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

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

$$r=radiusofsmallcircle$$$$R=radiusofbigsemicircle$$$$\pi r^2=4\pi \Rightarrow r=2$$$$R=\frac{3+x}{2}$$$${(R-r)}^2=r^2+{(R-3)}^2$$$$2R(3-r)=9$$$$R=\frac{9}{2(3-r)}=\frac{9}{2}$$$$\frac{3+x}{2}=\frac{9}{2}$$$$\Rightarrow x=6$$

Commented by Tawa11 last updated on 05/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by Shrinava last updated on 06/Sep/22

$$\mathrm{cool}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

Answered by HeferH last updated on 05/Sep/22

$$$$$$LetRbetheradiusofthesemicircle,Qthe$$$$centerofthissemicircleandrtheradius$$$$ofthetangentcircle:$$$$r^2\pi =4\pi$$$$r=2$$$$OC=2$$$$QC=R-3$$$$QO=R-2$$$$x=QC+R$$$$$$$$ThetriangleOCQisarighttriangle:$$$${(R-2)}^2={(R-3)}^2+2^2$$$${(R-2)}^2-{(R-3)}^2=4$$$$(R-2+R-3)(R-2-(R-3))=4$$$$(2R-5)\left(1\right)=4$$$$R=\frac{9}{2}$$$$x=\frac{9}{2}-3+\frac{9}{2}=6$$$$$$

Commented by Tawa11 last updated on 05/Sep/22

$$\mathrm{Great}\mathrm{sir}.$$

Commented by Shrinava last updated on 06/Sep/22

$$\mathrm{cool}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

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