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Question Number 89033 by M±th+et£s last updated on 14/Apr/20

Answered by ajfour last updated on 15/Apr/20

$$letcenterofsquarebeorigin.$$$$eq.ofleftcircle:$$$${(x+\frac{r}{2})}^2+y^2=r^2$$$$letsideofsquare=2s$$$$toprightcornerofsquare(s,s)$$$$thisliesonleftcircle$$$$\Rightarrow {(s+\frac{r}{2})}^2+s^2=r^2$$$$2s^2+rs-\frac{3r^2}{4}=0$$$$s=-\frac{r}{4}+\sqrt{\frac{r^2}{16}+\frac{3r^2}{8}}=\frac{r}{4}(\sqrt{7}-1)$$$${Area}_{sq}=4s^2=\frac{r^2}{2}(4-\sqrt{7}).$$

Commented by M±th+et£s last updated on 14/Apr/20

$$itsnotmyquestionbutithinkisolved$$$$theproblem$$

Commented by M±th+et£s last updated on 14/Apr/20

Commented by ajfour last updated on 15/Apr/20

$$sorry,ihadnotlookedcarefully.$$$$centerofonecircle,liesonthe$$$$circumferenceofother.$$

Commented by M±th+et£s last updated on 15/Apr/20

$$itsokandthankyouforyourtry$$

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