(R−r)^2 + R^2 = (R + r)^2 ⇔ R^2 + r^2 −2rR + R^2 = R^2 + r^2 +2rR ⇒ R^2 = 4rR ⇒ r = (R/4) = 1cm A


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Question Number 150793 by cherokeesay last updated on 15/Aug/21

${(R - r)}^{2} + R^{2} = {(R + r)}^{2} \Leftrightarrow$ $R^{2} + r^{2} - 2 r R + R^{2} = R^{2} + r^{2} + 2 r R \Rightarrow$ $R^{2} = 4 r R \Rightarrow r = \frac{R}{4} = 1 c m$ $A_{S} = π r^{2} = π \times 1 {c m}^{2} = π {c m}^{2}$
Commented by Rasheed.Sindhi last updated on 15/Aug/21

$N i c e!$ $P l w r i t e t h i s a s a n a n s w e r$ $o f y o u r q u e s t i o n . {D o n}^{'} t s t a r t n e w$ $t h r e a d f o r a n a n s w e r . T h a n k s .$
Commented by cherokeesay last updated on 15/Aug/21

the answer to this problem has already been processed by a member (see below) what I have established is my own resolution of this same exercise but accompanied by a diagram! that's all ! good to you.
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