Number 175709 by Shrinava last updated on 05/Sep/22

Answered by mr W last updated on 05/Sep/22

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

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

π r^{2} = 4 π \Rightarrow r = 2

R = \frac{3 + x}{2}

{(R - r)}^{2} = r^{2} + {(R - 3)}^{2}

2 R (3 - r) = 9

R = \frac{9}{2 (3 - r)} = \frac{9}{2}

\frac{3 + x}{2} = \frac{9}{2}

\Rightarrow x = 6

Commented by Tawa11 last updated on 05/Sep/22

Great sir

Commented by Shrinava last updated on 06/Sep/22

cool professor thank you

Answered by HeferH last updated on 05/Sep/22

L e t R b e t h e r a d i u s o f t h e s e m i c i r c l e, Q t h e

c e n t e r o f t h i s s e m i c i r c l e a n d r t h e r a d i u s

o f t h e t a n g e n t c i r c l e :

r^{2} π = 4 π

r = 2

O C = 2

Q C = R - 3

Q O = R - 2

x = Q C + R

T h e t r i a n g l e O C Q i s a r i g h t t r i a n g l e :

{(R - 2)}^{2} = {(R - 3)}^{2} + 2^{2}

{(R - 2)}^{2} - {(R - 3)}^{2} = 4

(R - 2 + R - 3) (R - 2 - (R - 3)) = 4

(2 R - 5) (1) = 4

R = \frac{9}{2}

x = \frac{9}{2} - 3 + \frac{9}{2} = 6

Commented by Tawa11 last updated on 05/Sep/22

Great sir .

Commented by Shrinava last updated on 06/Sep/22

cool professor thank you