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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

$$BLtanjentOLradius\therefore BL\mathrm{\perp }OL$$$$\therefore \mathrm{\angle }BLO=90\therefore \mathrm{\angle }BOL=90-60=30$$$$[\because △ABCequilateral\Rightarrow \mathrm{\angle }LBO=60$$$$letOL=r$$$$\therefore LM=rsin30=\frac{r}{2}$$$$MO=rcos30=\frac{\sqrt{3}r}{2}$$$$lengthofrectangle=2\times MO=\sqrt{3}r$$$$areaofrectangle=\frac{\sqrt{3}{r}^2}{2}squnit$$$$areaofhalfdisc=\frac{1}{2}\times \pi \times r^2$$$$\therefore \frac{\mathit{r}\mathit{e}\mathit{c}\mathit{t}\mathit{a}\mathit{g}\mathit{l}\mathit{e}\mathit{a}\mathit{r}\mathit{e}\mathit{a}}{\mathit{h}\mathit{a}\mathit{l}\mathit{f}\mathit{d}\mathit{i}\mathit{s}\mathit{c}\mathit{a}\mathit{r}\mathit{e}\mathit{a}}$$$$=\frac{\frac{\sqrt{3}{\mathit{r}}^2}{2}}{\frac{\mathit{\pi }{\mathit{r}}^2}{2}}=\frac{\sqrt{3}}{\mathit{\pi }}$$

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

$$\frac{rectanglearea}{halfdiscarea}=\frac{\sqrt{3}}{\pi }?$$

Commented by cherokeesay last updated on 21/Mar/22

$$yesSir!$$$$thanks.$$

Commented by Lakers last updated on 21/Mar/22

$$yourworkingplease$$

Commented by Tawa11 last updated on 22/Mar/22

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

Answered by cherokeesay last updated on 22/Mar/22

$$\frac{R}{CH}=cos30°=\frac{\sqrt{3}}{2}\iff R=\frac{CH\sqrt{3}}{2}$$$$\frac{a}{R}=cos30°=\frac{\sqrt{3}}{2}\iff a=\frac{3CH}{4}$$$$\frac{b}{R}=sin30°=\frac{1}{2}\iff b=\frac{CH\sqrt{3}}{4}$$$${\mathcal{A}}_{rec.}=2ab=\frac{6{CH}^2\sqrt{3}}{16}$$$$\frac{{\mathcal{A}}_{rec.}}{{\mathcal{A}}_{h.d}}=\frac{\frac{6{CH}^2\sqrt{3}}{16}}{\frac{\pi {\left(\frac{CH\sqrt{3}}{2}\right)}^2}{2}}=\frac{6{CH}^2\sqrt{3}}{16}\times \frac{8}{3\pi {CH}^2}=$$$$\frac{\sqrt{3}}{\pi }$$

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