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Question-194116




Question Number 194116 by universe last updated on 28/Jun/23
Answered by mr W last updated on 02/Jul/23
Commented by mr W last updated on 02/Jul/23
R=radius of semicircle  r=radius of small circle  AD=2R  OF=R−r  OE=(√((R−r)^2 −r^2 ))  AB=R−r+(√((R−r)^2 −r^2 ))  CD=R−r−(√((R−r)^2 −r^2 ))  AB×CD=r^2   AB+CD+2(√(AB×CD))=2(R−r)+2r=2R=AD  i.e. ((√(AB))+(√(CD)))^2 =((√(AD)))^2   ⇒(√(AD))=(√(AB))+(√(CD))
$${R}={radius}\:{of}\:{semicircle} \\ $$$${r}={radius}\:{of}\:{small}\:{circle} \\ $$$${AD}=\mathrm{2}{R} \\ $$$${OF}={R}−{r} \\ $$$${OE}=\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${AB}={R}−{r}+\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${CD}={R}−{r}−\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${AB}×{CD}={r}^{\mathrm{2}} \\ $$$${AB}+{CD}+\mathrm{2}\sqrt{{AB}×{CD}}=\mathrm{2}\left({R}−{r}\right)+\mathrm{2}{r}=\mathrm{2}{R}={AD} \\ $$$${i}.{e}.\:\left(\sqrt{{AB}}+\sqrt{{CD}}\right)^{\mathrm{2}} =\left(\sqrt{{AD}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{{AD}}=\sqrt{{AB}}+\sqrt{{CD}} \\ $$

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