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Question Number 3840 by Rasheed Soomro last updated on 22/Dec/15
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?
$$\mathcal{A}\:\:{semicircle}\:\:{contains}\:{a}\:{square}\:\:{of}\:\: \\ $$$${possible}\:{largest}\:{area}.{If}\:\:{s}\:\:{is}\:{the}\:{measure} \\ $$$${of}\:{the}\:{side}\:{of}\:{the}\:{square},{what}\:{is}\:{the} \\ $$$${radius}\:{of}\:{the}\:{semicircle}? \\ $$
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 △OBC. By virtue of the   symmetry of the square, area of the  square A_1 =4×{area of △OBC}. The  square has side length s   ∴ s^2 =4{area of △OBC}.  Let θ=∠BOC. ⇒OC=rcosθ.   ∴ area of △OBC=(1/2)OC×BC=((srcos)/2).  ∴ s^2 =2srcosθ⇒cosθ=(s/(2r)).   From △OBC, sinθ=(s/r).  ∵ sin^2 θ+cos^2 θ=1  ⇒(s^2 /r^2 )+(s^2 /(4r^2 ))=1⇒4s^2 +s^2 =4r^2   r^2 =(5/4)s^2 ⇒r=((√5)/2)s  (r,s>0)
$${Draw}\:{a}\:{semicircle}\:{of}\:{radius}\:{r}.\:{If}\:{the} \\ $$$${semicircle}\:{contains}\:{a}\:{square}\:{of}\: \\ $$$${maximum}\:{area},\:{the}\:{centre}\:{of}\:{the}\:{straight}\:{edge}\: \\ $$$${of}\:{the}\:{semicircle}\:{lies}\:{at}\:{the}\:{mid}−{point}\:{of} \\ $$$${the}\:{side}\:{of}\:{the}\:{square}\:{concurrent}\:{with} \\ $$$${the}\:{straight}\:{edge}\:{of}\:{the}\:{semicircle}. \\ $$$${Also},\:{the}\:{remaining}\:{two}\:{corners}\:{of}\:{the} \\ $$$${square}\:{touch}\:{the}\:{arc}\:{of}\:{the}\:{semicircle}. \\ $$$$ \\ $$$${Call}\:{the}\:{midpoint}\:{point}\:{O}\:{and}\:{label}\:{the}\:{square}\: \\ $$$${ABCD}.\:{CD}\:{is}\:{the}\:{side}\:{with}\:{O}\:{on}\:{it} \\ $$$${and}\:{AB}\:{is}\:{opposite}\:{CD}.\:{Form}\:{the} \\ $$$${triangle}\:\bigtriangleup{OBC}.\:{By}\:{virtue}\:{of}\:{the}\: \\ $$$${symmetry}\:{of}\:{the}\:{square},\:{area}\:{of}\:{the} \\ $$$${square}\:{A}_{\mathrm{1}} =\mathrm{4}×\left\{{area}\:{of}\:\bigtriangleup{OBC}\right\}.\:{The} \\ $$$${square}\:{has}\:{side}\:{length}\:{s}\: \\ $$$$\therefore\:{s}^{\mathrm{2}} =\mathrm{4}\left\{{area}\:{of}\:\bigtriangleup{OBC}\right\}. \\ $$$${Let}\:\theta=\angle{BOC}.\:\Rightarrow{OC}={rcos}\theta.\: \\ $$$$\therefore\:{area}\:{of}\:\bigtriangleup{OBC}=\frac{\mathrm{1}}{\mathrm{2}}{OC}×{BC}=\frac{{srcos}}{\mathrm{2}}. \\ $$$$\therefore\:{s}^{\mathrm{2}} =\mathrm{2}{srcos}\theta\Rightarrow{cos}\theta=\frac{{s}}{\mathrm{2}{r}}.\: \\ $$$${From}\:\bigtriangleup{OBC},\:{sin}\theta=\frac{{s}}{{r}}. \\ $$$$\because\:{sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\Rightarrow\frac{{s}^{\mathrm{2}} }{{r}^{\mathrm{2}} }+\frac{{s}^{\mathrm{2}} }{\mathrm{4}{r}^{\mathrm{2}} }=\mathrm{1}\Rightarrow\mathrm{4}{s}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}}{s}^{\mathrm{2}} \Rightarrow{r}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}{s}\:\:\left({r},{s}>\mathrm{0}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 24/Dec/15
G_O ^O D   _(Explanation)^(Logic) !
$$\mathcal{G}_{\mathcal{O}} ^{\mathcal{O}} \mathcal{D}\:\:\:_{\mathcal{E}{xplanation}} ^{\mathcal{L}{ogic}} ! \\ $$

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