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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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