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Question Number 191335 by Mingma last updated on 23/Apr/23

	Answered by mr W last updated on 23/Apr/23
	$radius of big quatercircle =a radius of big semicircle =b radius of small quatercircle =c b^2 +(2b)^2 =a^2 ⇒b^2 =(a^2 /5) c^2 +(c+b)^2 =a^2 2c^2 +2bc+b^2 =a^2 2c^2 +2(a/( (√5)))c+(a^2 /5)=a^2 ((c/a))^2 +(1/( (√5)))((c/a))−(2/5)=0 ⇒(c/a)=(1/2)(−(1/( (√5)))+(√((1/5)+((4×2)/5))))=(1/( (√5))) fraction of shaded area: (c^2 /a^2 )=((c/a))^2 =((1/( (√5))))^2 =(1/5)$
	$r a d i u s o f b i g q u a t e r c i r c l e = a$ $r a d i u s o f b i g s e m i c i r c l e = b$ $r a d i u s o f s m a l l q u a t e r c i r c l e = c$ $b^{2} + {(2 b)}^{2} = a^{2} \Rightarrow b^{2} = \frac{a^{2}}{5}$ $c^{2} + {(c + b)}^{2} = a^{2}$ $2 c^{2} + 2 b c + b^{2} = a^{2}$ $2 c^{2} + 2 \frac{a}{\sqrt{5}} c + \frac{a^{2}}{5} = a^{2}$ ${(\frac{c}{a})}^{2} + \frac{1}{\sqrt{5}} (\frac{c}{a}) - \frac{2}{5} = 0$ $\Rightarrow \frac{c}{a} = \frac{1}{2} (- \frac{1}{\sqrt{5}} + \sqrt{\frac{1}{5} + \frac{4 \times 2}{5}}) = \frac{1}{\sqrt{5}}$ $f r a c t i o n o f s h a d e d a r e a :$ $\frac{c^{2}}{a^{2}} = {(\frac{c}{a})}^{2} = {(\frac{1}{\sqrt{5}})}^{2} = \frac{1}{5}$
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