Question Number 3840 by Rasheed Soomro last updated on 22/Dec/15
$$\mathcal{A}\:\:{semicircle}\:\:{contains}\:{a}\:{square}\:\:{of}\:\: \\ $$$${possible}\:{largest}\:{area}.{If}\:\:{s}\:\:{is}\:{the}\:{measure} \\ $$$${of}\:{the}\:{side}\:{of}\:{the}\:{square},{what}\:{is}\:{the} \\ $$$${radius}\:{of}\:{the}\:{semicircle}? \\ $$
Answered by Yozzii last updated on 24/Dec/15
$${Draw}\:{a}\:{semicircle}\:{of}\:{radius}\:{r}.\:{If}\:{the} \\ $$$${semicircle}\:{contains}\:{a}\:{square}\:{of}\: \\ $$$${maximum}\:{area},\:{the}\:{centre}\:{of}\:{the}\:{straight}\:{edge}\: \\ $$$${of}\:{the}\:{semicircle}\:{lies}\:{at}\:{the}\:{mid}−{point}\:{of} \\ $$$${the}\:{side}\:{of}\:{the}\:{square}\:{concurrent}\:{with} \\ $$$${the}\:{straight}\:{edge}\:{of}\:{the}\:{semicircle}. \\ $$$${Also},\:{the}\:{remaining}\:{two}\:{corners}\:{of}\:{the} \\ $$$${square}\:{touch}\:{the}\:{arc}\:{of}\:{the}\:{semicircle}. \\ $$$$ \\ $$$${Call}\:{the}\:{midpoint}\:{point}\:{O}\:{and}\:{label}\:{the}\:{square}\: \\ $$$${ABCD}.\:{CD}\:{is}\:{the}\:{side}\:{with}\:{O}\:{on}\:{it} \\ $$$${and}\:{AB}\:{is}\:{opposite}\:{CD}.\:{Form}\:{the} \\ $$$${triangle}\:\bigtriangleup{OBC}.\:{By}\:{virtue}\:{of}\:{the}\: \\ $$$${symmetry}\:{of}\:{the}\:{square},\:{area}\:{of}\:{the} \\ $$$${square}\:{A}_{\mathrm{1}} =\mathrm{4}×\left\{{area}\:{of}\:\bigtriangleup{OBC}\right\}.\:{The} \\ $$$${square}\:{has}\:{side}\:{length}\:{s}\: \\ $$$$\therefore\:{s}^{\mathrm{2}} =\mathrm{4}\left\{{area}\:{of}\:\bigtriangleup{OBC}\right\}. \\ $$$${Let}\:\theta=\angle{BOC}.\:\Rightarrow{OC}={rcos}\theta.\: \\ $$$$\therefore\:{area}\:{of}\:\bigtriangleup{OBC}=\frac{\mathrm{1}}{\mathrm{2}}{OC}×{BC}=\frac{{srcos}}{\mathrm{2}}. \\ $$$$\therefore\:{s}^{\mathrm{2}} =\mathrm{2}{srcos}\theta\Rightarrow{cos}\theta=\frac{{s}}{\mathrm{2}{r}}.\: \\ $$$${From}\:\bigtriangleup{OBC},\:{sin}\theta=\frac{{s}}{{r}}. \\ $$$$\because\:{sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\Rightarrow\frac{{s}^{\mathrm{2}} }{{r}^{\mathrm{2}} }+\frac{{s}^{\mathrm{2}} }{\mathrm{4}{r}^{\mathrm{2}} }=\mathrm{1}\Rightarrow\mathrm{4}{s}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}}{s}^{\mathrm{2}} \Rightarrow{r}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}{s}\:\:\left({r},{s}>\mathrm{0}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 24/Dec/15
$$\mathcal{G}_{\mathcal{O}} ^{\mathcal{O}} \mathcal{D}\:\:\:_{\mathcal{E}{xplanation}} ^{\mathcal{L}{ogic}} ! \\ $$