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Question Number 167633 by cherokeesay last updated on 21/Mar/22

Commented by som(math1967) last updated on 22/Mar/22

Commented by som(math1967) last updated on 22/Mar/22

BL tanjent OL radius ∴BL⊥OL ∴∠BLO=90 ∴∠BOL=90−60=30 [∵ △ABC equilateral ⇒∠LBO=60 let OL=r ∴LM=rsin30=(r/2) MO=rcos30 =(((√3)r)/2) length of rectangle=2×MO=(√3)r area of rectangle=(((√3)r^2 )/2) squnit area of half disc=(1/2)×π×r^2 ∴ ((rectagle area)/(half disc area)) =((((√3)r^2 )/2)/((𝛑r^2 )/2))=((√3)/𝛑)

$${BL}\:{tanjent}\:{OL}\:\:{radius}\:\therefore{BL}\bot{OL} \\ $$$$\therefore\angle{BLO}=\mathrm{90}\:\therefore\angle{BOL}=\mathrm{90}−\mathrm{60}=\mathrm{30} \\ $$$$\left[\because\:\bigtriangleup{ABC}\:{equilateral}\:\Rightarrow\angle{LBO}=\mathrm{60}\right. \\ $$$${let}\:{OL}={r} \\ $$$$\therefore{LM}={rsin}\mathrm{30}=\frac{{r}}{\mathrm{2}} \\ $$$$\:{MO}={rcos}\mathrm{30}\:=\frac{\sqrt{\mathrm{3}}{r}}{\mathrm{2}} \\ $$$${length}\:{of}\:{rectangle}=\mathrm{2}×{MO}=\sqrt{\mathrm{3}}{r} \\ $$$${area}\:{of}\:{rectangle}=\frac{\sqrt{\mathrm{3}}{r}^{\mathrm{2}} }{\mathrm{2}}\:{squnit} \\ $$$${area}\:{of}\:{half}\:{disc}=\frac{\mathrm{1}}{\mathrm{2}}×\pi×{r}^{\mathrm{2}} \\ $$$$\therefore\:\:\frac{\boldsymbol{{rectagle}}\:\boldsymbol{{area}}}{\boldsymbol{{half}}\:\boldsymbol{{disc}}\:\boldsymbol{{area}}} \\ $$$$=\frac{\frac{\sqrt{\mathrm{3}}\boldsymbol{{r}}^{\mathrm{2}} }{\mathrm{2}}}{\frac{\boldsymbol{\pi{r}}^{\mathrm{2}} }{\mathrm{2}}}=\frac{\sqrt{\mathrm{3}}}{\boldsymbol{\pi}} \\ $$

Commented by som(math1967) last updated on 21/Mar/22

((rectangle area)/(half disc area))=((√3)/π) ?

$$\:\frac{{rectangle}\:{area}}{{half}\:{disc}\:{area}}=\frac{\sqrt{\mathrm{3}}}{\pi}\:\:\:? \\ $$

Commented by cherokeesay last updated on 21/Mar/22

yes Sir ! thanks.

$${yes}\:{Sir}\:! \\ $$$${thanks}. \\ $$

Commented by Lakers last updated on 21/Mar/22

your working please

$${your}\:{working}\:{please} \\ $$

Commented by Tawa11 last updated on 22/Mar/22

Nice sir

$$\mathrm{Nice}\:\mathrm{sir} \\ $$

Answered by cherokeesay last updated on 22/Mar/22

(R/(CH))=cos30°=((√3)/2) ⇔ R=((CH(√3))/2) (a/R)=cos30°=((√3)/2) ⇔ a = ((3CH)/4) (b/R)=sin30°=(1/2) ⇔ b = ((CH(√3))/4) A_(rec.) =2ab=((6CH^2 (√3))/(16)) (A_(rec.) /A_(h.d) )=(((6CH^2 (√3))/(16))/((π(((CH(√3))/2))^2 )/2))=((6CH^2 (√3))/(16))×(8/(3πCH^2 ))= ((√3)/π)

$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{R}}{{CH}}={cos}\mathrm{30}°=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Leftrightarrow\:{R}=\frac{{CH}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}}{{R}}={cos}\mathrm{30}°=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Leftrightarrow\:{a}\:=\:\frac{\mathrm{3}{CH}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{b}}{{R}}={sin}\mathrm{30}°=\frac{\mathrm{1}}{\mathrm{2}}\:\Leftrightarrow\:{b}\:=\:\frac{{CH}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathscr{A}_{{rec}.} =\mathrm{2}{ab}=\frac{\mathrm{6}{CH}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{16}} \\ $$$$\:\:\frac{\mathscr{A}_{{rec}.} }{\mathscr{A}_{{h}.{d}} }=\frac{\frac{\mathrm{6}{CH}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{16}}}{\frac{\pi\left(\frac{{CH}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}}=\frac{\mathrm{6}{CH}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{16}}×\frac{\mathrm{8}}{\mathrm{3}\pi{CH}^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{3}}}{\pi} \\ $$

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